SATHEE: Chapter 07 Coordinate Geometry

Chapter 07 Coordinate Geometry

Multiple Choice Questions (MCQs)

1 The distance of the point $P (2, 3)$ from the $X$ -axis is

(a) 2 (b) 3 (c) 1 (d) 5

Show Answer

Solution

(b) We know that, if $(x, y)$ is any point on the cartesian plane in first quadrant.

Then, $x =$ Perpendicular distance from $Y$ -axis

and $y =$ Perpendicular distance from $X$ -axis

Distance of the point $P (2, 3)$ from the $X$ -axis $=$ Ordinate of a point $P (2, 3) = 3$ .

2 The distance between the points

A (0, 6)

and

B (0, - 2)

(a) 6 (b) 8 (c) 4 (d) 2

Show Answer

Thinking Process

The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ . Use this formula and simplify it.

Solution

(b) $∵$ Distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ,

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Here,

$x_{1} = 0, y_{1} = 6 and x_{2} = 0, y_{2} = - 2$

$∴$ Distance between $A (0, 6)$ and $B (0, - 2)$ ,

$\begin{aligned} A B = \sqrt{(0 - 0)^{2} + (- 2 - 6)^{2}} \\ = \sqrt{0 + (- 8)^{2}} = \sqrt{8^{2}} = 8 \end{aligned}$

3 The distance of the point

P (- 6, 8)

from the origin is

(a) 8 (b) $2 \sqrt{7}$ (c) 10 (d) 6

Show Answer

Thinking Process

Coordinate of origin is $(0, 0)$

Solution

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Here, $x_{1} = - 6, y_{1} = 8$ and $x_{2} = 0, y_{2} = 0$

$∴$ Distance between $P (- 6, 8)$ and origin i.e., $O (0, 0)$ ,

$\begin{aligned} P O = \sqrt{[0 - (- 6)]^{2} + (0 - 8)^{2}} \\ = \sqrt{(6)^{2} + (- 8)^{2}} \\ = \sqrt{36 + 64} = \sqrt{100} = 10 \end{aligned}$

4 The distance between the points

(0, 5)

and

(- 5, 0)

(a) 5 (b) $5 \sqrt{2}$ (c) $2 \sqrt{5}$ (d) 10

Show Answer

Solution

(b) $∵$ Distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ,

$\begin{aligned} d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \\ Here, x_{1} = 0, y_{1} = 5 and x_{2} = - 5, y_{2} = 0 \\ ∴ Distance between the points (0, 5) and (- 5, 0) \\ \begin{array}{c} = \sqrt{(- 5 - 0)^{2} + (0 - 5)^{2}} \\ = \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2} \end{array} \end{aligned}$

5 If

A O B C

is a rectangle whose three vertices are

A (0, 3), 0 (0, 0)

and

B (5, 0)

, then the length of its diagonal is

(a) 5 (b) 3 (c) $\sqrt{34}$ (d) 4

Show Answer

Solution

(c)

Now, length of the diagonal $A B =$ Distance between the points $A (0, 3)$ and $B (5, 0)$ .

$∵$ Distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ,

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Here, $x_{1} = 0, y_{1} = 3$ and $x_{2} = 5, y_{2} = 0$

$∴$ Distance between the points $A (0, 3)$ and $B (5, 0)$

$\begin{aligned} A B = \sqrt{(5 - 0)^{2} + (0 - 3)^{2}} \\ = \sqrt{25 + 9} = \sqrt{34} \end{aligned}$

Hence, the required length of its diagonal is $\sqrt{34}$ .

6 The perimeter of a triangle with vertices

(0, 4), (0, 0)

and

(3, 0)

(a) 5 (b) 12 (c) 11 (d) $7 + \sqrt{5}$

Show Answer

Thinking Process

(i) Firstly, plot the given points on a graph paper and join them to get a triangle.

(ii) Secondly, determine the length of the each sides by using the distance formula,

$\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

(iii) Further, adding all the distance of a triangle to get the perimeter of a triangle.

Solution

(b) We plot the vertices of a triangle i.e., $(0, 4), (0, 0)$ and $(3, 0)$ on the paper shown as given below

Now, perimeter of $△ A O B =$ Sum of the length of all its sides $= d (A O) + d (O B) + d (A B)$

$∵$ Distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ,

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$=$ Distance between $A (0, 4)$ and $O (0, 0) +$ Distance between $O (0, 0)$ and $B (3, 0)$ + Distance between $A (0, 4)$ and $B (3, 0)$

$\begin{aligned} = \sqrt{(0 - 0)^{2} + (0 - 4)^{2}} + \sqrt{(3 - 0)^{2} + (0 - 0)^{2}} + \sqrt{(3 - 0)^{2} + (0 - 4)^{2}} \\ = \sqrt{0 + 16} + \sqrt{9 + 0} + \sqrt{(3)^{2} + (4)^{2}} = 4 + 3 + \sqrt{9 + 16} \\ = 7 + \sqrt{25} = 7 + 5 = 12 \end{aligned}$

Hence, the required perimeter of triangle is 12 .

7 The area of a triangle with vertices

A (3, 0), B (7, 0)

and

C (8, 4)

(a) 14 (b) 28 (c) 8 (d) 6

Show Answer

Thinking Process

The area of triangle, whose vertices are $A (x_{1}, y_{2}), B (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by $\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$ . Use this formula and simplify it to get the result.

Solution

(c) Area of $△ A B C$ whose Vertices $A \equiv (x_{1}, y_{1}), B \equiv (x_{2}, y_{2})$ and $C \equiv (x_{3}, y_{3})$ are given by

$Δ = | \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] |$

Here, $x_{1} = 3, y_{1} = 0, x_{2} = 7, y_{2} = 0, x_{3} = 8$ and $y_{3} = 4$

$∴ Δ = | \frac{1}{2} [3 (0 - 4) + 7 (4 - 0) + 8 (0 - 0)] | = | \frac{1}{2} (- 12 + 28 + 0) | = | \frac{1}{2} (16) | = 8$

Hence, the required area of $△ A B C$ is 8 .

8 The points

(- 4, 0), (4, 0)

and

(0, 3)

are the vertices of a

(a) right angled triangle (b) isosceles triangle

Show Answer

Solution

(b) Let $A (- 4, 0), B (4, 0), C (0, 3)$ are the given vertices.

Now, distance between $A (- 4, 0)$ and $B (4, 0)$ ,

$A B = \sqrt{[4 - (- 4)]^{2} + (0 - 0)^{2}}$

$∵$ distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$= \sqrt{(4 + 4)^{2}} = \sqrt{8^{2}} = 8$

Distance between $B (4, 0)$ and $C (0, 3)$ ,

$B C = \sqrt{(0 - 4)^{2} + (3 - 0)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$

Distance between $A (- 4, 0)$ and $C (0, 3)$ ,

$A C = \sqrt{[0 - (- 4)]^{2} + (3 - 0)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$

$∵ B C = A C$

Hence, $△ A B C$ is an isosceles triangle because an isosceles triangle has two sides equal.

9 The point which divides the line segment joining the points

(7, - 6)

and

(3, 4)

in ratio

1 : 2

internally lies in the

(a) I quadrant (b) II quadrant

Show Answer

Solution

(d) If $P (x, y)$ divides the line segment joining $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio

$m : n, then x = \frac{m x_{2} + n x_{1}}{m + n} and y = \frac{m y_{2} + n y_{1}}{m + n}$

Given that, $x_{1} = 7, y_{1} = - 6, x_{2} = 3, y_{2} = 4, m = 1$ and $n = 2$

$∴ x = \frac{1 (3) + 2 (7)}{1 + 2}, y = \frac{1 (4) + 2 (- 6)}{1 + 2}$ [by section formula]

$\Rightarrow x = \frac{3 + 14}{3}, y = \frac{4 - 12}{3}$

$\Rightarrow x = \frac{17}{3}, y = - \frac{8}{3}$

So, $(x, y) = \frac{17}{3}, - \frac{8}{3}$ lies in IV quadrant.

[since, in IV quadrant, $x$ -coordinate is positive and $y$ -coordinate is negative]

10 The point which lies on the perpendicular bisector of the line segment joining the points

A (- 2, - 5)

and

B (2, 5)

(a) $(0, 0)$ (b) $(0, 2)$ (c) $(2, 0)$ (d) $(- 2, 0)$

Show Answer

Solution

(a) We know that, the perpendicular bisector of the any line segment divides the line segment into two equal parts i.e., the perpendicular bisector of the line segment always passes through the mid-point of the line segment.

$∴$ Mid-point of the line segment joining the points $A (- 2, - 5)$ and $B (2, 5)$

$= \frac{- 2 + 2}{2}, \frac{- 5 + 5}{2} = (0, 0)$

since, mid-point of any line segment which passes through the points

$(x_{1}, y_{1}) and (x_{2}, y_{2}) = \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

Hence, $(0, 0)$ is the required point lies on the perpendicular bisector of the lines segment.

11 The fourth vertex

D

of a parallelogram

A B C D

whose three vertices are

A (- 2, 3), B (6, 7)

and

C (8, 3)

(a) $(0, 1)$ (b) $(0, - 1)$ (c) $(- 1, 0)$ (d) $(1, 0)$

Show Answer

Thinking Process

(i) Firstly, consider the fourth vertex of a parallelogram be $D (x_{4}, y_{4})$ .

(ii) Secondly, determine the mid point of $A C$ and $B D$ by using the formula

$\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

(iii) Further, equating both points and get the required coordinate of fourth vertex.

Solution

(b) Let the fourth vertex of parallelogram, $D \equiv (x_{4}, y_{4})$ and $L, M$ be the middle points of $A C$ and $B D$ , respectively.

Then, $L \equiv \frac{- 2 + 8}{2}, \frac{3 + 3}{2} \equiv (3, 3)$

since, mid - point of a line segment having points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$

$= \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

and

$M \equiv \frac{6 + x_{4}}{2}, \frac{7 + y_{4}}{2}$

Since, $A B C D$ is a parallelogram, therefore diagonals $A C$ and $B D$ will bisect each other. Hence, $L$ and $M$ are the same points.

$\begin{aligned} ∴ 3 = \frac{6 + x_{4}}{2} and 3 = \frac{7 + y_{4}}{2} \\ \Rightarrow 6 = 6 + x_{4} and 6 = 7 + y_{4} \\ \Rightarrow x_{4} = 0 and y_{4} = 6 - 7 \\ ∴ x_{4} = 0 and y_{4} = - 1 \end{aligned}$

Hence, the fourth vertex of parallelogram is $D \equiv (x_{4}, y_{4}) \equiv (0, - 1)$ .

12 If the point

P (2, 1)

lies on the line segment joining points

A (4, 2)

and

B (8, 4)

, then

(a) $A P = \frac{1}{3} A B$ (b) $A P = P B$

Show Answer

Solution

(d) Given that, the point $P (2, 1)$ lies on the line segment joining the points $A (4, 2)$ and $B (8, 4)$ , which shows in the figure below:

Now, distance between $A (4, 2)$ and $(2, 1), A P = \sqrt{(2 - 4)^{2} + (1 - 2)^{2}}$

$∵$ distance between two points two points $(x_{1}, y_{1})$ and $B (x_{2}, y_{2}), d$

$= \sqrt{(x_{2} - x_{1})^{2} + (x_{2} - y_{1})^{2}}$

$= \sqrt{(- 2)^{2} + (- 1)^{2}} = \sqrt{4 + 1} = \sqrt{5}$

Distance between $A (4, 2)$ and $B (8, 4)$ ,

$\begin{aligned} A B = \sqrt{(8 - 4)^{2} + (4 - 2)^{2}} \\ = \sqrt{(4)^{2} + (2)^{2}} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5} \end{aligned}$

Distance between $B (8, 4)$ and $P (2, 1), B P = \sqrt{(8 - 2)^{2} + (4 - 1)^{2}}$

$\begin{aligned} = \sqrt{6^{2} + 3^{2}} = \sqrt{36 + 9} = \sqrt{45} = 3 \sqrt{5} \\ ∴ A B = 2 \sqrt{5} = 2 A P \Rightarrow A P = \frac{A B}{2} \end{aligned}$

Hence, required condition is $A P = \frac{A B}{2}$ .

