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 $P(2,3)=3$.

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) $\because$ 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 \text{ and } x_2=0, y_2=-2 $

$\therefore \quad$ 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} $

Thinking Process

Coordinate of origin is $(0,0)$

Solution

(c) $\because$ 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, $\quad x_1=-6, y_1=8$ and $x_2=0, y_2=0$

$\therefore$ 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} $

Solution

(b) $\because$ 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}} \\ & \text{ Here, } x_1=0, y_1=5 \text{ and } x_2=-5, y_2=0 \\ & \therefore \text{ Distance between the points }(0,5) \text{ and }(-5,0) \\ & \begin{matrix} =\sqrt{(-5-0)^{2}+(0-5)^{2}} \\ =\sqrt{25+25}=\sqrt{50}=5 \sqrt{2} \end{matrix} \end{aligned} $

Solution

(c)

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

$\because \quad$ 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 $x_2=5, y_2=0$

$\therefore \quad$ 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 $\sqrt{34}$.

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 $\triangle A O B=$ Sum of the length of all its sides $=d(A O)+d(O B)+d(A B)$

$\because$ 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 .

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 $\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$. Use this formula and simplify it to get the result.

Solution

(c) Area of $\triangle A B C$ whose Vertices $A \equiv(x_1, y_1), B \equiv(x_2, y_2)$ and $C \equiv(x_3, y_3)$ are given by

$ \Delta=|\dfrac{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 $y_3=4$

$\therefore \quad \Delta=|\dfrac{1}{2}[3(0-4)+7(4-0)+8(0-0)]|=|\dfrac{1}{2}(-12+28+0)|=|\dfrac{1}{2}(16)|=8$

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

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}} $

$\because$ 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 $

$ \because \quad B C=A C $

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

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 \text{, then } x=\dfrac{m x_2+n x_1}{m+n} \text{ and } y=\dfrac{m y_2+n y_1}{m+n} $

Given that, $\quad x_1=7, y_1=-6, x_2=3, y_2=4, \quad m=1$ and $n=2$

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

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

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

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

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

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.

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

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

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

$ (x_1, y_1) \text{ and }(x_2, y_2)=\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} $

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

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

$ \dfrac{x_1+x_2}{2}, \dfrac{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, $L \equiv \dfrac{-2+8}{2}, \dfrac{3+3}{2} \equiv(3,3)$

since, mid - point of a line segment having points $(x_1, y_1)$ and $(x_2, y_2)$

$ =\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} $

and

$M \equiv \dfrac{6+x_4}{2}, \dfrac{7+y_4}{2}$

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} & \therefore \quad 3=\dfrac{6+x_4}{2} \text{ and } 3=\dfrac{7+y_4}{2} \\ & \Rightarrow \quad 6=6+x_4 \text{ and } 6=7+y_4 \\ & \Rightarrow \quad x_4=0 \quad \text{ and } y_4=6-7 \\ & \therefore \quad x_4=0 \quad \text{ and } y_4=-1 \end{aligned} $

Hence, the fourth vertex of parallelogram is $D \equiv(x_4, y_4) \equiv(0,-1)$.

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

$\because$ 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}} \\ & \quad=\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} \\ \therefore \quad A B = 2 \sqrt{5}=2 A P \Rightarrow A P=\dfrac{A B}{2} \end{aligned} $

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

Solution

(b) Given that, $P \dfrac{a}{3}, 4$ 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

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

since, mid-point of line segment having points $(x_1, y_1)$ and $(x_2, y_2)$

$ =\dfrac{(x_1+x_2)}{2}, \dfrac{(y_1+y_2)}{2} $

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

$ \therefore \quad \dfrac{a}{3}, 4=(-4,4) $

On comparing the coordinates, we get

$ \begin{matrix} & \dfrac{a}{3}=-4 \\ \therefore \quad & a=-12 \end{matrix} $

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

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

$\because$ mid-point of line segment passes through the points $(x_1, y_1)$ and $(x_2, y_2)$

$ =\dfrac{x_1+x_2}{2}, \dfrac{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

$$ \begin{equation*} (y-y_1)=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1) \tag{i} \end{equation*} $$

