CO-ORDINATE GEOMETRY

CO-ORDINATE GEOMETRY

7.1 RECTANGULAR CO-ORDINATES :

Take two perpendicular lines ${\mathbf{X}}^{\prime }\mathbf{O}\mathbf{X}$ and ${\mathbf{Y}}^{\prime }OY$ intersecting at the point $\mathbf{O}$. ${\mathbf{X}}^{\prime }\mathbf{O}\mathbf{X}$ and ${\mathbf{Y}}^{\prime }OY$ are called the co-ordinate axes. ${\mathbf{X}}^{\prime }\mathbf{O}\mathbf{x}$ is called the $\mathbf{X}$-axis, ${\mathbf{Y}}^{\prime }\mathbf{O}\mathbf{Y}$ is called the $\mathbf{Y}$-axis and $\mathbf{O}$ is called the origin. Lines ${\mathbf{X}}^{\prime }OX$ and ${\mathbf{Y}}^{\prime }OY$ are sometimes also called rectangular axes.

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7.1 (a) Co-ordinates of a Point :

Let $\mathbf{P}$ be any point as shown in figure. Draw PL and PM perpendiculars on $\mathbf{Y}$-axis and $\mathbf{X}$-axis, respectively. The length $\mathbf{L}\mathbf{P}$ (or $\mathbf{O}\mathbf{M}$ ) is called the $\mathbf{x}$ - coordinate of the abscissa of point $\mathbf{P}$ and MP $i$ called the $\mathbf{y}$ - coordinate or the ordinate of point $\mathbf{P}$. A point whose abscissa is $\mathbf{x}$ and ordinate is $\mathbf{y}$ named as the point $(\mathbf{x},\mathbf{y})$ or $\mathbf{P}(x,y)$.

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The two lines ${\mathbf{X}}^{\prime }\mathbf{O}\mathbf{X}$ and ${\mathbf{Y}}^{\prime }\mathbf{O}\mathbf{Y}$ divide the plane into four parts called quadrants. $XOY,YO{X}^{\prime }$ $X^{\prime }O{Y}^{\prime }$ and $Y^{\prime }OX$ are, respectively, called the first, second third and fourth quadrants. The following table shows the signs of the coordinates of pins situated in different quadrants :

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REMAKS

(i) Abscissa is the perpendicular distance of a point from $\mathbf{y}$-axis (i.e., positive to the right of $\mathbf{y}$-axis and negative to the left of $\mathbf{y}$ - axis)

(ii) Ordinate is positive above $\mathbf{x}$ - axis and negative below $\mathbf{x}$-axis.

(iii) Abscissa of any point on $y$-axis is zero.

(iv) Ordinate of any point of $\mathbf{x}$-axis is zero.

(v) Co-ordinates of the origin are $(0,0)$

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7.2 DISTACE BETWEEN TWO POINTS :

Let two points be $\mathbf{P}({\mathbf{x}}_1,{\mathbf{y}}_1)$ and $\mathbf{Q}({\mathbf{x}}_2,{\mathbf{y}}_2)$

Take two mutually perpendicular lines as the coordinate axis with $\mathbf{O}$ as origin. Mark the points $\mathbf{P}({\mathbf{x}}_1,{\mathbf{y}}_1)$ and $\mathbf{Q}({\mathbf{x}}_2,{\mathbf{y}}_2)$. Draw lines $\mathbf{P}\mathbf{A}$,

$\mathbf{Q}\mathbf{B}$ perpendicular to $\mathbf{X}$-axis from the points $\mathbf{P}$ and $\mathbf{Q}$, which meet the $\mathbf{X}$-axis in points $A$ and $B$, respectively.

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Draw lines PC and QD perpendicular to Y-axis, which meet the $\mathbf{Y}$ axis in $C$ and $D$, respectively. Produce $CP$ to meet $BQ$ in R. Now

$OA=$ abscissa of $P=x_1$

Similarly, $OB=x_2,OC=y_1$ and $OD=y_2$

Therefore, we have $PR=AB=OB-OA=x_2-x_1$

Similarly, $QR=QB-RB=QB-PA=y_2-y_1$

Now, using Pythagoras Theorem, in right angled triangle

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$PRQ$, we have $P{Q}^2=P{r}^2+R{Q}^2$

or $P{Q}^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2$

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Since the distance or length of the line-segment $PQ$ is always non-negative, on taking the positive square root, we get the distance as

$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

This result is known as distance formula.