13 If

P \frac{a}{3}, 4

is the mid-point of the line segment joining the points

Q (- 6, 5)

and

R (- 2, 3)

, then the value of

a

(a) -4 (b) -12 (c) 12 (d) -6

Show Answer

Solution

(b) Given that, $P \frac{a}{3}, 4$ is the mid-point of the line segment joining the points $Q (- 6, 5)$ and $R (- 2, 3)$ , which shows in the figure given below

$∴$ Mid-point of $Q R = P \frac{- 6 - 2}{2}, \frac{5 + 3}{2} = P (- 4, 4)$

since, mid-point of line segment having points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$

$= \frac{(x_{1} + x_{2})}{2}, \frac{(y_{1} + y_{2})}{2}$

But mid-point $P \frac{a}{3}, 4$ is given.

$∴ \frac{a}{3}, 4 = (- 4, 4)$

On comparing the coordinates, we get

$\begin{matrix} \frac{a}{3} = - 4 \\ ∴ & a = - 12 \end{matrix}$

Hence, the required value of $a$ is -12 .

14 The perpendicular bisector of the line segment joining the points

A (1, 5)

and

B (4, 6)

cuts the

Y

-axis at

(a) $(0, 13)$ (b) $(0, - 13)$ (c) $(0, 12)$ (d) $(13, 0)$

Show Answer

Solution

(a) Firstly, we plot the points of the line segment on the paper and join them.

We know that, the perpendicular bisector of the line segment $A B$ bisect the segment $A B$ , i.e., perpendicular bisector of line segment $A B$ passes through the mid-point of $A B$ .

$\begin{matrix} ∴ & Mid-point of & A B = \frac{1 + 4}{2}, \frac{5 + 6}{2} \\ \Rightarrow & P = \frac{5}{2}, \frac{11}{2} \end{matrix}$

$∵$ mid-point of line segment passes through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$

$= \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

Now, we draw a straight line on paper passes through the mid-point $P$ . We see that the perpendicular bisector cuts the $Y$ -axis at the point $(0, 13)$ .

Hence, the required point is $(0, 13)$ .

Alternate Method

We know that, the equation of line which passes through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is

$\begin{matrix} (i) & (y - y_{1}) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \end{matrix}$

Here,

$x_{1} = 1, y_{1} = 5 and x_{2} = 4, y_{2} = 6$

So, the equation of line segment joining the points $A (1, 5)$ and $B (4, 6)$ is

$\begin{matrix} (y - 5) = \frac{6 - 5}{4 - 1} (x - 1) \\ \Rightarrow & (y - 5) = \frac{1}{3} (x - 1) \\ \Rightarrow & 3 y - 15 = x - 1 \\ \Rightarrow & 3 y = x - 14 \Rightarrow & y = \frac{1}{3} x - \frac{14}{3} & \dots (ii) \end{matrix}$

$∴$ Slope of the line segment, $m_{1} = \frac{1}{3}$

If two lines are perpendicular to each other, then the relation between its slopes is

$\begin{matrix} (iii) & m_{1} \cdot m_{2} = - 1 \end{matrix}$

where, $m_{1} =$ Slope of line 1

and $=$ Slope of line 2

Also, we know that the perpendicular bisector of the line segment is perpendicular on the line segment.

Let slope of line segment is $m_{2}$ .

From Eq. (iii),

$\Rightarrow m_{2} = - 3$

$\begin{array}{r} m_{1} \cdot m_{2} = \frac{1}{3} \cdot m_{2} = - 1 \\ m_{2} = - 3 \end{array}$

Also we know that the perpendicular bisector is passes through the mid-point of line segment.

$∴$ Mid-point of line segment $= \frac{1 + 4}{2}, \frac{5 + 6}{2} = \frac{5}{2}, \frac{11}{2}$

Equation of perpendicular bisector, which has slope (-3) and passes through the point $\frac{5}{2}, \frac{11}{2}$ , is

$y - \frac{11}{2} = (- 3) x - \frac{5}{2}$

[since, equation of line passes through the point $(x_{1}, y_{1})$ and having slope $m$

$. (y - y_{1}) = m (x - x_{1})]$

$\begin{matrix} \Rightarrow & (2 y - 11) = - 3 (2 x - 5) \\ \Rightarrow & 2 y - 11 = - 6 x + 15 \\ \Rightarrow & 6 x + 2 y = 26 \\ \Rightarrow & 3 x + y = 13 & \dots (iv) \end{matrix}$

If the perpendicular bisector cuts the $Y$ -axis, then put $x = 0$ in Eq. (iv),

$3 \times 0 + y = 13 \Rightarrow y = 13$

So, the required point is $(0, 13)$ .

15 The coordinates of the point which is equidistant from the three vertices of the

△ A O B

as shown in the figure is

(a) $(x, y)$ (b) $(y, x)$

Show Answer

Thinking Process

(i) Firstly consider the new point be $P (h, k)$ .

(ii) Secondly, determine the distance $P O, P A$ and $P B$ by using the formula, $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ and equating them i. e., $P O = P A = P B$ .

(iii) Further, solving two-two terms at a time and solving them to get required point.

Solution

(a) Let the coordinate of the point which is equidistant from the three vertices $O (0, 0), A (0, 2 y)$ and $B (2 x, 0)$ is $P (h, k)$ .

Then, $P O = P A = P B$

$\Rightarrow$

$\begin{matrix} (i) & (P O)^{2} = (P A)^{2} = (P B)^{2} \end{matrix}$

By distance formula,

$\begin{matrix} [\sqrt{(h - 0)^{2} + (k - 0)^{2}}]^{2} = [\sqrt{(h - 0)^{2} + (k - 2 y)^{2}}]^{2} = [\sqrt{(h - 2 x)^{2} + (k - 0)^{2}}]^{2} \\ \Rightarrow h^{2} + k^{2} = h^{2} + (k - 2 y)^{2} = (h - 2 x)^{2} + k^{2} & \dots (ii) \end{matrix}$

Taking first two equations, we get

$\begin{matrix} h^{2} + k^{2} = h^{2} + (k - 2 y)^{2} \\ \Rightarrow & k^{2} = k^{2} + 4 y^{2} - 4 y k \Rightarrow 4 y (y - k) = 0 \\ \Rightarrow & y = k & [∵ y \neq 0] \end{matrix}$

Taking first and third equations, we get

$\begin{matrix} h^{2} + k^{2} = (h - 2 x)^{2} + k^{2} \\ \Rightarrow & h^{2} = h^{2} + 4 x^{2} - 4 x h \\ \Rightarrow & 4 x (x - h) = 0 \end{matrix}$

$\Rightarrow x = h [∵ x \neq 0]$

$∴$ Required points $= (h, k) = (x, y)$

16 If a circle drawn with origin as the centre passes through

\frac{13}{2}, 0

, then the point which does not lie in the interior of the circle is

(a) $\frac{- 3}{4}, 1$ (b) $2, \frac{7}{3}$

Show Answer

Solution

(d) It is given that, centre of circle in $(0, 0)$ and passes through the point $\frac{13}{2}, 0$ .

$∴$ Radius of circle $=$ Distance between $(0, 0)$ and $\frac{13}{2}, 0$

$= \sqrt{\frac{13}{2} - 0^{2} + (0 - 0)^{2}} = \sqrt{\frac{13^{2}}{2}} = \frac{13}{2} = 6.5$

A point lie outside on or inside the circles of the distance of it from the centre of the circle is greater than equal to or less than radius of the circle.

Now, to get the correct option we have to check the option one by one.

(a) Distance between $(0, 0)$ and $\frac{- 3}{4}, 1 = \sqrt{\frac{- 3}{4} - 0^{2} + (1 - 0)^{2}}$

$= \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4} = 1.25 < 6.5$

So, the point $(- \frac{3}{4}, 1)$ lies interior to the circle.

(b) Distance between $(0, 0)$ and $2, \frac{7}{3} = \sqrt{(2 - 0)^{2} + \frac{7}{3} - 0^{2}}$

$\begin{aligned} = \sqrt{4 + \frac{49}{9}} = \sqrt{\frac{36 + 49}{9}} \\ = \sqrt{\frac{85}{9}} = \frac{9.22}{3} = 3.1 < 6.5 \end{aligned}$

So, the point $2, \frac{7}{3}$ lies inside the circle.

$\begin{aligned} = \sqrt{25 + \frac{1}{4}} = \sqrt{\frac{101}{4}} = \frac{10.04}{2} \\ \Rightarrow = 5.02 < 6.5 \end{aligned}$

So, the point $5, - \frac{1}{2}$ lies inside the circle.

(d) Distance between $(0, 0)$ and $- 6, \frac{5}{2} = \sqrt{(- 6 - 0)^{2} + {\frac{5}{2} - 0^{2}}^{2}}$

$\begin{aligned} = \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{144 + 25}{4}} \\ = \sqrt{\frac{169}{4}} = \frac{13}{2} = 6.5 \end{aligned}$

So, the point $- 6, \frac{5}{2}$ lis an the circle i.e., does not lie interior to the circle.

17.

A

line intersects the

Y

-axis and

X

-axis at the points

P

and

Q

, respectively. If

(2, - 5)

is the mid-point of

P Q

, then the coordinates of

P

and

Q

are, respectively

(a) $(0, - 5)$ and $(2, 0)$ (b) $(0, 10)$ and $(- 4, 0)$

Show Answer

Solution

(d) Let the coordinates of $P$ and $Q (0, y)$ and $(x, 0)$ , respectively.

So, the mid-point of $P (0, y)$ and $Q (x, 0)$ is $M \frac{0 + x}{2}, \frac{y + 0}{2}$ .

$∵$ mid-point of a line segment having points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) = \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

But it is given that, mid-point of $P Q$ is $(2, - 5)$ .

So, the coordinates of $P$ and $Q$ are $(0, - 10)$ and $(4, 0)$ .

18 The area of a triangle with vertices

(a, b + c), (b, c + a)

and

(c, a + b)

(a) $(a + b + c)^{2}$ (b) 0 (c) $(a + b + c)$ (d) $a b c$

Show Answer

Solution

(b) Let the vertices of a triangle are, $A \equiv (x_{1}, y_{1}) \equiv (a, b + c)$

$\begin{array}{lr} B \equiv (x_{2}, y_{2}) \equiv (b, c + a) and C \equiv (x_{3}, y_{3}) \equiv (c, a + b) \\ ∵ Area of △ A B C = Δ = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \\ ∴ Δ = \frac{1}{2} [a (c + a - a - b) + b (a + b - b - c) + c (b + c - c - a)] \\ = \frac{1}{2} [a (c - b) + b (a - c) + c (b - a)] \\ = \frac{1}{2} (a c - a b + a b - b c + b c - a c) = \frac{1}{2} (0) = 0 \end{array}$

Hence, the required area of triangle is 0 .

19 If the distance between the points

(4, p)

and

(1, 0)

is 5 , then the value of

p

(a) 4 only (b) $\pm 4$ (c) -4 only (d) 0

Show Answer

Solution

(b) According to the question, the distance between the points $(4, p)$ and $(1, 0) = 5$ i.e.,

$\sqrt{(1 - 4)^{2} + (0 - p)^{2}} = 5$

$∵$ distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$\begin{matrix} \Rightarrow & \sqrt{(- 3)^{2} + p^{2}} = 5 \\ \Rightarrow & \sqrt{9 + p^{2}} = 5 \end{matrix}$

On squaring both the sides, we get

$\begin{array}{r} 9 + p^{2} = 25 \\ p^{2} = 16 \Rightarrow p = \pm 4 \end{array}$

$\Rightarrow$ $\pm 4$

20 If the points

A (1, 2), B (0, 0)

and

C (a, b)

are collinear, then

(a) $a = b$ (b) $a = 2 b$ (c) $2 a = b$ (d) $a = - b$

Show Answer

Solution

$B \equiv (x_{2}, y_{2}) \equiv (0, 0)$ and $C \equiv (x_{3}, y_{3}) \equiv (a, b)$ .

$∵$ Area of $△ A B C Δ = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$∴ Δ = \frac{1}{2} [1 (0 - b) + 0 (b - 2) + a (2 - 0)]$

$= \frac{1}{2} (- b + 0 + 2 a) = \frac{1}{2} (2 a - b)$

Since, the points $A (1, 2), B (0, 0)$ and $C (a, b)$ are collinear, then area of $△ A B C$ should be equal to zero.

i.e.,

area of $△ A B C = 0$

$\Rightarrow$

$\frac{1}{2} (2 a - b) = 0$

$\Rightarrow 2 a - b = 0$

$\Rightarrow 2 a = b$

Hence, the required relation is $2 a = b$ .