Here,

$ x_1=1, y_1=5 \text{ 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) =\dfrac{6-5}{4-1}(x-1) \\ \Rightarrow & (y-5) =\dfrac{1}{3}(x-1) \\ \Rightarrow & 3 y-15 =x-1 \\ \Rightarrow & 3 y=x-14 \Rightarrow & y =\dfrac{1}{3} x-\dfrac{14}{3} & \ldots \text{(ii)} \end{matrix} $

$\therefore$ Slope of the line segment, $m_1=\dfrac{1}{3}$

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

$$ \begin{equation*} m_1 \cdot m_2=-1 \tag{iii} \end{equation*} $$

where, $\quad m_1=$ Slope of line 1

and $\quad=$ 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),

$\Rightarrow \quad m_2=-3$

$ \begin{aligned} m_1 \cdot m_2 = \dfrac{1}{3} \cdot m_2=-1 \\ m_2 = -3 \end{aligned} $

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

$\therefore \quad$ Mid-point of line segment $=\dfrac{1+4}{2}, \dfrac{5+6}{2}=\dfrac{5}{2}, \dfrac{11}{2}$

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

$ y-\dfrac{11}{2}=(-3) \quad x-\dfrac{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 & \ldots \text{(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 $(0,13)$.

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, $\quad$ $ P O=P A=P B $

$\Rightarrow$

$$ \begin{equation*} (P O)^{2}=(P A)^{2}=(P B)^{2} \tag{i} \end{equation*} $$

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} & \ldots \text{(ii)} \end{matrix} $

Taking first two equations, we get

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

$\therefore \quad$ Required points $=(h, k)=(x, y)$

Solution

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

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

$ =\sqrt{\dfrac{13}{2}-0^{2}+(0-0)^{2}}=\sqrt{\dfrac{13^{2}}{2}}=\dfrac{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 $\dfrac{-3}{4}, 1=\sqrt{\dfrac{-3}{4}-0^{2}+(1-0)^{2}}$

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

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

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

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

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

(c) Distance between $(0,0)$ and 5, $\dfrac{-1}{2}=\sqrt{(5-0)^{2}+-\dfrac{1}{2}-0^{2}}$

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

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

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

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

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

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 $M \dfrac{0+x}{2}, \dfrac{y+0}{2}$.

$\because$ mid-point of a line segment having points $(x_1, y_1)$ and $(x_2, y_2)=\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

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

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

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) \text{ and } C \equiv(x_3, y_3) \equiv(c, a+b) \\ \because \text{ Area of } \triangle A B C = \Delta=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] \\ \therefore \quad \Delta = \dfrac{1}{2}[a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)] \\ =\dfrac{1}{2}[a(c-b)+b(a-c)+c(b-a)] \\ =\dfrac{1}{2}(a c-a b+a b-b c+b c-a c)=\dfrac{1}{2}(0)=0 \end{array} $

Hence, the required area of triangle is 0 .

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 $

$\because$ 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{aligned} 9+p^{2} = 25 \\ p^{2} = 16 \quad \Rightarrow \quad p= \pm 4 \end{aligned} $

$\Rightarrow$ $ \pm 4 $

Solution

(c) Let the given points are $A \equiv(x_1, y_1) \equiv(1,2)$,

$B \equiv(x_2, y_2) \equiv(0,0)$ and $C \equiv(x_3, y_3) \equiv(a, b)$.

$\because$ Area of $\triangle A B C \Delta=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\therefore \quad \Delta=\dfrac{1}{2}[1(0-b)+0(b-2)+a(2-0)]$

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

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

i.e.,

area of $\triangle A B C=0$

$\Rightarrow$

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

$\Rightarrow \quad 2 a-b=0$

$\Rightarrow \quad 2 a=b$

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

Solution

True

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

$[\because.$ 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 $\triangle A B C$, 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, $\quad \dfrac{A B}{O E}=\dfrac{4}{8}=\dfrac{1}{2}, \dfrac{A C}{D F}=\dfrac{2 \sqrt{2}}{4 \sqrt{2}}=\dfrac{1}{2}, \dfrac{B C}{E F}=\dfrac{2 \sqrt{2}}{4 \sqrt{2}}=\dfrac{1}{2}$

$\therefore \quad \dfrac{A B}{D E}=\dfrac{A C}{D F}=\dfrac{B C}{E F}$

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

Hence, both the triangles are similar.