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Corollary : The distance of a point $\mathbf{P}({\mathbf{x}}_1,{\mathbf{y}}_1)$ from the origin $(\mathbf{0},\mathbf{0})$ is given by

$OP=\sqrt{x_1^2+y_1^2}$

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Some useful points :

1. In questions relating to geometrical figures, take the given vertices in the given order and proceed as indicated.

(i) For an isosceles triangle - We have to prove that at least two sides are equal.

(ii) For an equilateral triangle - We have to prove that three sides are equal.

(iii) For a right -angled triangle - We have to prove that the sum of the squares of two sides is equal to the square of the third side.

(iv) for a square - We have to prove that the four sides are equal, two diagonals are equal.

(v) For a rhombus - We have to prove that four sides are equal (and there is no need to establish that two diagonals are unequal as the square is also a rhombus).

(vi) For a rectangle - We have to prove that the opposite sides are equal and two diagonals are equal.

(vii) For a Parallelogram - We have to prove that the opposite sides are equal (and there is no need to establish that two diagonals are unequal sat the rectangle is also a parallelogram).

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2. for three points to be collinear - We have to prove that the sum of the distances between two pairs of points is equal to the third pair of points.

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Ex. 1 Find the distance between the points $(8,-2)$ and $(3,-6)$.

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Sol. Let the points $(8,-2)$ and $(3,-6)$ be denoted by $P$ and $Q$, respectively.

Then, by distance formula, we obtain the distance $PQ$ as

$PQ=\sqrt{(3-8{)}^2+(-6+2{)}^2}=\sqrt{(-5{)}^2+(-4{)}^2}=\sqrt{41}$ unit

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Ex. 2 Prove that the points $(1,-1),(-\frac{1}{2},\frac{1}{2})$ and $(1,2)$ are the vertices of an isosceles triangle.

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Sol. Let the point $(1,-1),(-\frac{1}{2},\frac{1}{2})$ and $(1,2)$ be denoted by $P,Q$ and $R$, respectively. Now

$PQ=\sqrt{(-\frac{1}{2}-{)}^2+(\frac{1}{2}+1{)}^2}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$

$QR=\sqrt{(1+\frac{1}{2}{)}^2+(2-\frac{1}{2}{)}^2}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$

$PR=\sqrt{(1-1{)}^2+(2+1{)}^2}=\sqrt{9}=3$

From the above, we see that $PQ=QR\therefore$ The triangle is isosceles.

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Ex. 3 Using distance formula, show that the points $(-3,2),(1,-2)$ and $(9,-10)$ are collinear.

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Sol. Let the given points $(-3,2),(1,-2)$ and $(9,-10)$ be denoted by A, B and C, respectively. Points A, B and C will be collinear, if the sum of the lengths of two line-segments is equal to the third.

Now, $AB=\sqrt{(1+3{)}^2+(-2-2{)}^2}=\sqrt{16+16}=4\sqrt{2}$

$BC=\sqrt{(9-1{)}^2+(-10+2{)}^2}=\sqrt{64+64}=8\sqrt{2}$

$AC=\sqrt{(9+3{)}^2+(-10-2{)}^2}=\sqrt{144+144}=12\sqrt{2}$

Since, $AB+BC=4\sqrt{2}+8\sqrt{2}=12\sqrt{2}=AC$, the, points $A,B$ and $C$ are collinear.

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Ex. 4 Find a point on the $X$-axis which is equidistant from the points $(5,4)$ and $(-2,3)$.

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Sol. Since the required point (say $P$ ) is on the $X$-axis, its ordinate will be zero. Let the abscissa of the point be $x$.

Therefore, coordinates of the point $P$ are $(x,0)$.

Let $A$ and $B$ denote the points $(5,4)$ and $(-2,3)$, respectively.

Since we are given that $AP=BP$, we have $A{P}^2=B{P}^2$

i.e., $(x-5{)}^2+(0-4{)}^2=(x+2{)}^2+(0-3{)}^2$

or $x^2+25-10x+16=x^2+4+4x+9$ or $-14x=-28$ or $x=2$

Thus, the required point is $(2,0)$.