Very Short Answer Type Questions

Write whether True or False and justify your answer

1. $△ A B C$ with vertices $A (0 - 2, 0), B (2, 0)$ and $C (0, 2)$ is similar to $△ D E F$ with vertices $D (- 4, 0), E (4, 0)$ and $F (0, 4)$ .

Show Answer

Solution

True

$∴$ Distance between $A (2, 0)$ and $B (2, 0), A B = \sqrt{[2 - (2)]^{2} + (0 - 0)^{2}} = 4$

$[∵ .$ distance between the points $(x_{1}, y_{1})$ and $. (x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}]$

Similarly, distance between $B (2, 0)$ and $C (0, 2), B C = \sqrt{(0 - 2)^{2} + (2 - 0)^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}$ In $△ A B C$ , distance between $C (0, 2)$ and $A - (2, 0)$ ,

$C A = \sqrt{[0 - (2)^{2}] + (2 - 0)^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}$

Distance between $F (0, 4)$ and $D (- 4, 0), F D = \sqrt{(0 + 4)^{2} + (0 - 4)^{2}} = \sqrt{4^{2} + (- 4)^{2}} = 4 \sqrt{2}$

Distance between $F (0, 4)$ and $E (4, 0), F E = \sqrt{(4 - 0)^{2} + (0 - 4)^{2}} = \sqrt{4^{2} + 4^{2}} = 4 \sqrt{2}$ and distance between $E (4, 0)$ and $D (- 4, 0), E D = \sqrt{[4 - (- 4)]^{2} + (0)^{2}} = \sqrt{8^{2}} = 8$

Now, $\frac{A B}{O E} = \frac{4}{8} = \frac{1}{2}, \frac{A C}{D F} = \frac{2 \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2}, \frac{B C}{E F} = \frac{2 \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2}$

$∴ \frac{A B}{D E} = \frac{A C}{D F} = \frac{B C}{E F}$

Here, we see that sides of $△ A B C$ and $△ F D E$ are propotional.

Hence, both the triangles are similar.

2 The point

P (- 4, 2)

lies on the line segment joining the points

A (- 4, 6)

and

B (- 4, - 6)

Show Answer

Solution

True

We plot all the points $P (- 4, 2), A (- 4, 6)$ and $B (- 4, - 6)$ on the graph paper.

From the figure, point $P (- 4, 2)$ lies on the line segment joining the points $A (- 4, 6)$ and $B (- 4, - 6)$ .

3 The points

(0, 5), (0, - 9)

and

(3, 6)

are collinear.

Show Answer

Solution

False

Here, $x_{1} = 0, x_{2} = 0, x_{3} = 3$ and $y_{1} = 5, y_{2} = - 9, y_{3} = 6$

$∵$ Area of triangle $Δ = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$∴ Δ = \frac{1}{2} [0 (- 9 - 6) + 0 (6 - 5) + 3 (5 + 9)]$

$= \frac{1}{2} (0 + 0 + 3 \times 14) = 21 \neq 0$

If the area of triangle formed by the points $(0, 5), (0 - 9)$ and $(3, 6)$ is zero, then the points are collinear.

Hence, the points are non-collinear.

4 Point

P (0, 2)

is the point of intersection of

Y

-axis and perpendicular bisector of line segment joining the points

A (- 1, 1)

and

B (3, 3)

Show Answer

Solution

False

We know that, the points lies on perpendicular bisector of the line segment joining the two points is equidistant from these two points.

$\begin{aligned} ∴ P A = \sqrt{[- 4 - (4)]^{2} + (6 - 2)^{2}} \\ = \sqrt{(0)^{2} + (4)^{2}} = 4 \\ P B = \sqrt{[- 4 - 4]^{2} + (- 6 - 2)^{2}} \sqrt{(0)^{2} + (- 8)^{2}} = 8 \\ ∵ P A \neq P B \end{aligned}$

So, the point $P$ does not lie on the perpendicular bisctor of $A B$ .

Alternate Method

Slope of the line segment joining the points $A (- 1, 1)$ and $B (3, 3), m_{1} = \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}$

$∵ m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Since, the perpendicular bisector is perpendicular to the line segment, so its slope,

$m_{2} = - \frac{1}{(y_{2})} = - 2$

[by perpendicularity condition, $m_{1} m_{2} = - 1$ ]

Also, the perpendicular bisector passing through the mid-point of the line segment joining the points $A (- 1, 1)$ and $B (3, 3)$ .

$∴ Mid-point = \frac{- 1 + 3}{2}, \frac{1 + 3}{2} = (1, 2)$

[since, mid-point of the line segment joining the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is

$\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

Now, equation of perpendicular bisector have slope $(- 2)$ and passes through the point $(1, 2)$ is

$\begin{matrix} (y - 2) = (- 2) (x - 1) \\ \Rightarrow & y - 2 = - 2 x + 2 \\ \Rightarrow & 2 x + y = 4 & \dots (i) \end{matrix}$

[since, the equation of line is $(y - y_{1}) = m (x - x_{1})$ ]

If the perpendicular bisector cuts the $Y$ -axis, then put $x = 0$ in Eq. (i), we get

$2 \times 0 + y = 4$

$\Rightarrow y = 4$

Hence, the required intersection point is $(0, 4)$ .

5 The points

A (3, 1), B (12, - 2)

and

C (0, 2)

cannot be vertices of a triangle.

Show Answer

Solution

True

Let $A \equiv (x_{1}, y_{1}) \equiv (3, 1), B \equiv (x_{2}, y_{2}) \equiv (12, - 2)$

and $C \equiv (x_{3}, x_{3}) = (0, 2)$

$∴$ Area of $△ A B C Δ = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \frac{1}{2} [3 - (2 - 2) + 12 (2 - 1) + 0 1 - (- 2)]$

$= \frac{1}{3} [3 (- 4) + 12 (1) + 0]$

$= \frac{1}{3} (- 12 + 12) = 0$

$∵$ Area of $△ A B C = 0$

Hence, the points $A (3, 1), B (12, - 2)$ and $C (0, 2)$ are collinear. So, the points $A (3, 1)$ , $B (12, - 2)$ and $C (0, 2)$ cannot be the vertices of a triangle.

6 The points

A (4, 3), B (6, 4), C (5, - 6)

and

D (- 3, 5)

are vertices of a parallelogram.

Show Answer

Solution

False

Now, distance between $A (4, 3)$ and $B (6, 4), A B = \sqrt{(6 - 4)^{2} + (4 - 3)^{2}} = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$

$[∵$ distance between the points $(x_{1}, y_{1})$ and $. (x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}]$

Distance between $B (6, 4)$ and $C (5, - 6), B C = \sqrt{(5 - 6)^{2} + (- 6 - 4)^{2}}$

$\begin{aligned} = \sqrt{(- 1)^{2} + (- 10)^{2}} \\ = \sqrt{1 + 100} = \sqrt{101} \end{aligned}$

Distance between $C (5, - 6)$ and $D (- 3, 5), C D = \sqrt{(- 3 - 5)^{2} + (5 + 6)^{2}}$

$\begin{aligned} = \sqrt{(- 8)^{2} + 11^{2}} \\ = \sqrt{64 + 121} = \sqrt{185} \end{aligned}$

Distance between $D (- 3, 5)$ and $A (4, 3), D A = \sqrt{(4 + 3)^{2} + (3 - 5)^{2}}$

$\begin{aligned} = \sqrt{7^{2} + (- 2)^{2}} \\ = \sqrt{49 + 4} = \sqrt{53} \end{aligned}$

In parallelogram, opposite sides are equal. Here, we see that all sides $A B, B C, C D$ and $D A$ are different.

Hence, given vertices are not the vertices of a parallelogram.

7 A circle has its centre at the origin and a point

P (5, 0)

lies on it. The point

Q (6, 8)

lies outside the circle.

Show Answer

Thinking Process

Firstly, we find the distance between $Q (6, 8)$ and origin $O (0, 0)$ by distance formula and check $O Q$ is greater than the length of radius, i.e., OP or not.

Solution

True

First, we draw a circle and a point from the given information.

Now, distance between origin i.e., $O (0, 0)$ and $P (5, 0), O P = \sqrt{(5 - 0)^{2} + (0 - 0)^{2}}$

$∵$ Distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$= \sqrt{5^{2} + 0^{2}} = 5 = Radius of circle and distance between origin O (0, 0)$

and $Q (6, 8), O Q = \sqrt{(6 - 0)^{2} + (8 - 0)^{2}} = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$

We know that, if the distance of any point from the centre is less than/equal to/ more than the radius, then the point is inside/on/outside the circle, respectively.

Here, we see that, $O Q > O P$

Hence, it is true that point $Q (6, 8)$ , lies outside the circle.

8 The point

A (2, 7)

lies on the perpendicular bisector of the line segment joining the points

P (5, - 3)

and

Q (0, - 4)

Show Answer

Solution

False

If $A (2, 7)$ lies on perpendicular bisector of $P (6, 5)$ and $Q (0, - 4)$ , then $A P = A Q$

$\begin{aligned} ∴ \\ A P = \sqrt{(6 - 2)^{2} + (5 - 7)^{2}} \\ = \sqrt{(4)^{2} + (- 2)^{2}} \\ = \sqrt{16 + 4} = \sqrt{20} \\ A = \sqrt{(0 - 2)^{2} + (- 4 - 7)^{2}} \\ = \sqrt{(- 2)^{2} + (- 11)^{2}} \\ = \sqrt{4 + 121} = \sqrt{125} \end{aligned}$

So, $A$ does not lies on the perpendicular bisector of $P Q$ .

Alternate Method

If the point $A (2, 7)$ lies on the perpendicular bisector of the line segment, then the point $A$ satisfy the equation of perpendicular bisector.

Now, we find the equation of perpendicular bisector. For this, we find the slope of perpendicular bisector.

$\begin{aligned} ∴ Slope of perpendicular bisector = & \frac{- 1}{Slope of line segment joining} \\ the points (5, - 3) and (0, - 4) \\ = \frac{- 1}{\frac{- 4 - (- 3)}{0 - 5}} = 5 ∵ slope = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}$

[since, perpendicular bisector is perpendicular to the line segment, so its slopes have the condition, $. m_{1} \cdot m_{2} = - 1]$

Since, the perpendicular bisector passes through the mid-point of the line segment joining the points $(5, - 3)$ and $(0, - 4)$ .

$∴$ Mid-point of $P Q = \frac{5 + 0}{2}, \frac{- 3 - 4}{2} = \frac{5}{2}, \frac{- 7}{2}$

So, the equation of perpendicular bisector having slope $\frac{1}{3}$ and passes through the mid-point $\frac{5}{2}, \frac{- 7}{2}$ is,

$\begin{matrix} \Rightarrow & 2 y + 7 = 10 x - 25 \\ \Rightarrow & 10 x - 2 y - 32 = 0 \\ \Rightarrow & 10 x - 2 y = 32 \\ \Rightarrow & 5 x - y = 16 & \dots (i) \end{matrix}$

Now, check whether the point $A (2, 7)$ lie on the Eq. (i) or not.

$5 \times 2 - 7 = 10 - 7 = 3 \neq 16$

Hence, the point $A (2, 7)$ does not lie on the perpendicular bisector of the line segment.

9 The point

P (5, - 3)

is one of the two points of trisection of line segment joining the points

A (7, - 2)

and

B (1, - 5)

Show Answer

Solution

True

Let $P (5, - 3)$ divides the line segment joining the points $A (7, - 2)$ and $B (1, - 5)$ in the ratio $k : 1$ internally.

By section formula, the coordinate of point $P$ will be

$\begin{matrix} \frac{k (1) + (1) (7)}{k + 1}, \frac{k (- 5) + 1 (- 2)}{k + 1} \\ i.e., & \frac{k + 7}{k + 1}, \frac{- 5 k - 2}{k + 1} \\ Now, & (5, - 3) = \frac{k + 7}{k + 1}, \frac{- 5 k - 2}{k + 1} \\ \Rightarrow & \frac{k + 7}{k + 1} = 5 \\ \Rightarrow & k + 7 = 5 k + 5 \\ \Rightarrow & - 4 k = - 2 \\ ∴ & k = \frac{1}{2} \end{matrix}$

So the point $P$ divides the line segment $A B$ in ratio $1 : 2$ .