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 $B(-4,-6)$.

Solution

False

Here, $x_1=0, x_2=0, x_3=3$ and $y_1=5, y_2=-9, y_3=6$

$\because \quad$ Area of triangle $\Delta=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\therefore \quad \Delta=\dfrac{1}{2}[0(-9-6)+0(6-5)+3(5+9)]$

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

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.

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} & \therefore \quad 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 \\ & \because \quad 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 $B(3,3), m_1=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}$

$ \because m=\dfrac{y_2-y_1}{x_2-x_1} $

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

$ m_2=-\dfrac{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 $B(3,3)$.

$ \therefore \quad \text{ Mid-point }=\dfrac{-1+3}{2}, \dfrac{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

$ \dfrac{x_1+x_2}{2}, \dfrac{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 & \ldots \text{(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 $

$\Rightarrow \quad y=4$

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

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) $

$\therefore \quad$ Area of $\triangle A B C \Delta=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

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

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

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

$\because \quad$ Area of $\triangle A B C=0$

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.

Solution

False

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

$[\because$ 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.

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 $P(5,0), O P=\sqrt{(5-0)^{2}+(0-0)^{2}}$

$\because$ 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=\text{ 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.

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} & \therefore \\ & 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.

$\begin{aligned} \therefore \text{ Slope of perpendicular bisector }= & \dfrac{-1}{\text{ Slope of line segment joining }} \\ & \text{ the points }(5,-3) \text{ and }(0,-4) \\ = \dfrac{-1}{\dfrac{-4-(-3)}{0-5}}=5 \quad \because \text{ slope }=\dfrac{y_2-y_1}{x_2-x_1}\end{aligned}$

[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 $(0,-4)$.

$\therefore \quad$ Mid-point of $P Q=\dfrac{5+0}{2}, \dfrac{-3-4}{2}=\dfrac{5}{2}, \dfrac{-7}{2}$

So, the equation of perpendicular bisector having slope $\dfrac{1}{3}$ and passes through the mid-point $\dfrac{5}{2}, \dfrac{-7}{2}$ 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 & \ldots \text{(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.

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} & \dfrac{k(1)+(1)(7)}{k+1}, \dfrac{k(-5)+1(-2)}{k+1} \\ \text{ i.e., } & \dfrac{k+7}{k+1}, \dfrac{-5 k-2}{k+1} \\ \text{ Now, } & (5,-3)=\dfrac{k+7}{k+1}, \dfrac{-5 k-2}{k+1} \\ \Rightarrow & \dfrac{k+7}{k+1}=5 \\ \Rightarrow & k+7=5 k+5 \\ \Rightarrow & -4 k=-2 \\ \therefore & k=\dfrac{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 $A B$.

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.

$\because \quad$ Area of triangle $=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here, $x_1=-6, x_2=-4, x_3=3$ and $y_1=10, y_2=6, y_3=-8$

$\therefore \quad$ Area of $\triangle A B C=\dfrac{1}{2}[-6{6-(-8)}+(-4)(-8-10)+3(10-6)]$

$ \begin{aligned} & =\dfrac{1}{2}[-6(14)+(-4)(-18)+3(4)] \\ & =\dfrac{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} $

$\because$ 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} \\ \therefore \quad A B = \dfrac{2}{9} A C \end{aligned} $

which is the required relation.

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} $

$\because$ 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.

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} $

$\because$ 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 \text{ and } A D=B C $

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

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

$ =\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.

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.

(c) If all three sides of a triangle are equal, then it is an equilateral 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{aligned} A B = \sqrt{(-4+5)^{2}+(-2-6)^{2}} \\ =\sqrt{(1)^{2}+(-8)^{2}} \\ =\sqrt{1+64}=\sqrt{65} \quad[\because 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{aligned} $

We see that,

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

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

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

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 $Q(7,-4)$.