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Ex. 5 The vertices of a triangle are $(-2,0),(2,3)$ and $(1,-3)$. Is the triangle equilateral, isosceles or scalene?

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Sol. Let the points $(-2,0),(2,3)$ and $(1,-3)$ be denoted by $A,B$ and $C$ respectively. Then,

$AB=\sqrt{(2+2{)}^2+(3-0{)}^2}=5BC=\sqrt{(1-2{)}^2+(-3-3{)}^2}=\sqrt{37}$

and $AC=\sqrt{(1+2{)}^2+(-0-0{)}^2}=3\sqrt{2}$ Clearly, $AB\ne BC\ne AC$.

Therefore, $ABC$ is a scalene triangle.

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Ex. 6 The length of a line-segments is 10. If one end is at $(2,-3)$ and the abscissa of the second end is 10, show that its ordinate is either 3 or -9 .

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Sol. Let $(2,-3)$ be the point $A$. let the ordinate of the second end $B$ be $y$. Then its coordinates will be $(10,y)$.

$\therefore AB=\sqrt{(10-2{)}^2+(y+3{)}^2}=10$ (Given)

$\begin{array}{ccc}\text{ or }64+9+y^2+6y=100& \text{ or }& y^2+6y+73-100=0\\ \text{ or }{y}^2+6y-27=0& \text{ or }& (y+9)(y-3)=0\\ \text{ Therefore, }y=9& \text{ or }& y=3.\end{array}$

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Ex. 7 Show that the points $(-2,5),(3,-4)$ and $(7,10)$ are the vertices of a right triangle.

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Sol. Let the three points be $A(-2,5),B(3,-4)$ and $C(7,10)$.

Then $A{B}^2=(3+2{)}^2+(-4-5{)}^2=106$

$B{C}^2=(7-3{)}^2+(10+4{)}^2=212$

$A{C}^2=(7+2{)}^2+(10-5{)}^2=106$ We see that $B{C}^2=A{B}^{2}1+A{C}^2$

$212=106+106$ $212=212$

$\therefore \mathrm{\angle }A={90}^{\circ }$ Thus, $ABC$ is a right triangle, right angled at $A$.

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Ex. 8 If the distance of $P(x,y)$ from $A(5,1)$ and $B(-1,5)$ are equal, prove that $3x=2y$.

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Sol. $P(x,y),A(5,1)$ and $B(-1,5)$ are the given points.

$AP=BP$ (Given)

$\begin{array}{cc}\therefore & A{P}^2=B{P}^2\text{ or }A{P}^2-B{P}^2=0\\ \text{ or }& {(x-5{)}^2+(y-1)}^2-(x+1{)}^2+(y-5{)}^2=0\\ \text{ or }& x^2+25-10x+y^2+1-2y-x^2-1-2x-y^2-25+10y=0\\ \text{ or }& -12x+8y=0\text{ or }3xx=2y.\end{array}$

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7.3 SECTION FORMULAE :

7.3 (a) Formula for Internal Division :

The coordinates of the pint which divided the line segment joining the pints $(x_1,y_1)$ and $(x_2,y_2)$

internally in the ratio $\mathbf{m}:\mathbf{n}$ are given by $x=\frac{m{x}_2+n{x}_1}{m+n},y=\frac{m{y}_2+m{y}_1}{m+n}$

Proof :Let $O$ be the origin and let $OX$ and $OY$ be the $X$-axis and $Y$-axis respectively. Let $\mathbf{A}(x_1,y_1)$ and $\mathbf{B}(x_2,y_2)$ bet the given points. Let $(\mathbf{x},\mathbf{y})$ be the coordinates of the point $\mathbf{p}$ which divides $\mathbf{A}\mathbf{B}$ internally in the ratio $\mathbf{m}:\mathbf{n}$ Draw $AL\perp OX,BM\perp OX,PN\perp Ox$. Also, draw $\mathbf{A}\mathbf{H}$ and $\mathbf{P}\mathbf{K}$ perpendicular from $\mathbf{A}$ and $\mathbf{P}$ on $\mathbf{P}\mathbf{N}$ and $\mathbf{B}\mathbf{M}$ respectively. Then

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$\begin{array}{rl}& OL=x_1,ON=x,OM=x_2,AL=y_1,PN=y\text{ and }BM=y_2.\\ & \therefore AH=LN=ON-OL=x-x_1,PH=PH-HN\\ & =PN-AL=y-y_1,PK=NM=OM-ON=x_2-x\\ & \text{ and }BK=BM-MK=BM-PN=y_2-y.\end{array}$

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Clearly, $\mathrm{△}AHP$ and $\mathrm{△}PKB$ are similar.