Hence, point $P$ in the point of trisection of $A B$ .

10 The points

A (- 6, 10), B (- 4, 6)

and

C (3, - 8)

are collinear such that

A B = \frac{2}{9} A C

Show Answer

Solution

True

If the area of triangle formed by the points $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is zero, then the points are collinear.

$∵$ Area of triangle $= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

Here, $x_{1} = - 6, x_{2} = - 4, x_{3} = 3$ and $y_{1} = 10, y_{2} = 6, y_{3} = - 8$

$∴$ Area of $△ A B C = \frac{1}{2} [- 6 6 - (- 8) + (- 4) (- 8 - 10) + 3 (10 - 6)]$

$\begin{aligned} = \frac{1}{2} [- 6 (14) + (- 4) (- 18) + 3 (4)] \\ = \frac{1}{2} (- 84 + 72 + 12) = 0 \end{aligned}$

So, given points are collinear.

Now, distance between $A (- 6, 10)$ and $B (- 4, 6), A B = \sqrt{(- 4 + 6)^{2} + (6 - 10)^{2}}$

$= \sqrt{2^{2} + 4^{2}} = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$

$∵$ distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Distance between $A (- 6, 10)$ and $C (3, - 8), A C = \sqrt{(3 + 6)^{2} + (- 8 - 10)^{2}}$

$\begin{aligned} = \sqrt{9^{2} + 18^{2}} = \sqrt{81 + 324} \\ = \sqrt{405} = \sqrt{81 \times 5} = 9 \sqrt{5} \\ ∴ A B = \frac{2}{9} A C \end{aligned}$

which is the required relation.

11 The point

P (- 2, 4)

lies on a circle of radius 6 and centre

(3, 5)

Show Answer

Solution

False

If the distance between the centre and any point is equal to the radius, then we say that point lie on the circle.

Now, distance between $P (- 2, 4)$ and centre $(3, 5)$

$\begin{aligned} = \sqrt{(3 + 2)^{2} + (5 - 4)^{2}} \\ = \sqrt{5^{2} + 1^{2}} \\ = \sqrt{25 + 1} = \sqrt{26} \end{aligned}$

$∵$ distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

which is not equal to the radius of the circle.

Hence, the point $P (- 2, 4)$ does not lies on the circle.

12 The points

A (- 1, - 2), B (4, 3), C (2, 5)

and

D (- 3, 0)

in that order form a rectangle.

Show Answer

Solution

True

Distance between $A (- 1, - 2)$ and $B (4, 3)$ ,

$\begin{aligned} A B = \sqrt{(4 + 1)^{2} + (3 + 2)^{2}} \\ = \sqrt{5^{2} + 5^{2}} = \sqrt{25 + 25} = 5 \sqrt{2} \end{aligned}$

Distance between $C (2, 5)$ and $D (- 3, 0)$ ,

$\begin{aligned} C D = \sqrt{(- 3 - 2)^{2} + (0 - 5)^{2}} \\ = \sqrt{(- 5)^{2} + (- 5)^{2}} \\ = \sqrt{25 + 25} = 5 \sqrt{2} \end{aligned}$

$∵$ distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Distance between $A (- 1, - 2)$ and $D (- 3, 0)$ ,

$A D = \sqrt{(- 3 + 1)^{2} + (0 + 2)^{2}} = \sqrt{(- 2)^{2} + 2^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}$

and distance between $B (4, 3)$ and $C (2, 5), B C = \sqrt{(4 - 2)^{2} + (3 - 5)^{2}}$

$= \sqrt{2^{2} + (- 2)^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}$

We know that, in a rectangle, opposite sides and equal diagonals are equal and bisect each other.

Since,

$A B = C D and A D = B C$

Also, distance between $A (- 1, - 2)$ and $C (2, 5), A C = \sqrt{(2 + 1)^{2} + (5 + 2)^{2}}$

$\begin{array}{r} = \sqrt{3^{2} + 7^{2}} = \sqrt{9 +} \\ and distance between D (- 3, 0) and B (4, 3), D B = \sqrt{(4 + 3)^{2} + (3 - 0)^{2}} \end{array}$

$= \sqrt{7^{2} + 3^{2}} = \sqrt{49 + 9} = \sqrt{58}$

Since, diagonals $A C$ and $B D$ are equal.

Hence, the points $A (- 1, - 2), B (4, 3), C (2, 5)$ and $D (- 3, 0)$ form a rectangle.

Short Answer Type Questions

1 Name the type of triangle formed by the points $A (- 5, 6), B (- 4, - 2)$ and $C (7, 5)$ .

Show Answer

Thinking Process

(i) Firstly, determine the distances $A B, B C$ and $C A$ by using the distance formula,

$\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

(ii) Secondly, check the condition for types of a triangles

(a) If any two sides are equal, then it is an isosceles triangle.

(b) If sides of a triangle satisfy the phythagoras theorem, then it is a right angled triangle.

(d) If none of the side of a triangle are equal, then it is an scalene triangle.

Solution

To find the type of triangle, first we determine the length of all three sides and see whatever condition of triangle is satisfy by these sides.

Now, using distance formula between two points,

and

$\begin{array}{r} A B = \sqrt{(- 4 + 5)^{2} + (- 2 - 6)^{2}} \\ = \sqrt{(1)^{2} + (- 8)^{2}} \\ = \sqrt{1 + 64} = \sqrt{65} [∵ d = \sqrt{(x_{2} - x_{1}) + (y_{2} - y_{1})^{2}}] \\ B C = \sqrt{(7 + 4)^{2} + (5 + 2)^{2}} = \sqrt{(11)^{2} + (7)^{2}} \\ = \sqrt{121 + 49} = \sqrt{170} \\ C A = \sqrt{(- 5 - 7)^{2} + (6 - 5)^{2}} = \sqrt{(- 12)^{2} + (1)^{2}} \\ = \sqrt{144 + 1} = \sqrt{145} \\ A B \neq B C \neq C A \end{array}$

We see that,

and not hold the condition of Pythagoras in a $△ A B C$ .

i.e., $(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}$

Hence, the required triangle is scalene because all of its sides are not equal i.e., different to each other.

2 Find the points on the

X

-axis which are at a distance of

2 \sqrt{5}

from the point

(7, - 4)

. How many such points are there?

Show Answer

Solution

We know that, every point on the $X$ -axis in the form $(x, 0)$ . Let $P (x, 0)$ the point on the $X$ -axis have $2 \sqrt{5}$ distance from the point $Q (7, - 4)$ .

$By given condition, P Q = 2 \sqrt{5} [∵ distance formula = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}]$

$\begin{array}{r} \Rightarrow (P Q)^{2} = 4 \times 5 \\ \Rightarrow (x - 7)^{2} + (0 + 4)^{2} = 20 \\ \Rightarrow x^{2} + 49 - 14 x + 16 = 20 \\ \Rightarrow x^{2} - 14 x + 65 - 20 = 0 \\ \Rightarrow x^{2} - 14 x + 45 = 0 \\ \Rightarrow x^{2} - 9 x - 5 x + 45 = 0 [by factorisation method] \\ \Rightarrow x (x - 9) - 5 (x - 9) = 0 \\ \Rightarrow (x - 9) (x - 5) = 0 \\ ∴ x = 5, 9 \end{array}$

Hence, there are two points lies on the axis, which are $(5, 0)$ and $(9, 0)$ , have $2 \sqrt{5}$ distance from the point $(7, - 4)$ .

3 What type of quadrilateral do the points

A (2, - 2), B (7, 3) C (11, - 1)

and

D (6, - 6)

taken in that order form?

Show Answer

Thinking Process

(i) Firstly, determine the distances $A B, B C, C D, D A, A C$ and $B D$ by using the distance formula $= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .

(ii) Secondly, check the condition for types of a quadrilaterals

(a) If all four sides and also diagonals are equal, then quadrilateral is a square.

(b) If all four sides are equal but diagonal are not equal, then quadrilateral is a rhombus.

Solution

To find the type of quadrilateral, we find the length of all four sides as well as two diagonals and see whatever condition of quadrilateral is satisfy by these sides as well as diagonals. Now, using distance formula between two points,

$sides, \begin{aligned} A B = \sqrt{(7 - 2)^{2} + (3 + 2)^{2}} \\ = \sqrt{(5)^{2} + (5)^{2}} = \sqrt{25 + 25} \\ = \sqrt{50} = 5 \sqrt{2} \end{aligned}$

since, distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) = \sqrt{(x_{2} - x_{1}) + (y_{2} - y_{1})^{2}}$

and

$\begin{aligned} B C = & \sqrt{(11 - 7)^{2} + (- 1 - 3)^{2}} = \sqrt{(4)^{2} + (- 4)^{2}} \\ = & \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2} \\ C D = & \sqrt{(6 - 11)^{2} + (- 6 + 1)^{2}} \\ = & \sqrt{(- 5)^{2} + (- 5)^{2}} \\ = & \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2} \\ D A = & \sqrt{(2 - 6)^{2} + (- 2 + 6)^{2}} \\ = & \sqrt{(- 4)^{2} + (4)^{2}} = \sqrt{16 + 16} \\ = & \sqrt{32} = 4 \sqrt{2} \\ Diagonals, A C = & \sqrt{(11 - 2)^{2} + (- 1 + 2)^{2}} \\ = & \sqrt{(9)^{2} + (1)^{2}} = \sqrt{81 + 1} = \sqrt{82} \\ B D = & \sqrt{(6 - 7)^{2} + (- 6 - 3)^{2}} \\ = & \sqrt{(- 1)^{2} + (- 9)^{2}} \\ = & \sqrt{1 + 81} = \sqrt{82} \end{aligned}$

and

Here, we see that the sides $A B = C D$ and $B C = D A$

Also, diagonals are equal i.e., $A C = B D$

which shows the quadrilateral is a rectangle.

4 Find the value of

a

, if the distance between the points

A (- 3, - 14)

and

B (a, - 5)

is 9 units.

Show Answer

Solution

According to the question,

Distance between $A (- 3, - 14)$ and $B (a, - 5), A B = 9$

$[∵$ distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1}) + (y_{2} - y_{1})^{2}}]$

$\begin{matrix} \Rightarrow & \sqrt{(a + 3)^{2} + (- 5 + 14)^{2}} = 9 \\ \Rightarrow & \sqrt{(a + 3)^{2} + (9)^{2}} = 9 \end{matrix}$

On squaring both the sides, we get

$(a + 3)^{2} + 81 = 81$

$\Rightarrow (a + 3)^{2} = 0 \Rightarrow a = - 3$

Hence, the required value of $a$ is -3 .

5 Find a point which is equidistant from the points

A (- 5, 4)

and

B (- 1, 6)

. How many such points are there?

Show Answer

Solution

Let $P (h, k)$ be the point which is equidistant from the points $A (- 5, 4)$ and $B (- 1, 6)$ .

$\begin{aligned} ∴ P A = P B [∵ by distance formula, distance = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}] \\ \Rightarrow (P A)^{2} = (P B)^{2} \\ \Rightarrow (- 5 - h)^{2} + (4 - k)^{2} = (- 1 - h)^{2} + (6 - k)^{2} \\ \Rightarrow 25 + h^{2} + 10 h + 16 + k^{2} - 8 k = 1 + h^{2} + 2 h + 36 + k^{2} - 12 k \\ \Rightarrow 25 + 10 h + 16 - 8 k = 1 + 2 h + 36 - 12 k \\ \Rightarrow 8 h + 4 k + 41 - 37 = 0 \\ \Rightarrow 8 h + 4 k + 4 = 0 \\ \Rightarrow 2 h + k + 1 = 0 \\ Mid-point of A B = \frac{- 5 - 1}{2}, \frac{4 + 6}{2} = (- 3, 5) \\ ∵ mid-point = \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2} \end{aligned}$

At point (-3, 5), from Eq. (i),

$\begin{aligned} 2 h + k = 2 (- 3) + 5 \\ = - 6 + 5 = - 1 \\ \Rightarrow 2 h + k + 1 = 0 \end{aligned}$

So, the mid-point of $A B$ satisfy the Eq. (i). Hence, infinite number of points, in fact all points which are solution of the equation $2 h + k + 1 = 0$ , are equidistant from the points $A$ and $B$ .