$\text{ By given condition, } \quad P Q=2 \sqrt{5} \quad[\because \text{ distance formula }=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}]$

$ \begin{aligned} \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 \quad \text{ [by factorisation method] } \\ \Rightarrow x(x-9)-5(x-9)=0 \\ \Rightarrow (x-9)(x-5)=0 \\ \therefore x=5,9 \end{aligned} $

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

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.

(c) If opposite sides of a quadrilateral are equal and diagonals are also equal then quadrilateral is a rectangle.

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,

$ \text{ 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{array}{rl} 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} \\ \text{ 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{array} $

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.

Solution

According to the question,

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

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

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

Solution

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

$ \begin{aligned} & \therefore \quad P A=P B \quad[\because \text{ by distance formula, distance }=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}] \\ & \Rightarrow \quad(P A)^{2}=(P B)^{2} \\ & \Rightarrow \quad(-5-h)^{2}+(4-k)^{2}=(-1-h)^{2}+(6-k)^{2} \\ & \Rightarrow \quad 25+h^{2}+10 h+16+k^{2}-8 k=1+h^{2}+2 h+36+k^{2}-12 k \\ & \Rightarrow \quad 25+10 h+16-8 k=1+2 h+36-12 k \\ & \Rightarrow \quad 8 h+4 k+41-37=0 \\ & \Rightarrow \quad 8 h+4 k+4=0 \\ & \Rightarrow \quad 2 h+k+1=0 \\ & \text{ Mid-point of } A B=\dfrac{-5-1}{2}, \dfrac{4+6}{2}=(-3,5) \\ & \because \text{ mid-point }=\dfrac{x_1+x_2}{2}, \dfrac{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 \quad 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 $2 x+y+1=0$

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} \therefore & & \text{ Mid-point of } A B = \dfrac{-5+4}{2}, \dfrac{-2-2}{2} \\ \Rightarrow & R = -\dfrac{1}{2},-2 \end{matrix} $

$[\because.$ mid-point of a line segment passes through the points $(x_1, y_1)$ and $(x_2, y_2)$ is

$ \dfrac{x_1+x_2}{2}, \dfrac{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 $Q-\dfrac{1}{2}, 0$.

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

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{aligned} m_1=\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{-2-(-2)}{4-(-5)}=\dfrac{-2+2}{4+5}=\dfrac{0}{9} \\ m_1 = 0 \end{aligned} $

$ \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{aligned} m_1 \cdot m_2 = -1 \\ m_2 = \dfrac{-1}{m_1}=\dfrac{-1}{0} \\ m_2 = \infty \end{aligned} $

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

$ \begin{aligned} \therefore \quad \text{ Mid-point }=\dfrac{-5+4}{2}, \dfrac{-2-2}{2}= & \dfrac{-1}{2},-2 \\ & \because \text{ mid-point }=\dfrac{x_1+x_2}{2}, \dfrac{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 $\dfrac{-1}{2},-2$ is

$ \begin{aligned} (y+2) = \infty \quad x+\dfrac{1}{2} \\ \Rightarrow \quad \dfrac{y+2}{x+\dfrac{1}{2}} = \infty=\dfrac{1}{0} \Rightarrow x+\dfrac{1}{2}=0 \\ \therefore \quad x = \dfrac{-1}{2} \end{aligned} $

So, the coordinates of the point $Q$ is $\dfrac{-1}{2}, 0$ 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 $

$[\because.$ 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{\dfrac{-1}{2}-4^{2}+(0+2)^{2}} \\ & =\sqrt{\dfrac{-9^{2}}{2}+(2)^{2}}=\sqrt{\dfrac{81}{4}+} \\ Q A = \sqrt{-5+\dfrac{1}{2}+(-2-0)^{2}} \\ & =\sqrt{\dfrac{-9^{2}}{2}+(2)^{2}} \\ & =\sqrt{\dfrac{81}{4}+4}=\sqrt{\dfrac{97}{4}}=\sqrt{\dfrac{97}{2}} \end{aligned} $

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

Thinking Process

(i) First, using the condition of collinearity

$ \dfrac{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.