$\therefore \frac{AP}{BP}=\frac{AH}{PK}=\frac{PH}{BK}$

$\Rightarrow \frac{m}{n}=\frac{x-x_1}{x_2-x}=\frac{y-y_1}{y_2-y}$

Now, $\frac{m}{n}=\frac{x-x_1}{x_2-x}$

$\Rightarrow m{x}_2-mx=nx-n{x}_1\Rightarrow mx+nx=m{x}_2+n{x}_1$

$\Rightarrow x=\frac{m{x}_2+n{x}_1}{m+n}$ and $\frac{m}{n}=\frac{y-y_1}{y_2-y}$

$\Rightarrow m{y}_2-my=ny-n{y}_1\Rightarrow my+ny=m{y}_2+n{y}_1$

$\Rightarrow y=\frac{m{y}_2+n{y}_1}{m+n}$

Thus, the coordinates of $P$ are $(\frac{m{x}_2+n{x}_1}{m+n},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{m+n})$

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REMARKS

If $\mathbf{P}$ is the mid-point of $\mathbf{A}\mathbf{B}$, then it divides $\mathbf{A}\mathbf{B}$ in the ratio $\mathbf{1}:\mathbf{1}$, so its coordinates are $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

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7.3 (b) Formula for External Division :

The coordinates of the points which divides the line segment joining the points $(x_1,{\mathbf{y}}_1)$ and $(x_2,y_2)$ externally in the ratio $\mathbf{m}:\mathbf{n}$ are given by

$x=\frac{m{x}_2-n{x}_1}{m-n},y=\frac{m{y}_2-n{y}_1}{m-n}$

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Ex. 9 Find the coordinates of the point which divides the line segment joining the points $(6,3)$ and $(-4,5)$ in the ratio $3:2$ (i) internally (ii) externally.

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Sol. Let $P(x,y)$ be the required point.

(i) For internal division, we have

$\begin{array}{rl}x& =\frac{3x-4+2\times 6}{3+2}\\ \text{ and }y& =\frac{3\times 5+2\times 3}{3+2}\\ \Rightarrow x& =0\text{ and }y=\frac{21}{5}\end{array}$

So the coordinates of $P$ are $(0,21/5)$

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(ii) For external division, we have

$\begin{array}{rl}& x=\frac{3x-4-2\times 6}{3-2}\\ & \text{ any }y=\frac{3\times 5-2\times 3}{3-2}\\ & \Rightarrow x=-24\text{ and }y=9\end{array}$

So the coordinates of $P$ are $(-24,9)$.

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Ex. 10 In which ratio does the point $(-1,-1)$ divides the line segment joining the pints $(4,4)$ and $(7,7)$ ?

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Sol. Suppose the point $C(-1,-1)$ divides the line joining the points $A(4,4)$ and $B(7,7)$ in the ratio $k:1$ Then, the coordinates of $C$ are $(\frac{7k+4}{k+1},\frac{7k+4}{k+1})$

But, we are given that the coordinates of the points $C$ are $(-1,-1).\therefore \frac{7k+4}{k+1}=-1\Rightarrow k=-\frac{5}{8}$

Thus, $C$ divides $AB$ externally in the ratio $5:8$.

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Ex. 11 In what ratio does the X-axis divide the line segment joining the points $(2,-3)$ and $(5,6)$ ?

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Sol. Let the required ratio be $k:1$. Then the coordinates of the point of division are $(\frac{5\lambda +2}{k+1},\frac{6\lambda -3}{k+1})$. But, it is a point on $X$-axis on which $y$-coordinate of every point is zero.

$\therefore \frac{6\lambda -3}{k+1}=0\Rightarrow k=\frac{1}{2}$ Thus, the required ratio is $\frac{1}{2}:1$ or $1:2$.

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Ex. 12 $A(1,1)$ and $B(2,-3)$ are two points and $D$ is a point on $AB$ produced such that $AD=3AB$. Find the coordinates of $D$.