Replacing $h, k$ by $x$ , $y$ in above equation, we have $2 x + y + 1 = 0$

6 Find the coordinates of the point

Q

on the

X

-axis which lies on the perpendicular bisector of the line segment joining the points

A (- 5, - 2)

and B

(4, - 2)

. Name the type of triangle formed by the point

Q, A

and B.

Show Answer

Solution

Firstly, we plot the points of the line segment on the paper and join them.

We know that, the perpendicular bisector of the line segment $A B$ bisect the segment $A B$ , i.e., perpendicular bisector of the line segment $A B$ passes through the mid-point of $A B$ .

$\begin{matrix} ∴ & Mid-point of A B = \frac{- 5 + 4}{2}, \frac{- 2 - 2}{2} \\ \Rightarrow & R = - \frac{1}{2}, - 2 \end{matrix}$

$[∵ .$ mid-point of a line segment passes through the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is

$\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

Now, we draw a straight line on paper passes through the mid-point $R$ . We see that perpendicular bisector cuts the $X$ -axis at the point $Q - \frac{1}{2}, 0$ .

Hence, the required coordinates of $Q \equiv - \frac{1}{2}, 0$

Alternate Method

(i) To find the coordinates of the point of $Q$ on the $X$ -axis. We find the equation of perpendicular bisector of the line segment $A B$ .

Now, slope of line segment $A B$ ,

Let

$\begin{array}{r} m_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 2 - (- 2)}{4 - (- 5)} = \frac{- 2 + 2}{4 + 5} = \frac{0}{9} \\ m_{1} = 0 \end{array}$

$\Rightarrow$

Let the slope of perpendicular bisector of line segment is $m_{2}$ .

Since, perpendicular bisector is perpendicular to the line segment $A B$ .

By perpendicularity condition of two lines,

$\begin{array}{r} m_{1} \cdot m_{2} = - 1 \\ m_{2} = \frac{- 1}{m_{1}} = \frac{- 1}{0} \\ m_{2} = \infty \end{array}$

Also, we know that, the perpendicular bisector is always passes through the mid-point of the line segment.

$\begin{aligned} ∴ Mid-point = \frac{- 5 + 4}{2}, \frac{- 2 - 2}{2} = & \frac{- 1}{2}, - 2 \\ ∵ mid-point = \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2} \end{aligned}$

To find the equation of perpendicular bisector of line segment, we find the slope and a point through which perpendicular bisector is pass.

Now, equation of perpendicular bisector having slope $\infty$ and passing through the point $\frac{- 1}{2}, - 2$ is

$\begin{array}{r} (y + 2) = \infty x + \frac{1}{2} \\ \Rightarrow \frac{y + 2}{x + \frac{1}{2}} = \infty = \frac{1}{0} \Rightarrow x + \frac{1}{2} = 0 \\ ∴ x = \frac{- 1}{2} \end{array}$

So, the coordinates of the point $Q$ is $\frac{- 1}{2}, 0$ on the $X$ -axis which lies on the perpendicular bisector of the line segment joining the point $A B$ .

To know the type of triangle formed by the points $Q, A$ and $B$ . We find the length of all three sides and see whatever condition of triangle is satisfy by these sides.

Now, using distance formula between two points,

$A B = \sqrt{(4 + 5)^{2} + (- 2 + 2)^{2}} = \sqrt{(9)^{2} + 0} = 9$

$[∵ .$ distance between two points $(x_{1}, y_{1})$ and $. (x_{2}, y_{2}) = \sqrt{(x_{2} - x_{1}) + (y_{2} - y_{1})^{2}}]$

and

$\begin{aligned} B Q = \sqrt{\frac{- 1}{2} - 4^{2} + (0 + 2)^{2}} \\ = \sqrt{\frac{- 9^{2}}{2} + (2)^{2}} = \sqrt{\frac{81}{4} +} \\ Q A = \sqrt{- 5 + \frac{1}{2} + (- 2 - 0)^{2}} \\ = \sqrt{\frac{- 9^{2}}{2} + (2)^{2}} \\ = \sqrt{\frac{81}{4} + 4} = \sqrt{\frac{97}{4}} = \sqrt{\frac{97}{2}} \end{aligned}$

$= \sqrt{\frac{- 9^{2}}{2} + (2)^{2}} = \sqrt{\frac{81}{4} + 4} = \sqrt{\frac{97}{4}} = \sqrt{\frac{97}{2}}$

We see that, $B Q = Q A \neq A B$

which shows that the triangle formed by the points $Q, A$ and $B$ is an isosceles.

7 Find the value of

m

, if the points

(5, 1), (- 2, - 3)

and

(8, 2 m)

are collinear.

Show Answer

Thinking Process

(i) First, using the condition of collinearity

$\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0$

(ii) Simplify it and get the result.

Solution

Let $A \equiv (x_{1}, y_{1}) \equiv (5, 1), B \equiv (x_{2}, y_{2}) \equiv (- 2, - 3), C \equiv (x_{3}, y_{3}) \equiv (8, 2 m)$ Since, the points $A \equiv (5, 1), B \equiv (- 2, - 3)$ and $C \equiv (8, 2 m)$ are collinear.

$∴$ Area of $△ A B C = 0$

$\Rightarrow$

$\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0$

$\Rightarrow \frac{1}{2} [5 (- 3 - 2 m) + (- 2) (2 m - 1) + 8 1 - (- 3)] = 0$

$\Rightarrow$ $\frac{1}{2} (- 15 - 10 m - 4 m + 2 + 32) = 0$

$\Rightarrow \frac{1}{2} (- 14 m + 19) = 0 \Rightarrow m = \frac{19}{14}$

Hence, the required value of $m$ is $\frac{19}{14}$ .

8 If the point

A (2, - 4)

is equidistant from

P (3, 8)

and

Q (- 10, y)

, then find the value of

y

. Also, find distance

P Q

Show Answer

Solution

According to the question,

$A (2, - 4)$ is equidistant from $P (3, 8) = Q (- 10, y)$ is equidistant from $A (2, - 4)$

i.e.,

$\Rightarrow$ $\begin{array}{r} P A = Q A \\ \sqrt{(2 - 3)^{2} + (- 4 - 8)^{2}} = \sqrt{(2 + 10)^{2} + (- 4 - y)^{2}} \end{array}$

$[∵ .$ distance between two points $(x_{1}, y_{1})$ and $. (x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1}) + (y_{2} - y_{1})^{2}}]$

$\begin{matrix} \Rightarrow & \sqrt{(- 1)^{2} + (- 12)^{2}} = \sqrt{(12)^{2} + (4 + y)^{2}} \\ \Rightarrow & \sqrt{1 + 144} = \sqrt{144 + 16 + y^{2} + 8 y} \\ \Rightarrow & \sqrt{145} = \sqrt{160 + y^{2} + 8 y} \end{matrix}$

On squaring both the sides, we get

$\begin{matrix} 145 = 160 + y^{2} + 8 y \\ 2^{2} + 8 y + 160 - 145 = 0 \\ y^{2} + 8 y + 15 = 0 \\ y^{2} + 5 y + 3 y + 15 = 0 \\ y (y + 5) + 3 (y + 5) = 0 \\ (y + 5) (y + 3) = 0 \end{matrix}$

$\begin{matrix} \Rightarrow & y^{2} + 8 y + 160 - 145 = 0 \\ \Rightarrow & y^{2} + 8 y + 15 = 0 \\ \Rightarrow & y^{2} + 5 y + 3 y + 15 = 0 \\ \Rightarrow & y (y + 5) + 3 (y + 5) = 0 \end{matrix}$

If $y + 5 = 0$ , then $y = - 5$

If $y + 3 = 0$ , then $y = - 3$

$∴ y = - 3, - 5$

Now, distance between $P (3, 8)$ and $Q (- 10, y)$ ,

$\begin{matrix} P Q = \sqrt{(- 10 - 3)^{2} + (y - 8)^{2}} \\ = \sqrt{(- 13)^{2} + (- 3 - 8)^{2}} \\ = \sqrt{169 + 121} = \sqrt{290} \end{matrix}$

Again, distance between $P (3, 8)$ and $(- 10, y), P Q = \sqrt{(- 13)^{2} + (- 5 - 8)^{2}}$ [putting $y = - 5$ ]

$= \sqrt{169 + 169} = \sqrt{338}$

Hence, the values of $y$ are $- 3, - 5$ and corresponding values of $P Q$ are $\sqrt{290}$ and $\sqrt{338} = 13 \sqrt{2}$ , respectively.

9 Find the area of the triangle whose vertices are

(- 8, 4), (- 6, 6)

and

(- 3, 9)

Show Answer

Solution

Given that, the vertices of triangles

Let

$\begin{aligned} (x_{1}, y_{1}) & \to (- 8, 4) \\ (x_{2}, y_{2}) & \to (- 6, 6) \\ (x_{3}, y_{3}) & \to (- 3, 9) \end{aligned}$

and

We know that, the area of triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$

$\begin{array}{r} Δ = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} + (y_{1} - y_{2})] \\ ∴ = \frac{1}{2} [- 8 (6 - 9) - 6 (9 - 4) + (- 3) (4 - 6)] \\ = \frac{1}{2} [- 8 (- 3) - 6 (5) - 3 (- 2)] = \frac{1}{2} (24 - 30 + 6) \\ = \frac{1}{2} (30 - 30) = \frac{1}{2} (0) = 0 \end{array}$

Hence, the required area of triangle is 0 .

10 In what ratio does the

X

-axis divide the line segment joining the points

(- 4, - 6)

and

(- 1, 7)

? Find the coordinates of the points of division.

Show Answer

Thinking Process

(i) Firstly, determine the ratio by using the formula $\frac{λ x_{2} + x_{1}}{λ + 1}, \frac{λ y_{2} + y_{1}}{λ + 1}$ .

(ii) Further, put the y coordinate of above formula is zero and get the value of $λ$ .

(iii) Finally, put the value of $λ$ in the internally section formula and get the required result.

Solution

Let the required ratio be $λ : 1$ . So, the coordinates of the point $M$ of division $A (- 4, - 6)$ and $B (- 1, 7)$ are

$\frac{λ x_{2} + 1 \cdot x_{1}}{λ + 1}, \frac{λ y_{2} + 1 \cdot y_{1}}{λ + 1}$

Here, $x_{1} = - 4, x_{2} = - 1$ and $y_{1} = - 6, y_{2} = 7$

i.e., $\frac{λ (- 1) + 1 (- 4)}{λ + 1}, \frac{λ (7) + 1 \cdot (- 6)}{λ + 1} = \frac{- λ - 4}{λ + 1}, \frac{7 λ - 6}{λ + 1}$

But according to the question, line segment joining $A (- 4, - 6)$ and $B (- 1, 7)$ is divided by the $X$ -axis. So, $y$ -coordinate must be zero.

$\begin{aligned} ∴ \frac{7 λ - 6}{λ + 1} \Rightarrow 7 λ - 6 = 0 \\ ∴ λ = \frac{6}{7} \end{aligned}$

So, the required ratio is $6 : 7$ and the point of division $M$ is $\frac{- \frac{6}{7} - 4}{\frac{6}{7} + 1}, \frac{7 \times \frac{6}{7} - 6}{\frac{6}{7} + 1}$

i.e., $\frac{- 34}{\frac{13}{7}}, \frac{6 - 6}{\frac{13}{7}}$ i.e., $\frac{- 34}{13}, 0$ .

Hence, the required point of division is $\frac{- 34}{13}, 0$ .

11 Find the ratio in which the point

P \frac{3}{4}, \frac{5}{12}

divides the line segment joining the points

A \frac{1}{2}, \frac{3}{2}

and

B (2, - 5)

Show Answer

Solution

Let $P \frac{3}{4}, \frac{5}{12}$ divide $A B$ internally in the ratio $m : n$ .