$\therefore$ Area of $\triangle A B C=0$

$\Rightarrow$

$\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$

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

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

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

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

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

$[\because.$ 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 $y=-3$

$\therefore \quad y=-3,-5$

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 $(-10, y), P Q=\sqrt{(-13)^{2}+(-5-8)^{2}} \quad$ [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.

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{aligned} \Delta = \dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3+(y_1-y_2)] \\ \therefore \quad = \dfrac{1}{2}[-8(6-9)-6(9-4)+(-3)(4-6)] \\ =\dfrac{1}{2}[-8(-3)-6(5)-3(-2)]=\dfrac{1}{2}(24-30+6) \\ =\dfrac{1}{2}(30-30)=\dfrac{1}{2}(0)=0 \end{aligned} $

Hence, the required area of triangle is 0 .

Thinking Process

(i) Firstly, determine the ratio by using the formula $\dfrac{\lambda x_2+x_1}{\lambda+1}, \dfrac{\lambda y_2+y_1}{\lambda+1}$.

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

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

Solution

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

$ \dfrac{\lambda x_2+1 \cdot x_1}{\lambda+1}, \dfrac{\lambda y_2+1 \cdot y_1}{\lambda+1} $

Here, $x_1=-4, x_2=-1$ and $y_1=-6, y_2=7$

i.e., $\quad \dfrac{\lambda(-1)+1(-4)}{\lambda+1}, \dfrac{\lambda(7)+1 \cdot(-6)}{\lambda+1}=\dfrac{-\lambda-4}{\lambda+1}, \dfrac{7 \lambda-6}{\lambda+1}$

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} & \therefore \quad \dfrac{7 \lambda-6}{\lambda+1} \Rightarrow 7 \lambda-6=0 \\ & \therefore \quad \lambda=\dfrac{6}{7} \end{aligned} $

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

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

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

Solution

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

Using the section formula, we get

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

$\because$ 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 $\dfrac{m_2 x_1+m_1 x_2}{m_1+m_2}, \dfrac{m_2 y_1+m_1 y_2}{m_1+m_2}$

On equating, we get

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

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

Solution

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

By section formula,

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

$\because$ 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 $\dfrac{m_2 x_1+m_1 x_2}{m_1+m_2}, \dfrac{m_2 y_1+m_1 y_2}{m_1+m_2}$

and

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

and

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

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

and

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

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

and

$ b=-3 $

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

$\Rightarrow \quad 9 a-9=0$

$\therefore \quad a=1$

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

Solution

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

since, mid-point of a line segment having points $(x_1, y_1)$ and $(x_2, y_2)=\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

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

Now, equating coordinates on both sides, we get

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

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

$\Rightarrow \quad k=22$

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

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

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

$[\because.$ 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 $2 \sqrt{61}$.

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

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

$[\because.$ 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, $\quad$ length of diameter $=10 \sqrt{2}$

$ \begin{aligned} \therefore \quad \text{ Length of radius } = \dfrac{\text{ Length of diameter }}{2} \\ & =\dfrac{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 \\ \therefore & a =3,5 \end{matrix} $

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

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

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} \therefore \quad \text{ Coordinate of point } P & \equiv \dfrac{5(1)+3(2)}{1+2}, \dfrac{1(1)+2(2)}{1+2} \\ & \equiv \dfrac{5+6}{3}, \dfrac{1+4}{3} \equiv \dfrac{11}{3}, \dfrac{5}{3} \\ & \because \text{ by section formula for internal ratio } \equiv \dfrac{m_1 x_2+m_2 x_1}{m_1+m_2}, \dfrac{m_1 y_2+m_2 y_1}{m_1+m_2} \end{aligned} $

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

[given]

$ \begin{aligned} \therefore & 3 \dfrac{11}{3}-18 \dfrac{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 .

Thinking Process

(i) Firstly, consider the vertices of of $\triangle A B C$ 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,

$ |\dfrac{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 $\triangle A B C$.