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Sol. We have, $AD=3AB$. Therefore, $BD=2AB$. Thus $D$ divides $AB$ externally in the ratio $AD:BD=3:2$ Hence, the coordinates of $D$ are

$\therefore (\frac{3\times 2-2\times 1}{3-2},\frac{3x-3-2\times 1}{3-2})$

$=(4,-11)$.

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Ex. 13 Determine the ratio in which the line $3x+y-9=0$ divides the segment joining the pints $(1,3)$ and $(2,7)$.

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Sol. Suppose the line $3x+y-9=0$ divides the line segment joining $A(1,3)$ and $B(2,7)$ in the ratio $k:1$ at point C. The, the coordinates of $C$ are $(\frac{2k+1}{k+1},\frac{7k+3}{k+1})$ But, $C$ lies on $3x+y-9=0$, therefore

$3(\frac{2k+1}{k+1})+\frac{7k+3}{k+1}-9=0\Rightarrow 6k+3+7k+3-9k-9=0\Rightarrow k=\frac{3}{4}$

So, the required ratio is $3:4$ internally.

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7.4 CENTROID OF A TRIANGLE :

Prove that the coordinates of the triangle whose vertices are $(x_1,y_1),(x_2,y_2)$ and $(y_3,y_3)$ are $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$. Also, deduce that the medians of a triangle are concurrent.

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Proof :

Let $\mathbf{A}({\mathbf{x}}_1,{\mathbf{y}}_1,\mathbf{B}(x_2,y_2).$ and $\mathbf{C}(x_3,y_3)$ be the vertices of $\mathrm{△}ABC$ whose medians are $AD,BE$ and $CF$ respectively. So. $D,E$ and $F$ are respectively the mid-points of $BC,CA$ and $AB$.

Coordinates of $\mathbf{D}$ are $(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$. Coordinates of a point dividing $AD$ in the ratio $\mathbf{2}:\mathbf{1}$ are

$(\frac{1.x_1+2(\frac{x_2+x_3}{2})}{1+2},\frac{1\cdot y_1+(\frac{y_2+y_3}{2})}{1+2})=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

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The coordinates of $E$ are $(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})$. The coordinates of a point dividing BE in the ratio $2:1$ are $(\frac{1.x_2+\frac{2(x_1+x_3)}{2}}{1+2},\frac{1.y_2+\frac{2(y_1+y_3)}{2}}{1+2})=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Similarly the coordinates of a point dividing CF in the ratio $2:1$ are $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Thus, the point having coordinates $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$ is common to $AD,BE$ and $CF$ and divides them in the ratio $1:2$.

Hence, medians of a triangle are concurrent and the coordinates of the centroid are $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$.

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7.5 AREA OF A TRIANGLE :

Let $\mathbf{A}\mathbf{B}\mathbf{C}$ be any triangle whose vertices are $\mathbf{A}({\mathbf{x}}_1,{\mathbf{y}}_1)\mathbf{B}({\mathbf{x}}_2,{\mathbf{y}}_3)$. Draw BL, AM and $CN$ perpendicular from $B,A$ and $C$ respectively, to the $X$-axis. ABLM, AMNC and BLNC are all trapeziums.

Area of $\mathrm{△}ABC=$ Area of trapezium ABLM + Area of trapezium AMNC - Area of trapezium BLNC We know that, Area of trapezium $=\frac{1}{2}$ (Sum of parallel sides) $($ distance $b/w$ them)

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Therefore

Area of $\mathrm{△}ABC=\frac{1}{2}(BL+AM)(LM)+\frac{1}{2}(AM+CN)MN-\frac{1}{2}(BL+CN)(LN)$

Area of $\mathrm{△}ABC=\frac{1}{2}(y_2+y_1)x_1-x_2)+\frac{1}{2}(y_1+y_3)(x_3-x_1)-\frac{1}{2}(y_2+y_3)(x_3-x_2)$

Area of $\mathrm{△}ABC=\frac{1}{2}|[x_1(y_2-y_3)+x_2(y_3-y)+x_3(y_1-y_2)]|$

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7.5 (a) Condition for collinearity :

Three points $A(x_1,y_1)B(x_2,y_2)$ and $C(x_3,y_3)$ are collinear if Area of $\mathrm{△}ABC=0$.