Using the section formula, we get

$\frac{3}{4}, \frac{5}{12} = \frac{2 m - \frac{n}{2}}{m + n}, \frac{- 5 m + \frac{3}{2} n}{m + n}$

$∵$ internal section formula, the coordinates of point $P$ divides the line segment joining the point $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in the ratio $m_{1} : m_{2}$ internally is $\frac{m_{2} x_{1} + m_{1} x_{2}}{m_{1} + m_{2}}, \frac{m_{2} y_{1} + m_{1} y_{2}}{m_{1} + m_{2}}$

On equating, we get

$\begin{aligned} \frac{3}{4} = \frac{2 m - \frac{n}{2}}{m + n} and \frac{5}{12} = \frac{- 5 m + \frac{3}{2} n}{m + n} \\ \Rightarrow \frac{3}{4} = \frac{4 m - n}{2 (m + n)} and \frac{5}{12} = \frac{- 10 m + 3 n}{2 (m + n)} \\ \Rightarrow \frac{3}{2} = \frac{4 m - n}{m + n} and \frac{5}{6} = \frac{- 10 m + 3 n}{m + n} \\ \Rightarrow 3 m + 3 n = 8 m - 2 n and 5 m + 5 n = - 60 m + 18 n \\ \Rightarrow 5 n - 5 m = 0 and 65 m - 13 n = 0 \\ \Rightarrow n = m and 13 (5 m - n) = 0 \\ \Rightarrow n = m and 5 m - n = 0 \\ Since, m = n does not satisfy. \\ ∴ 5 m - n = 0 \\ \Rightarrow 5 m = n \\ ∴ \frac{m}{n} = \frac{1}{5} \end{aligned}$

Hence, the required ratio is $1 : 5$ ,

12 If

P (9 a - 2, - b)

divides line segment joining

A (3 a + 1, - 3)

and

B (8 a, 5)

in the ratio

3 : 1

, then find the values of

a

and

b

Show Answer

Solution

Let $P (9 a - 2, - b)$ divides $A B$ internally in the ratio $3 : 1$ .

By section formula,

$9 a - 2 = \frac{3 (8 a) + 1 (3 a + 1)}{3 + 1}$

and

$\Rightarrow 9 a - 2 = \frac{24 a + 3 a + 1}{4}$

and

$- b = \frac{15 - 3}{4}$

$\Rightarrow$ $9 a - 2 = \frac{27 a + 1}{4}$

and

$- b = \frac{12}{4}$

$\Rightarrow 36 a - 8 = 27 a + 1$

and

$b = - 3$

$\Rightarrow 36 a - 27 a - 8 - 1 = 0$

$\Rightarrow 9 a - 9 = 0$

$∴ a = 1$

Hence, the required values of $a$ and $b$ are 1 and -3 .

13. If

(a, b)

is the mid-point of the line segment joining the points

A (10, - 6), B (k, 4)

and

a - 2 b = 18

, then find the value of

k

and the distance AB.

Show Answer

Solution

Since, $(a, b)$ is the mid-point of line segment $A B$ . $∴$ $(a, b) = \frac{10 + k}{2}, \frac{- 6 + 4}{2}$

since, mid-point of a line segment having points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) = \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

$\Rightarrow (a, b) = \frac{10 + k}{2}, - 1$

Now, equating coordinates on both sides, we get

$\begin{matrix} ∴ & a = \frac{10 + k}{2} \\ and & b = - 1 \\ Given, & a - 2 b = 18 \\ From Eq. (ii), & a - 2 (- 1) = 18 \\ \Rightarrow & a + 2 = 18 \Rightarrow a = 16 \\ From Eq. (i), & 16 = \frac{10 + k}{2} \\ \Rightarrow & 32 = 10 + k \Rightarrow k = 22 \end{matrix}$

Hence, the required value of $k$ is 22 .

$\Rightarrow k = 22$

$∴ A \equiv (10, - 6), B \equiv (22, 4)$

Now, distance between $A (10, - 6)$ and $B (22, 4)$ ,

$A B = \sqrt{(22 - 10)^{2} + (4 + 6)^{2}}$

$[∵ .$ distance between the points $(x_{1}, y_{1})$ and $. (x_{2}, y_{2}), d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}]$

$\begin{aligned} = \sqrt{(12)^{2} + (10)^{2}} = \sqrt{144 + 100} \\ = \sqrt{244} = 2 \sqrt{61} \end{aligned}$

Hence, the required distance of $A B$ is $2 \sqrt{61}$ .

14 If the centre of a circle is

(2 a, a - 7)

, then Find the values of

a

, if the circle passes through the point

(11, - 9)

and has diameter

10 \sqrt{2}

units.

Show Answer

Thinking Process

(i) Firstly, determine the distance between centre and point on a circle by using the distance Formula $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ , which is equal to the radius of circle.

(ii) Further, using the given condition and simply it.

Solution

By given condition,

Distance between the centre $C (2 a, a - 7)$ and the point $P (11, - 9)$ , which lie on the circle $=$ Radius of circle

$∴$ Radius of circle $= \sqrt{(11 - 2 a)^{2} + (- 9 - a + 7)^{2}}$

$[∵ .$ distance between two points $(x_{1}, y_{1})$ and $. (x_{2}, y_{2}) = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}]$

Given that, length of diameter $= 10 \sqrt{2}$

$\begin{aligned} ∴ Length of radius = \frac{Length of diameter}{2} \\ = \frac{10 \sqrt{2}}{2} = 5 \sqrt{2} \end{aligned}$

Put this value in Eq. (i), we get

$5 \sqrt{2} = \sqrt{(11 - 2 a)^{2} + (- 2 - a)^{2}}$

Squaring on both sides, we get

$\begin{matrix} 50 = (11 - 2 a)^{2} + (2 + a)^{2} \\ \Rightarrow & 50 = 121 + 4 a^{2} - 44 a + 4 + a^{2} + 4 a \\ \Rightarrow & 5 a^{2} - 40 a + 75 = 0 \\ \Rightarrow & a^{2} - 8 a + 15 = 0 \\ \Rightarrow & a^{2} - 5 a - 3 a + 15 = 0 \\ \Rightarrow & a (a - 5) - 3 (a - 5) = 0 \\ \Rightarrow & (a - 5) (a - 3) = 0 \\ ∴ & a = 3, 5 \end{matrix}$

$\Rightarrow a^{2} - 5 a - 3 a + 15 = 0 [by factorisation method]$

Hence, the required values of $a$ are 5 and 3 .

15 The line segment joining the points

A (3, 2)

and

B (5, 1)

is divided at the point

P

in the ratio

1 : 2

and it lies on the line

3 x - 18 y + k = 0

. Find the value of

k

Show Answer

Solution

Given that, the line segment joining the points $A (3, 2)$ and $B (5, 1)$ is divided at the point $P$ in the ratio $1 : 2$ .

$\begin{aligned} ∴ Coordinate of point P & \equiv \frac{5 (1) + 3 (2)}{1 + 2}, \frac{1 (1) + 2 (2)}{1 + 2} \\ \equiv \frac{5 + 6}{3}, \frac{1 + 4}{3} \equiv \frac{11}{3}, \frac{5}{3} \\ ∵ by section formula for internal ratio \equiv \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{aligned}$

But the point $P \frac{11}{3}, \frac{5}{3}$ lies on the line $3 x - 18 y + k = 0$

[given]

$\begin{aligned} ∴ & 3 \frac{11}{3} - 18 \frac{5}{3} + k = 0 \\ \Rightarrow & 11 - 30 + k = 0 \\ \Rightarrow & k - 19 = 0 \Rightarrow k = 19 \end{aligned}$

Hence, the required value of $k$ is 19 .

16 If

D - \frac{1}{2}, \frac{5}{2}, E (7, 3)

and

F \frac{7}{2}, \frac{7}{2}

are the mid-points of sides of

△ A B C

, then find the area of the

△ A B C

Show Answer

Thinking Process

(i) Firstly, consider the vertices of of $△ A B C$ be $A (x_{1} y_{y}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ .

(ii) With the help of mid-point formula form the equations in terms of $x_{1}, y_{1}, x_{2}, y_{2}, x_{3}$ and $y_{3}$ and solve them to get the values of vertices.

(iii) Further, determine the area of triangle by using the formula,

$| \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] |$

Solution

Let $A \equiv (x_{1}, y_{1}), B \equiv (x_{2}, y_{2})$ and $C \equiv (x_{3}, y_{3})$ are the vertices of the $△ A B C$ .

Gives, $D - \frac{1}{2}, \frac{5}{2}, E (7, 3)$ and $F \frac{7}{2}, \frac{7}{2}$ be the mid-points of the sides $B C, C A$ and $A B$ , respectively.

Since, $D - \frac{1}{2}, \frac{5}{2}$ is the mid-point of $B C$ .

$∴ \frac{x_{2} + x_{3}}{2} = - \frac{1}{2}$

since, mid-point of a line segment having points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

$\begin{matrix} and & \frac{y_{2} + y_{3}}{2} = \frac{5}{2} \\ \Rightarrow & x_{2} + x_{3} = - 1 \\ and & y_{2} + y_{3} = 5 & \dots (ii) \\ As E (7, 3) is the mid-point of C A . \end{matrix}$

$∴$ $\frac{x_{3} + x_{1}}{2} = 7$

and $\frac{y_{3} + y_{1}}{2} = 3$

$\Rightarrow$ $x_{3} + x_{1} = 14$

and $y_{3} + y_{1} = 6$

Also, $F \frac{7}{2}, \frac{7}{2}$ is the mid-point of $A B$ .

$\begin{matrix} ∴ & \frac{x_{1} + x_{2}}{2} = \frac{7}{2} \\ and & \frac{y_{1} + y_{2}}{2} = \frac{7}{2} \\ \Rightarrow & x_{1} + x_{2} = 7 \\ and & y_{1} + y_{2} = 7 & \dots (iv) \end{matrix}$

On adding Eqs. (i), (iii) and (v), we get

$\begin{matrix} 2 (x_{1} + x_{2} + x_{3}) = 20 \\ \Rightarrow & x_{1} + x_{2} + x_{3} = 10 & \dots (vii) \end{matrix}$

On subtracting Eqs. (i), (iii) and (v) from Eq. (vii) respectively, we get

$x_{1} = 11, x_{2} = - 4, x_{3} = 3$

On adding Eqs. (ii), (iv) and (vi), we get

$\begin{matrix} 2 (y_{1} + y_{2} + y_{3}) = 18 \\ \Rightarrow & y_{1} + y_{2} + y_{3} = 9 \end{matrix}$

On subtracting Eqs. (ii), (iv) and (vi) from Eq. (viii) respectively, we get

$y_{1} = 4, y_{2} = 3, y_{3} = 2$

Hence, the vertices of $△ A B C$ are $A (11, 4), B (- 4, 3)$ and $C (3, 2)$ .

$\begin{aligned} ∵ Area of △ A B C = Δ = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \\ ∴ Δ = \frac{1}{2} [11 (3 - 2) + (- 4) (2 - 4) + 3 (4 - 3)] \\ = \frac{1}{2} [11 \times 1 + (- 4) (- 2) + 3 (1)] \\ = \frac{1}{2} (11 + 8 + 3) = \frac{22}{2} = 11 \end{aligned}$

$∴$ Required area of $△ A B C = 11$

17 If the points

A (2, 9), B (a, 5)

and

C (5, 5)

are the vertices of a

△ A B C

right angled at

B

, then find the values of

a

and hence the area of

△ A B C

Show Answer

Solution

Given that, the points $A (2, 9), B (a, 5)$ and $C (5, 5)$ are the vertices of a $△ A B C$ right angled at $B$ . By Pythagoras theorem,

$\begin{matrix} A C^{2} = A B^{2} + B C^{2} & \dots (i) \end{matrix}$

Now, by distance formula, $A B = \sqrt{(a - 2)^{2} + (5 - 9)^{2}}$

$∵$ distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) = \sqrt{(x_{2} - x_{1}^{2}) + (y_{2} - y_{1})^{2}}$

$\begin{aligned} = \sqrt{a^{2} + 4 - 4 a + 16} = \sqrt{a^{2} - 4 a + 20} \\ B C = \sqrt{(5 - a)^{2} + (5 - 5)^{2}} \\ = \sqrt{(5 - a)^{2} + 0} = 5 - a \\ A C = \sqrt{(2 - 5)^{2} + (9 - 5)^{2}} \\ = \sqrt{(- 3)^{2} + (4)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 \end{aligned}$

and

Put the values of $A B, B C$ and $A C$ in Eq. (i), we get

$\begin{matrix} (5)^{2} = (\sqrt{a^{2} - 4 a + 20})^{2} + (5 - a)^{2} \\ \Rightarrow & 25 = a^{2} - 4 a + 20 + 25 + a^{2} - 10 a \\ \Rightarrow & 2 a^{2} - 14 a + 20 = 0 \\ \Rightarrow & a^{2} - 7 a + 10 = 0 \\ \Rightarrow & a^{2} - 2 a - 5 a + 10 = 0 \\ \Rightarrow & a (a - 2) - 5 (a - 2) = 0 \\ \Rightarrow & (a - 2) (a - 5) = 0 \\ ∴ & a = 2, 5 \end{matrix}$

$\Rightarrow a^{2} - 2 a - 5 a + 10 = 0 [by factorisation method]$

Here, $a \neq 5$ , since at $a = 5$ , the length of $B C = 0$ . It is not possible because the sides $A B, B C$ and $C A$ form a right angled triangle.