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

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

$ \therefore \quad \dfrac{x_2+x_3}{2}=-\dfrac{1}{2} $

since, mid-point of a line segment having points $(x_1, y_1)$ and $(x_2, y_2)$ is $\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

$ \begin{matrix} \text{ and } & \dfrac{y_2+y_3}{2}=\dfrac{5}{2} \\ \Rightarrow & x_2+x_3=-1 \\ \text{ and } & y_2+y_3=5 & \ldots \text{(ii)}\\ \text{ As } E(7,3) \text{ is the mid-point of } C A . & \end{matrix} $

$\therefore$ $\dfrac{x_3+x_1}{2}=7$

and $\dfrac{y_3+y_1}{2}=3$

$\Rightarrow$ $x_3+x_1=14$

and $y_3+y_1=6$

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

$ \begin{matrix} \therefore & \dfrac{x_1+x_2}{2}=\dfrac{7}{2} \\ \text{ and } & \dfrac{y_1+y_2}{2}=\dfrac{7}{2} \\ \Rightarrow & x_1+x_2=7 \\ \text{ and } & y_1+y_2=7 & \ldots \text{(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 & \ldots \text{(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 $\triangle A B C$ are $A(11,4), B(-4,3)$ and $C(3,2)$.

$ \begin{aligned} \because \quad \text{ Area of } \triangle A B C = \Delta=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] \\ \therefore \quad \Delta = \dfrac{1}{2}[11(3-2)+(-4)(2-4)+3(4-3)] \\ & =\dfrac{1}{2}[11 \times 1+(-4)(-2)+3(1)] \\ & =\dfrac{1}{2}(11+8+3)=\dfrac{22}{2}=11 \end{aligned} $

$\therefore$ Required area of $\triangle A B C=11$

Solution

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

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

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

$\because$ 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 \\ \therefore & a=2,5 \end{matrix} $

$ \Rightarrow \quad a^{2}-2 a-5 a+10=0 \quad \text{ [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} \because \quad \text{ Area of } \triangle A B C = \dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] \\ \therefore \quad \Delta = \dfrac{1}{2}[2(5-5)+2(5-9)+5(9-5)] \\ & =\dfrac{1}{2}[2 \times 0+2(-4)+5(4)] \\ & =\dfrac{1}{2}(0-8+20)=\dfrac{1}{2} \times 12=6 \end{aligned} $

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

Solution

According to the question,

$ \begin{matrix} \text{ Given that, } & P R = \dfrac{3}{5} P Q \\ \Rightarrow & \dfrac{P Q}{P R} = \dfrac{5}{3} \\ \Rightarrow & \dfrac{P R+R Q}{P R} = \dfrac{5}{3} \\ \Rightarrow & 1+\dfrac{R Q}{P R} = \dfrac{5}{3} \\ \Rightarrow & \dfrac{P Q}{P R} = \dfrac{5}{3}-1=\dfrac{2}{3} \\ \therefore & R Q: P R = 2: 3 \\ \text{ 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} \therefore \quad(x, y) = \dfrac{3(2)+2(-1)}{3+2}, \dfrac{3(5)+2(3)}{3+2} \\ & \because \text{ by internal section formula, } \dfrac{m_2 x_1+m_1 x_2}{m_1+m_2}, \dfrac{m_2 y_1+m_1 y_2}{m_1+m_2} \\ & =\dfrac{6-2}{5}, \dfrac{15+6}{5}=\dfrac{4}{5}, \dfrac{21}{5} \end{aligned} $

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

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 $\triangle A B C=0$

$ \begin{matrix} & \dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2]. =0 \\ \text{ Here, } & x_1=k+1, x_2=3 k, x_3=5 k-1 \text{ and } y_1=2 k_1, y_2=2 k+3, y_3 =5 k \\ \Rightarrow & \dfrac{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 & \dfrac{1}{2}[(k+1)(-3 k+3)+3 k(3 k)+(5 k-1)(2 k-2 k-3)] =0 \\ \Rightarrow & \dfrac{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 & \text{ [by factorisation method] } \\ \Rightarrow 2 k^{2}-5 k+2=0 & \text{ [divide by 3 ] } \\ \Rightarrow 2 k^{2}-4 k-k+2=0 \\ \Rightarrow 2 k(k-2)-1(k-2)=0 \\ \text{ If } k-2=0 \text{, then } k=2 & (k-2)(2 k-1)=0 \\ \text{ If } 2 k-1=0 \text{, then } k=\dfrac{1}{2} & \\ \therefore k=2, \dfrac{1}{2} & \\ \text{ Hence, the required values of } k \text{ are } 2 \text{ and } \dfrac{1}{2} \text{. } \end{array} $

Thinking Process

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

Then, coordinate of $P$ be $\dfrac{\lambda x_2+x_1}{\lambda+1}, \dfrac{\lambda y_2+y_1}{\lambda+1}$.