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7.6 AREA OF QUADRILATERAL :

Let the vertices of Quadrilateral $ABCD$ are $\mathbf{A}({\mathbf{x}}_1,{\mathbf{y}}_1),\mathbf{B}({\mathbf{x}}_2,{\mathbf{y}}_2,\mathbf{C}({\mathbf{x}}_3,{\mathbf{y}}_3).$ and $\mathbf{D}({\mathbf{x}}_4,{\mathbf{y}}_4)$

So, Area of quadrilateral $ABCD=$ Area of $\mathrm{△}ABC+$ Area of $\mathrm{△}ACD$

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Ex. 14 The vertices of $\mathrm{△}ABC$ are $(-2,1),(5,4)$ and $(2,-3)$ respectively. Find the area of triangle.

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Sol. $A(-2,1),B(-2,1)$ and $C(2,-3)$ be the vertices of triangle.

So, $x_1=-2,y_1=1;x_2=5,y_2=4;x_3=2y_3=-3$

$\begin{array}{rl}\text{ Area of }\mathrm{\Delta }ABC& .=\frac{1}{2}|[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\ & =\frac{1}{2}|[(-2)(4+3)+(5)(-3-1)+2(1-4)]=\frac{1}{2}[-14+(-20)+(-6)].\\ & =\frac{1}{2}|-40|=20\text{ sq. unit. }\end{array}$

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Ex. 15 The area of a triangle is 5. Two of its vertices area $(2,1)$ and $(3,-2)$. The third vertex lies on $y=x+3$. Find the third vertex.

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Sol. Let the third vertex be $(x_3,y_3)$ area of triangle

$\begin{array}{rl}& =\frac{1}{2}|[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\ & \text{ As }{x}_1=2y_1=1;x_2=3,y_2=-2;\text{ Area of }\mathrm{\Delta }=5\text{ sq. unit }\\ & \Rightarrow 5=\frac{1}{2}|2(-2-y_3)+3(y_3-1)+x_3(1+2)|\Rightarrow 10=|3x_3+y_3-7|\\ & \Rightarrow 3x_3+y_3-7=\pm 10\end{array}$

Taking positive sign

$3x_3+y_3-7=10\Rightarrow 3x\therefore +y_3=17\dots \dots \dots \dots .(i)$

Taking negative sign

$\Rightarrow 3x_3+y_3-7=-10\Rightarrow 3x_{\therefore }+y_3=-3\dots \dots \dots \dots (ii)$

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Given that $(x_3,y_3)$ lies on $y=x+3$

So, $-x\therefore +y_3=3$

Solving eq. (i) & (iii)

$x_3=\frac{7}{2},y_3=\frac{13}{2}$

Solving eq (ii) & (iii)

$x_3=\frac{-3}{2},y_3=\frac{3}{2}$

So the third vertex are $(\frac{7}{2},\frac{13}{2})$ or $(\frac{-3}{2},\frac{3}{2})$

Ex. 16 Find the area of quadrilateral whose vertices, taken in order, are (-3, 2), B(5, 4), (7, -6) and D (-5, -4).

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Sol. Area of quadrilateral $=$ Area of $\mathrm{△}ABC+$ Area of $\mathrm{△}ACD$

So, $$ Area of $\mathrm{△}ABC=\frac{1}{2}|(-3)(4+6)+5(-6-2)+7(2-4)|$

$=\frac{1}{2}|-30-40-14|$

$=\frac{1}{2}|-84|=42$ Sq. units

Area of $\mathrm{△}ACD$

$=\frac{1}{2}|-3(-6+4)+7(-4-2)+(-5)(2+6)|$

$=\frac{1}{2}|+6-42-40|=\frac{1}{2}|-76|=38$ Sq. units

So, $$ Area of quadrilateral $ABCD=42+38=80$ Sq. units.