So,

$a = 2$

Now, the coordinate of $A, B$ and $C$ becomes $(2, 9), (2, 5)$ and $(5, 5)$ , respectively.

$\begin{aligned} ∵ Area of △ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \\ ∴ Δ = \frac{1}{2} [2 (5 - 5) + 2 (5 - 9) + 5 (9 - 5)] \\ = \frac{1}{2} [2 \times 0 + 2 (- 4) + 5 (4)] \\ = \frac{1}{2} (0 - 8 + 20) = \frac{1}{2} \times 12 = 6 \end{aligned}$

Hence, the required area of $△ A B C$ is 6 sq units.

18 Find the coordinates of the point

R

on the line segment joining the points

P (- 1, 3)

and

Q (2, 5)

such that

P R = \frac{3}{5} P Q

Show Answer

Solution

According to the question,

$\begin{matrix} Given that, & P R = \frac{3}{5} P Q \\ \Rightarrow & \frac{P Q}{P R} = \frac{5}{3} \\ \Rightarrow & \frac{P R + R Q}{P R} = \frac{5}{3} \\ \Rightarrow & 1 + \frac{R Q}{P R} = \frac{5}{3} \\ \Rightarrow & \frac{P Q}{P R} = \frac{5}{3} - 1 = \frac{2}{3} \\ ∴ & R Q : P R = 2 : 3 \\ or & P R : R Q = 3 : 2 \end{matrix}$

Suppose, $R (x, y)$ be the point which divides the line segment joining the points $P (- 1, 3)$ and $Q (2, 5)$ in the ratio $3 : 2$ .

$\begin{aligned} ∴ (x, y) = \frac{3 (2) + 2 (- 1)}{3 + 2}, \frac{3 (5) + 2 (3)}{3 + 2} \\ ∵ by internal section formula, \frac{m_{2} x_{1} + m_{1} x_{2}}{m_{1} + m_{2}}, \frac{m_{2} y_{1} + m_{1} y_{2}}{m_{1} + m_{2}} \\ = \frac{6 - 2}{5}, \frac{15 + 6}{5} = \frac{4}{5}, \frac{21}{5} \end{aligned}$

Hence, the required coordinates of the point $R$ is $\frac{4}{5}, \frac{21}{5}$ .

19 Find the values of

k

, if the points

A (k + 1, 2 k), B (3 k, 2 k + 3)

and

C (5 k - 1, 5 k)

are collinear.

Show Answer

Solution

We know that, if three points are collinear, then the area of triangle formed by these points is zero.

Since, the points $A (k + 1, 2 k), B (3 k, 2 k + 3)$ and $C (5 k - 1, 5 k)$ are collinear.

Then, area of $△ A B C = 0$

$\begin{matrix} \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}] . = 0 \\ Here, & x_{1} = k + 1, x_{2} = 3 k, x_{3} = 5 k - 1 and y_{1} = 2 k_{1}, y_{2} = 2 k + 3, y_{3} = 5 k \\ \Rightarrow & \frac{1}{2} [(k + 1) (2 k + 3 - 5 k) + 3 k (5 k - 2 k) + (5 k - 1) (2 k - (2 k + 3))] = 0 \\ \Rightarrow & \frac{1}{2} [(k + 1) (- 3 k + 3) + 3 k (3 k) + (5 k - 1) (2 k - 2 k - 3)] = 0 \\ \Rightarrow & \frac{1}{2} [- 3 k^{2} + 3 k - 3 k + 3 + 9 k^{2} - 15 k + 3] = 0 \end{matrix}$

$\begin{array}{lcr} \Rightarrow 6 k^{2} - 15 k + 6 = 0 & [by factorisation method] \\ \Rightarrow 2 k^{2} - 5 k + 2 = 0 & [divide by 3 ] \\ \Rightarrow 2 k^{2} - 4 k - k + 2 = 0 \\ \Rightarrow 2 k (k - 2) - 1 (k - 2) = 0 \\ If k - 2 = 0, then k = 2 & (k - 2) (2 k - 1) = 0 \\ If 2 k - 1 = 0, then k = \frac{1}{2} \\ ∴ k = 2, \frac{1}{2} \\ Hence, the required values of k are 2 and \frac{1}{2} . \end{array}$

20 Find the ratio in which the line

2 x + 3 y - 5 = 0

divides the line segment joining the points

(8, - 9)

and

(2, 1)

. Also, find the coordinates of the point of division.

Show Answer

Thinking Process

(i) Firstly, consider the given line divides the line segment $A B$ in the ratio $λ : 1$ .

Then, coordinate of $P$ be $\frac{λ x_{2} + x_{1}}{λ + 1}, \frac{λ y_{2} + y_{1}}{λ + 1}$ .

(ii) Substitute the coordinate in the given equation of line and get the value of $λ$ .

(iii) Further, substitute the value of $λ$ in Point $P$ .

Solution

Let the line $2 x + 3 y - 5 = 0$ divides the line segment joining the points $A (8, - 9)$ and $B (2, 1)$ in the ratio $λ$ : 1 at point $P$ .

$\begin{aligned} ∴ Coordinates of P \equiv \frac{2 λ + 8}{λ + 1}, \frac{λ - 9}{λ + 1} \\ ∵ internal division = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{aligned}$

But $P$ lies on $2 x + 3 y - 5 = 0$ .

$\begin{matrix} ∴ & 2 \frac{2 λ + 8}{λ + 1} + 3 \frac{λ - 9}{λ + 1} - 5 = 0 \\ \Rightarrow & 2 (2 λ + 8) + 3 (λ - 9) - 5 (λ + 1) = 0 \\ \Rightarrow & 4 λ + 16 + 3 λ - 27 - 5 λ - 5 = 0 \\ \Rightarrow & λ = 8 \Rightarrow λ : 1 = 8 : 1 \end{matrix}$

So, the point $P$ divides the line in the ratio $8 : 1$ .

$\begin{aligned} Point of division P & \equiv \frac{2 (8) + 8}{8 + 1}, \frac{8 - 9}{8 + 1} \\ \equiv \frac{16 + 8}{9}, - \frac{1}{9} \\ \equiv \frac{24}{9}, \frac{- 1}{9} \equiv \frac{8}{3}, \frac{- 1}{9} \end{aligned}$

Hence, the required point of division is $\frac{8}{3}, \frac{- 1}{9}$ .

Long Answer Type Questions

1 If $(- 4, 3)$ and $(4, 3)$ are two vertices of an equilateral triangle, then find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Show Answer

Solution

Let the third vertex of an equilateral triangle be $(x, y)$ . Let $A (- 4, 3), B (4, 3)$ and $C (x, y)$ . We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal.

$∴$ $\begin{matrix} A B = B C = C A \\ (i) & A B^{2} = B C^{2} = C A^{2} \end{matrix}$

$\Rightarrow$

Now, taking first two parts

$\begin{matrix} A B^{2} = B C^{2} \\ \Rightarrow & (4 + 4)^{2} + (3 - 3)^{2} = (x - 4)^{2} + (y - 3)^{2} \\ \Rightarrow & 64 + 0 = x^{2} + 16 - 8 x + y^{2} + 9 - 6 y \\ \Rightarrow & x^{2} + y^{2} - 8 x - 6 y = 39 & \dots (ii) \end{matrix}$

Now, taking first and third parts,

$\begin{matrix} A B^{2} = C A^{2} \\ \Rightarrow & (4 + 4)^{2} + (3 - 3)^{2} = (- 4 - x)^{2} + (3 - y)^{2} \\ \Rightarrow & 64 + 0 = 16 + x^{2} + 8 x + 9 + y^{2} - 6 y \\ \Rightarrow & x^{2} + y^{2} + 8 x - 6 y = 39 & \dots (iii) \end{matrix}$

On subtracting Eq. (ii) from Eq. (iii), we get

$\begin{aligned} x^{2} + y^{2} + 8 x - 6 y = 39 \\ \frac{\frac{x^{2} + y^{2} - 8 x - 6 y = 39}{+}}{16 x = 0} \\ x = 0 \end{aligned}$

Now, put the value of $x$ in Eq. (ii), we get

$\begin{aligned} 0 + y^{2} - 0 - 6 y = 39 \\ \Rightarrow y^{2} - 6 y - 39 = 0 \\ ∴ y = \frac{6 \pm \sqrt{(- 6)^{2} - 4 (1) (- 39)}}{2 \times 1} \\ ∵ solution of a x^{2} + b x + c = 0 is x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ \Rightarrow y = \frac{6 \pm \sqrt{36 + 156}}{2} \\ \Rightarrow y = \frac{6 \pm \sqrt{192}}{2} \\ \Rightarrow y = \frac{6 \pm 2 \sqrt{48}}{2} = 3 \pm \sqrt{48} \\ \Rightarrow y = 3 \pm 4 \sqrt{3} \\ \Rightarrow y = 3 + 4 \sqrt{3} or 3 - 4 \sqrt{3} \end{aligned}$

So, the points of third vertex are $(0, 3 + 4 \sqrt{3})$ or $(3 - 4 \sqrt{3})$

But given that, the origin lies in the interior of the $△ A B C$ and the $x$ -coordinate of third vertex is zero. Then, $y$ -coordinate of third vertex should be negative.

Hence, the required coordinate of third vertex, $C \equiv (0, 3 - 4 \sqrt{3}) . [∵ c \equiv (0, 3 + 4 \sqrt{3})]$

2. $A (6, 1), B (8, 2)$ and $C (9, 4)$ are three vertices of a parallelogram $A B C D$ . If $E$ is the mid-point of $D C$ , then find the area of $△ A D E$ .

Show Answer

Thinking Process

(i) Firstly, consider the fourth vertex of a parallelogram be $D (x, y)$ .

(ii) Using the concept that mid-point of both diagonals are coincide, determine the coordinate of fourth vertex.

(iii) Also, determine the coordinate of E by using mid point formula,

$\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

(iv) Further, determine the required area of triangle by using the formula,

and simplify it to get the result.

$| \frac{1}{2} [x_{1} (y_{1} - y_{2}) + x_{2} (y_{2} - y_{3}) + x_{3} (y_{3} - y_{1})] |$

Solution

Given that, $A (6, 1), B (8, 2)$ and $C (9, 4)$ are three vertices of a parallelogram $A B C D$ .

Let the fourth vertex of parallelogram be $(x, y)$ .

We know that, the diagonals of a parallelogram bisect each other.

$\begin{matrix} ∴ & Mid-point of B D = Mid-point of A C \\ \Rightarrow & \frac{8 + x}{2}, \frac{2 + y}{2} = \frac{6 + 9}{2}, \frac{1 + 4}{2} \end{matrix}$

$∵$ mid-point of a line segment joining the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) = \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

$\Rightarrow \frac{8 + x}{2}, \frac{2 + y}{2} = \frac{15}{2}, \frac{5}{2}$

$\begin{matrix} ∴ & \frac{8 + x}{2} = \frac{15}{2} \\ \Rightarrow & 8 + x = 15 \Rightarrow x = 7 \\ and & \frac{2 + y}{2} = \frac{5}{2} \\ \Rightarrow & 2 + y = 5 \Rightarrow y = 3 \end{matrix}$

So, fourth vertex of a parallelogram is $D (7, 3)$ .