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

(iii) Further, substitute the value of $\lambda$ 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 $\lambda$ : 1 at point $P$.

$ \begin{aligned} & \therefore \quad \text{ Coordinates of } P \equiv \dfrac{2 \lambda+8}{\lambda+1}, \dfrac{\lambda-9}{\lambda+1} \\ & \because \text{ internal division }=\dfrac{m_1 x_2+m_2 x_1}{m_1+m_2}, \dfrac{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} \therefore & 2 \dfrac{2 \lambda+8}{\lambda+1}+3 \dfrac{\lambda-9}{\lambda+1}-5=0 \\ \Rightarrow & 2(2 \lambda+8)+3(\lambda-9)-5(\lambda+1)=0 \\ \Rightarrow & 4 \lambda+16+3 \lambda-27-5 \lambda-5=0 \\ \Rightarrow & \lambda=8 \Rightarrow \lambda: 1=8: 1 \end{matrix} $

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

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

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

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.

$\therefore$ $$ \begin{gather*} A B=B C=C A \\ A B^{2}=B C^{2}=C A^{2} \tag{i} \end{gather*} $$

$ \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 & \ldots \text{(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 & \ldots \text{(iii)} \end{matrix} $

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

$ \begin{aligned} & x^{2}+y^{2}+8 x-6 y=39 \\ & \dfrac{\dfrac{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 \quad y^{2}-6 y-39=0 \\ & \therefore \quad y=\dfrac{6 \pm \sqrt{(-6)^{2}-4(1)(-39)}}{2 \times 1} \\ & \because \text{solution of } a x^{2}+b x+c=0 \text{ is } x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ & \Rightarrow \quad y=\dfrac{6 \pm \sqrt{36+156}}{2} \\ & \Rightarrow \quad y=\dfrac{6 \pm \sqrt{192}}{2} \\ & \Rightarrow \quad y=\dfrac{6 \pm 2 \sqrt{48}}{2}=3 \pm \sqrt{48} \\ & \Rightarrow \quad y=3 \pm 4 \sqrt{3} \\ & \Rightarrow \quad y=3+4 \sqrt{3} \text{ 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 $\triangle A B C$ 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, $C \equiv(0,3-4 \sqrt{3}) . \quad[\because c \equiv(0,3+4 \sqrt{3})]$

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,

$ \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} $

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

and simplify it to get the result.

$ |\dfrac{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} \therefore & \text{ Mid-point of } B D=\text{ Mid-point of } A C \\ \Rightarrow & \dfrac{8+x}{2}, \dfrac{2+y}{2}=\dfrac{6+9}{2}, \dfrac{1+4}{2} \end{matrix} $

$\because$ mid-point of a line segment joining the points $(x_1, y_1)$ and $(x_2, y_2)=\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

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

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

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

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

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

$\because$ area of $\triangle A B C$ with vertices $(x_1, y_1),(x_2, y_2)$ and $(x_3, y_3)=\dfrac{1}{2}[x_1(y_2-y_3).$

$ +x_2(y_3-y_1)+x_3(y_1-y_2) $

$\therefore$ Area of $\triangle A D E$ with vertices $A(6,1), D(7,3)$ and $E 8, \dfrac{7}{2}$,

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

$ =\dfrac{-3}{4} \quad \text{ [but area cannot be negative] } $

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

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 $\triangle A B C$.