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DAILY PRACTIVELY PROBLEMS 7

OBJECTIVE DPP - 7.1

1. The points $(-a,-b),(0,0),(a,b)$ and $(a^2,ab)$ are

(A) Collinear

(B) Vertices of a parallelogram

(C) Vertices of a rectangle

(D) None of these

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2. If the points $(5,1),(1,p)$ & $(4,2)$ are collinear then the value of $p$ will be

(A) 1

(B) 5

(C) 2

(D) -2

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3. Length of the median from $B$ on $AC$ where $A(-1,3),B(1,-1),(5,1)$ is

(A) $\sqrt{18}$

(B) $\sqrt{10}$

(C) $2\sqrt{3}$

(D) 4

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4. The points $(0,-1),(-2,3),(6,7)$ and $(8,3)$ are -

(A) Collinear

(B) Vertices of a parallelogram which is not a rectangle

(C) Verticals of a rectangle, which is not a square

(D) None of these

CO-ORDINATE GEOMETRY

CO-ORDINATE GEOMETRY

5. If $(3,-4)$ and $(-6,5)$ are the extremities of the diagonal of a parallelogram and $(-2,1)$ is third vertex, then its fourth vertex is

(A) $(-1,0)$

(B) $(0,-1)$

(C) $(-1,1)$

(D) None of these

CO-ORDINATE GEOMETRY

CO-ORDINATE GEOMETRY

6. The area of a triangle whose vertices are $(a,c+a),(a,c)$ and $(-a,c-a)$ are

(A) $a^2$

(B) $b^2$

(C) $c^2$

(D) $a^2+c^2$

CO-ORDINATE GEOMETRY

CO-ORDINATE GEOMETRY

7. The are of the quadrilateral’s the coordinates of whose verticals are $(1,-2),(6,2),(5,3)$ and $(3,4)$ are

(A) $\frac{9}{2}$

(B) 5

(C) $\frac{11}{2}$

(D) 11

CO-ORDINATE GEOMETRY

CO-ORDINATE GEOMETRY

SUBJECTIVE DPP - 7.2

1. Find the distance between the points :

(i) $P(-6,7)$ and $Q(-1,-5)$.

(ii) $A(a{t}_1{}^2,2a{t}_1)$ and $B(a{t}_2{}^2,2a{t}_2)$.

CO-ORDINATE GEOMETRY

Sol. 1. (i) $13$

(ii) $a(t_2-t_1)\sqrt{(t_2+t_1{)}^2+4}$

CO-ORDINATE GEOMETRY

2. If the point $(x,y)$ is equidistant from the points $(a+b,b-a)$ and $(a-b,a+b)$, prove that $bx=ay$.

CO-ORDINATE GEOMETRY

3. Find the value of $x$, if the distance between the points $(x,-1)$ and $(3,2)$ is 5 .

CO-ORDINATE GEOMETRY

Sol. 3. $x=7$ or $-1$

CO-ORDINATE GEOMETRY

4. Show that the points $(a,a),(-a,-a)$ and $-\sqrt{3a},\sqrt{3a})$ are the vertices of an equilateral triangle.

CO-ORDINATE GEOMETRY

5. Show that the points $(1,1),(-2,7)$ and $(3,-3)$ are collinear.

CO-ORDINATE GEOMETRY

6. Prove that $(2,-2),(-2,1)$ and $(5,2)$ are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.

CO-ORDINATE GEOMETRY

Sol. 6. $\frac{25}{2}$ sq. units , $5\sqrt{2}$

CO-ORDINATE GEOMETRY

7. If $A(-1,3),B(1,-1)$ and $C(5,1)$ are the vertices of a triangle $ABC$, find the length of the median passing through the vertex $A$.

CO-ORDINATE GEOMETRY

Sol. 7. $5$ units

CO-ORDINATE GEOMETRY

8. Show that the points $A(1,2),B(5,4),C(3,8)$ and $D(-1,6)$ are the vertices of a square.

CO-ORDINATE GEOMETRY

9. The abscissa of a point is twice its ordinate and the sum of the abscissa and the ordinate is -6 . What are the coordinates of the point?

CO-ORDINATE GEOMETRY

Sol. 9. $(-4,-2)$

CO-ORDINATE GEOMETRY

10. If two vertices of triangle are $(3,7)$ an $(-1,5)$ and its centroid is $(1,3)$, find the coordinates of the third vertex.

CO-ORDINATE GEOMETRY

Sol. 10. $(1,-3)$

CO-ORDINATE GEOMETRY

11. If the mid point of the line-segment joining the points $(-7,14)$ and $(K,4)$ is $(a,b)$, where $2a+3b=5$, find the value of $K$.