Now, mid-point of side $D C \equiv \frac{7 + 9}{2}, \frac{3 + 4}{2}$

$E \equiv 8, \frac{7}{2}$

$∵$ area of $△ A B C$ with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3}) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) .$

$+ x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})$

$∴$ Area of $△ A D E$ with vertices $A (6, 1), D (7, 3)$ and $E 8, \frac{7}{2}$ ,

$\begin{aligned} Δ = \frac{1}{2} 63 - \frac{7}{2} + 7 \frac{7}{2} - 1 + 8 (1 - 3) \\ = \frac{1}{2} 6 \times \frac{- 1}{2} + 7 \frac{5}{2} + 8 (- 2) \\ = \frac{1}{2} - 3 + \frac{35}{2} - 16 \\ = \frac{1}{2} \frac{35}{2} - 19 = \frac{1}{2} \frac{- 3}{2} \end{aligned}$

$= \frac{- 3}{4} [but area cannot be negative]$

Hence, the required area of $△ A D E$ is $\frac{3}{4}$ sq units.

3 The points

A (x_{1}, y_{1}), B (x_{2}, y_{2})

and

C (x_{3}, y_{3})

are the vertices of

△ A B C

(i) The median from $A$ meets $B C$ at $D$ . Find the coordinates of the point $D$ .

(ii) Find the coordinates of the point $P$ on $A D$ such that $A P : P D = 2 : 1$ .

(iii) Find the coordinates of points $Q$ and $R$ on medians $B E$ and $C F$ , respectively such that $B Q : Q E = 2 : 1$ and $C R : R F = 2 : 1$ .

(iv) What are the coordinates of the centroid of the $△ A B C$ ?

Show Answer

Solution

Given that, the points $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ are the vertices of $△ A B C$ .

(i) We know that, the median bisect the line segment into two equal parts i.e., here $D$ is the mid-point of $B C$ .

$∴$ Coordinate of mid-point of $B C = \frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}$

$\Rightarrow D \equiv \frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}$

(ii) Let the coordinates of a point $P$ be $(x, y)$ .

Given that, the point $P (x, y)$ , divide the line joining $A (x_{1}, y_{1})$ and $D \frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}$ in the ratio $2 : 1$ , then the coordinates of $P$

$\equiv \frac{2 \cdot \frac{x_{2} + x_{3}}{2} + 1 \cdot x_{1}}{2 + 1}, \frac{\frac{y_{2} + y_{3}}{2} + 1 \cdot y_{1}}{2 + 1}$

$∵$ internal section formula $= \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}}$

$\equiv \frac{x_{2} + x_{3} + x_{1}}{3}, \frac{y_{2} + y_{3} + y_{1}}{2}$

$∴$ So, required coordinates of point $P \equiv \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}$

(iii) Let the coordinates of a point $Q$ be $(p, q)$

Given that, the point $Q (p, q)$ , divide the line joining $B (x_{2}, y_{2})$ and $E \frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2}$ in the ratio $2 : 1$ , then the coordinates of $Q$

$\begin{aligned} \equiv \frac{2 \cdot \frac{x_{1} + x_{3}}{2} + 1 \cdot x_{2}}{2 + 1}, \frac{\frac{y_{1} + y_{2}}{2} + 1 \cdot y_{2}}{2 + 1} \\ = \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3} \end{aligned}$

since, $B E$ is the median of side $C A$ , so $B E$ divides $A C$ in to two equal parts.

$∴ mid-point of A C = Coordinate of E \Rightarrow E = \frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2}$

So, the required coordinate of point $Q \equiv \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}$

Now, let the coordinates of a point $E$ be $(α, β)$ . Given that, the point $R (α, β)$ , divide the line joining $C (x_{3}, y_{3})$ and $F \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$ in the ratio $2 : 1$ , then the coordinates of $R$

$\begin{aligned} \equiv \frac{2 \cdot \frac{x_{1} + x_{2}}{2} + 1 \cdot x_{3} 2 \cdot \frac{y_{1} + y_{2}}{2} + 1 \cdot y_{3}}{2 + 1}, \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3} \end{aligned}$

since, $C F$ is the median of side $A B$ . So, $C F$ divides $A B$ in to two equal parts.

$∴ mid-point of A B = coordinate of F \Rightarrow F = \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

So, the required coordinate of point $R \equiv \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}$

(iv) Coordinate of the centroid of the $△ A B C$

$\begin{aligned} = \frac{Sum of abscissa of all vertices}{3}, \frac{Sum of ordinate of all vertices}{3} \\ = \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3} \end{aligned}$

4 If the points

A (1, - 2), B (2, 3), C (a, 2)

and

D (- 4, - 3)

form a parallelogram, then find the value of

a

and height of the parallelogram taking

A B

as base.

Show Answer

Solution

In parallelogram, we know that, diagonals are bisects each other i.e., mid-point of $A C =$ mid-point of $B D$

$(1, - 2)$

$\begin{matrix} \Rightarrow & \frac{1 + a}{2}, \frac{- 2 + 2}{2} = \frac{2 - 4}{2}, \frac{3 - 3}{2} \\ \Rightarrow & \frac{1 + a}{2} = \frac{2 - 4}{2} = \frac{- 2}{2} = - 1 \end{matrix}$

since, mid-point of a line segment having points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$

$\begin{matrix} \Rightarrow & 1 + a = - 2 \\ \Rightarrow & a = - 3 \end{matrix}$

So, the required value of $a$ is -3 .

Given that, $A B$ as base of a parallelogram and drawn a perpendicular from $D$ to $A B$ which meet $A B$ at $P$ . So, $D P$ is a height of a parallelogram.

$\begin{matrix} \Rightarrow & (y - y_{1}) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ \Rightarrow & (y + 2) = \frac{3 + 2}{2 - 1} (x - 1) \\ \Rightarrow & (y + 2) = 5 (x - 1) \\ \Rightarrow & 5 x - y = 7 & \dots (i) \\ Slope of A B, say m_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3 + 2}{2 - 1} = 5 \end{matrix}$

Let the slope of $D P$ be $m_{2}$ .

Since, $D P$ is perpendicular to $A B$ .

By condition of perpendicularity,

$\begin{array}{r} m_{1} \cdot m_{2} = - 1 = \\ m_{2} = - \frac{1}{5} \end{array}$

Now, Eq. of $D P$ , having slope $- \frac{1}{5}$ and passing the point $(- 4, - 3)$ is

$\begin{matrix} (y - y_{1}) = m_{2} (x - x_{1}) \\ \Rightarrow & (y + 3) = - \frac{1}{5} (x + 4) \\ \Rightarrow & 5 y + 15 = - x - 4 & \dots (ii) \\ \Rightarrow & x + 5 y = - 19 \end{matrix}$

On adding Eqs. (i) and (ii), then we get the intersection point $P$ .

Put the value of $y$ from Eq. (i) in Eq. (ii), we get

$x + 5 (5 x - 7) = - 19$

$\begin{matrix} \Rightarrow & x + 25 x - 35 = - 19 \\ \Rightarrow & 26 x = 16 \\ ∴ & x = \frac{8}{13} \end{matrix}$

Put the value of $x$ in Eq. (i), we get

$\Rightarrow y = \frac{40 - 91}{13} \Rightarrow y = \frac{- 51}{13}$

$∴$ Coordinates of point $P \equiv \frac{8}{13}, \frac{- 51}{13}$

So, length of the height of a parallelogram,

$D P = \sqrt{\frac{8}{13} + 4^{2} + \frac{- 51}{13} + 3^{2}}$

$[∵ .$ by distance formula, distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , is

$. d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}]$

$\Rightarrow$ $\begin{aligned} D P = \sqrt{\frac{60^{2}}{13} + \frac{- 12^{2}}{13}} \\ = \frac{1}{13} \sqrt{3600 + 144} \\ = \frac{1}{13} \sqrt{3744} = \frac{12 \sqrt{26}}{13} \end{aligned}$

Hence, the required length of height of a parallelogram is $\frac{12 \sqrt{26}}{13}$ .

5 Students of a school are standing in rows and columns in their playground for a drill practice.

A, B, C

and

D

are the positions of four students as shown in figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students

A, B, C

and

D

? If so, what should be his position?

Show Answer

Solution

Yes, from the figure we observe that the positions of four students $A, B, C$ and $D$ are $(3, 5), (7, 9), (11, 5)$ and $(7, 1)$ respectively i.e., these are four vertices of a quadrilateral. Now, we will find the type of this quadrilateral. For this, we will find all its sides.

Now,

$A B = \sqrt{(7 - 3)^{2} + (9 - 5)^{2}}$

by distance formula, $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$\begin{aligned} A B = \sqrt{(4)^{2} + (4)^{2}} = \sqrt{16 + 16} \\ A B = 4 \sqrt{2} \\ B C = \sqrt{(11 - 7)^{2} + (5 - 9)^{2}} = \sqrt{(4)^{2} + (- 4)^{2}} \\ = \sqrt{16 + 16} = 4 \sqrt{2} \\ C D = \sqrt{(7 - 11)^{2} + (1 - 5)^{2}} = \sqrt{(- 4)^{2} + (- 4)^{2}} \\ = \sqrt{16 + 16} = 4 \sqrt{2} \\ D A = \sqrt{(3 - 7)^{2} + (5 - 1)^{2}} = \sqrt{(- 4)^{2} + (4)^{2}} \\ = \sqrt{16 + 16} = 4 \sqrt{2} \end{aligned}$

and

We see that, $A B = B C = C D = D A$ i.e., all sides are equal.

Now, we find length of both diagonals.

and

$\begin{aligned} A C = \sqrt{(11 - 3)^{2} + (5 - 5)^{2}} = \sqrt{(8)^{2} + 0} = 8 \\ B D = \sqrt{(7 - 7)^{2} + (1 - 9)^{2}} = \sqrt{0 + (- 8)^{2}} = 8 \\ A C = B D \end{aligned}$

Here,

Since,

Which represent a square. Also known the diagonals of a square bisect each other. So, $P$ be position of Jaspal in which he is equidistant from each of the four students $A, B, C$ and $D$ .

$∴$ Coordinates of point $P \equiv$ Mid-point of $A C$

$\equiv \frac{3 + 11}{2}, \frac{5 + 5}{2} \equiv \frac{14}{2}, \frac{10}{2} \equiv (7, 5)$

since, mid-point of a line segment having points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) = \frac{x_{1} + y_{1}}{2}, \frac{x_{2} + y_{2}}{2}$

Hence, the required position of Jaspal is $(7, 5)$ .

6. Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distance covered are in straight lines). If the house is situated at

(2, 4)

, bank at

(5, 8)

, school at

(13, 14)

and office at

(13, 26)

and coordinates are in

k m

Show Answer

Solution

By given condition, we drawn a figure in which every place are indicated with his coordinates and direction also.

We know that,

distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ,

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Now, distance between house and bank $= \sqrt{(5 - 2)^{2} + (8 - 4)^{2}}$

$= \sqrt{(3)^{2} + (4)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

Distance between bank and daughter’s school

$\begin{aligned} = \sqrt{(13 - 5)^{2} + (14 - 8)^{2}} = \sqrt{(8)^{2} + (6)^{2}} \\ = \sqrt{64 + 36} = \sqrt{100} = 10 \end{aligned}$

Distance between daughter’s school and office $= \sqrt{(13 - 13)^{2} + (26 - 14)^{2}}$

$= \sqrt{0 + (12)^{2}} = 12$

Total distance (House + Bank + School + Office) travelled $= 5 + 10 + 12 = 27$ units

Distance between house to offices $= \sqrt{(13 - 2)^{2} + (26 - 4)^{2}}$

$\begin{aligned} = \sqrt{(11)^{2} + (22)^{2}} = \sqrt{121 + 484} \\ = \sqrt{605} = 24.59 \approx 24.6 k m \end{aligned}$

So, extra distance travelled by Ayush in reaching his office $= 27 - 24.6 = 2.4 k m$ Hence, the required extra distance travelled by Ayush is $2.4 k m$ .

Mathematics 10

Chapter 07 Coordinate Geometry

Multiple Choice Questions (MCQs)

Very Short Answer Type Questions

Short Answer Type Questions

Long Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

Mathematics 10

Chapter 07 Coordinate Geometry

Multiple Choice Questions (MCQs)

Very Short Answer Type Questions

Short Answer Type Questions

Long Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

Menu