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

$\therefore$ Coordinate of mid-point of $B C=\dfrac{x_2+x_3}{2}, \dfrac{y_2+y_3}{2}$

$ \Rightarrow \quad D \equiv \dfrac{x_2+x_3}{2}, \dfrac{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 $D \dfrac{x_2+x_3}{2}, \dfrac{y_2+y_3}{2}$ in the ratio $2: 1$, then the coordinates of $P$

$ \equiv \dfrac{2 \cdot \dfrac{x_2+x_3}{2}+1 \cdot x_1}{2+1}, \dfrac{\dfrac{y_2+y_3}{2}+1 \cdot y_1}{2+1} $

$\because$ internal section formula $=\dfrac{m_1 x_2+m_2 x_1}{m_1+m_2}, \dfrac{m_1 y_2+m_2 y_1}{m_1+m_2}$

$\equiv \dfrac{x_2+x_3+x_1}{3}, \dfrac{y_2+y_3+y_1}{2}$

$\therefore$ So, required coordinates of point $P \equiv \dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}$

(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 $E \dfrac{x_1+x_3}{2}, \dfrac{y_1+y_3}{2}$ in the ratio $2: 1$, then the coordinates of $Q$

$ \begin{aligned} & \equiv \dfrac{2 \cdot \dfrac{x_1+x_3}{2}+1 \cdot x_2}{2+1}, \dfrac{\dfrac{y_1+y_2}{2}+1 \cdot y_2}{2+1} \\ & =\dfrac{x_1+x_2+x_3}{3}, \dfrac{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.

$ \therefore \text{ mid-point of } A C=\text{ Coordinate of } E \Rightarrow E=\dfrac{x_1+x_3}{2}, \dfrac{y_1+y_3}{2} $

So, the required coordinate of point $Q \equiv \dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}$

Now, let the coordinates of a point $E$ be $(\alpha, \beta)$. Given that, the point $R(\alpha, \beta)$, divide the line joining $C(x_3, y_3)$ and $F \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$ in the ratio $2: 1$, then the coordinates of $R$

$ \begin{aligned} & \equiv \dfrac{2 \cdot \dfrac{x_1+x_2}{2}+1 \cdot x_3 2 \cdot \dfrac{y_1+y_2}{2}+1 \cdot y_3}{2+1}, \dfrac{x_1+x_2+x_3}{3}, \dfrac{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.

$ \therefore \text{ mid-point of } A B=\text{ coordinate of } F \Rightarrow F=\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} $

So, the required coordinate of point $R \equiv \dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}$

(iv) Coordinate of the centroid of the $\triangle A B C$

$ \begin{aligned} & =\dfrac{\text{ Sum of abscissa of all vertices }}{3}, \dfrac{\text{ Sum of ordinate of all vertices }}{3} \\ & =\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3} \end{aligned} $

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

since, mid-point of a line segment having points $(x_1, y_1)$ and $(x_2, y_2)$ is $\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

$ \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) = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1) \\ \Rightarrow & (y+2) = \dfrac{3+2}{2-1}(x-1) \\ \Rightarrow & (y+2) = 5(x-1) \\ \Rightarrow & 5 x-y = 7 & \ldots \text{(i)}\\ & \text{ Slope of } A B, \text{ say } m_1 = \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{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{aligned} m_1 \cdot m_2 = -1= \\ m_2 = -\dfrac{1}{5} \end{aligned} $

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

$ \begin{matrix} & (y-y_1) =m_2(x-x_1) \\ \Rightarrow & (y+3) =-\dfrac{1}{5}(x+4) \\ \Rightarrow & 5 y+15 =-x-4 & \ldots \text{(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 \\ \therefore & x = \dfrac{8}{13} \end{matrix} $

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

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

$\therefore \quad$ Coordinates of point $P \equiv \dfrac{8}{13}, \dfrac{-51}{13}$

So, length of the height of a parallelogram,

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

$[\because.$ 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{\dfrac{60^{2}}{13}+\dfrac{-12^{2}}{13}} \\ & =\dfrac{1}{13} \sqrt{3600+144} \\ & =\dfrac{1}{13} \sqrt{3744}=\dfrac{12 \sqrt{26}}{13} \end{aligned} $

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

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 $D$.

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

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

since, mid-point of a line segment having points $(x_1, y_1)$ and $(x_2, y_2)=\dfrac{x_1+y_1}{2}, \dfrac{x_2+y_2}{2}$

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

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 km \end{aligned} $

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