CO-ORDINATE GEOMETRY

Sol. 11. $K=-15$

CO-ORDINATE GEOMETRY

12. Prove hat the points $(a,0),(0,b)$ and $(1,1)$ are collinear if $\frac{1}{a}+\frac{1}{b}=1$.

CO-ORDINATE GEOMETRY

13. The co-ordinates of two points $A$ & $B$ are $(3,4)$ and $(5,-2)$ respectively. Find the co-ordinate of point $P$ if $PA=PB$, the area of $\mathrm{△}APB=10$.

CO-ORDINATE GEOMETRY

Sol. 13. $(7,2)$ or $(1,0)$

CO-ORDINATE GEOMETRY

14. Four points $A(6,3),B(-3,5)C(4,-2)$ and $D(x,3x)$ are given in such a way that $\frac{\text{ Area }(\mathrm{△}DBC)}{\text{ Area }(\mathrm{△}ABC)}=\frac{1}{2}$ find $x$.

CO-ORDINATE GEOMETRY

Sol. 14. $\frac{11}{8},-\frac{3}{8}$

CO-ORDINATE GEOMETRY

15. Show that the points $A(2,-2),B(14,10),C(11,13)$ and $D(-1,1)$ are the vertices of a rectangle.

[CBSE-2004]

CO-ORDINATE GEOMETRY

16. Determine the ratio in which the point $(-6$, a) divides the join of $A(-3,-1)$ and $B(-8,9)$. Also find the value of a.

[CBSE 2004]

CO-ORDINATE GEOMETRY

Sol. 16. $3:2,a=5$

CO-ORDINATE GEOMETRY

17. Find a pint on $X$-axis which is equidistant from the points $(7,6)$ and $(-3,4)$.

[CBSE - 2005]

CO-ORDINATE GEOMETRY

Sol. 17. $(3,0)$

CO-ORDINATE GEOMETRY

18. The line segment joining the points $(3,-4)$ and $(1,2)$ is trisected at the pints $P$ and $Q$. if the coordinates of $P$ and $Q$ are $(p,-2)$ and $(5/3$,$)$ respectively. Finds the value of $p$ and $q$.

[CBSE 2005]

CO-ORDINATE GEOMETRY

Sol. 18. $p=7/3,q=0$

CO-ORDINATE GEOMETRY

19. If $A(-2,-1),B(a,0),C(4,b)$ and $D(1,2)$ are the verities of a parallelogram, find the values of $a$ and $b$.

[ -2006]

CO-ORDINATE GEOMETRY

Sol. 19. $a=1,b=3$

CO-ORDINATE GEOMETRY

20. The coordinates of one end point of a diameter of a circle are ( $4,-1)$ and the coordinates of the centre of the circle are $(1,-3)$. Find the coordinates of the other end of the diameter.

[CBSE-2007]

CO-ORDINATE GEOMETRY

Sol. 20. $(-2,-5)$

CO-ORDINATE GEOMETRY

21. The pint $R$ divides the line segment $AB$, where $A(-4,0)$ and $B(0,6)$ are such that $AR=\frac{3}{4}AB$. Find the coordinates or R.

[CBSE - 2008]

CO-ORDINATE GEOMETRY

Sol. 21. $(-1,\frac{9}{2})$

CO-ORDINATE GEOMETRY

22. For what value of $k$ are the pints $(1,1),(3,k)$ and $(-1,4)$ collinear ?

[CBSE - 2008]

CO-ORDINATE GEOMETRY

Sol. 22. $k=-2$

CO-ORDINATE GEOMETRY

23. Find the area of the $\mathrm{△}ABC$ with vertices $A(-5,7),B(-4,-5)$ and $C(4,5)$.

[CBSE - 2008]

CO-ORDINATE GEOMETRY

Sol. 23. $53$ sq. units

CO-ORDINATE GEOMETRY

24. If the point $P(x,y)$ is equidistant from the points $A(3,6)$ and $B(-3,4)$ prove that $3x+y-5=0$.

[CBSE - 2008]

CO-ORDINATE GEOMETRY

25. If $A(4-8),B(3,6)$ and $C(5,-4)$ are the vertices of a $\mathrm{△}ABC,D$ is the mid-point of $BC$ and is $P$ is point on $AD$ joined such that $\frac{AP}{PD}=2$ find the coordinates of $P$.

[CBSE - 2008]

CO-ORDINATE GEOMETRY

Sol. 25. $(4,-2)$