CO-ORDINATE GEOMETRY

CO-ORDINATE GEOMETRY

7.1 RECTANGULAR CO-ORDINATES :

Take two perpendicular lines $\mathbf{X}^{\prime} \mathbf{O X}$ and $\mathbf{Y}^{\prime} OY$ intersecting at the point $\mathbf{O}$. $\mathbf{X}^{\prime} \mathbf{O X}$ and $\mathbf{Y}^{\prime} OY$ are called the co-ordinate axes. $\mathbf{X}^{\prime} \mathbf{O x}$ is called the $\mathbf{X}$-axis, $\mathbf{Y}^{\prime} \mathbf{O Y}$ 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 $\mathbf{L P}$ (or $\mathbf{O M}$ ) 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 $\mathbf{X}^{\prime} \mathbf{O X}$ and $\mathbf{Y}^{\prime} \mathbf{O Y}$ divide the plane into four parts called quadrants. $XOY, YOX’$ $X^{\prime} O Y^{\prime}$ and $Y^{\prime} O X$ 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 $\mathbf{P A}$,

$\mathbf{Q B}$ 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 $C P$ to meet $B Q$ in R. Now

$OA=$ abscissa of $P=x_1$

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

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

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 $\quad PQ^{2}=Pr^{2}+RQ^{2}$

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

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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 $P Q$ as

$PQ=\sqrt{(3-8)^{2}+(-6+2)^{2}} \quad=\sqrt{(-5)^{2}+(-4)^{2}}=\sqrt{41}$ 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 $PQ=QR \quad \therefore \quad$ 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, $\quad A B=\sqrt{(1+3)^{2}+(-2-2)^{2}}=\sqrt{16+16}=4 \sqrt{2}$

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

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

Since, $A B+B C=4 \sqrt{2}+8 \sqrt{2}=12 \sqrt{2}=A C$, 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 $\quad AP^{2}=BP^{2}$

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

or $\quad x^{2}+25-10 x+16=x^{2}+4+4 x+9 \quad$ or $\quad-14 x=-28$ 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,

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

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

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 $(10, y)$.

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

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

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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 $\quad AB^{2}=(3+2)^{2}+(-4-5)^{2}=106$

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

$AC^{2}=(7+2)^{2}+(10-5)^{2}=106 \quad$ We see that $\quad BC^{2}=AB^{2} 1+AC^{2}$

$212=106+106 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ $212=212$

$\therefore \quad \angle A=90^{\circ} \quad$ 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 $3 x=2 y$.

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

$AP=BP \quad$ (Given)

$\begin{matrix} \therefore & AP^{2}=BP^{2} \quad \text { or } \quad AP^{2}-BP^{2}=0 \\ \text { or } & {(x-5)^{2}+(y-1)}^{2}-{(x+1)^{2}+(y-5)^{2}}=0 \\ \text { or } & x^{2}+25-10 x+y^{2}+1-2 y-x^{2}-1-2 x-y^{2}-25+10 y=0 \\ \text { or } & -12 x+8 y=0 \quad \text { or } \quad 3 xx=2 y .\end{matrix} $

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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 $x=\frac{mx_2+nx_1}{m+n}, y=\frac{my_2+my_1}{m+n}$

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 $\mathbf{A B}$ internally in the ratio $\mathbf{m}: \mathbf{n}$ Draw $AL \perp OX, BM \perp OX, PN \perp Ox$. Also, draw $\mathbf{A H}$ and $\mathbf{P K}$ perpendicular from $\mathbf{A}$ and $\mathbf{P}$ on $\mathbf{P N}$ and $\mathbf{B M}$ respectively. Then

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

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Clearly, $\triangle AHP$ and $\triangle PKB$ are similar.

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

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

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

$\Rightarrow \quad mx_2-mx=nx-nx_1 \quad \Rightarrow \quad mx+nx=mx_2+nx_1$

$\Rightarrow \quad x=\frac{mx_2+nx_1}{m+n}$ and $\quad \frac{m}{n}=\frac{y-y_1}{y_2-y}$

$\Rightarrow \quad my_2-my=ny-ny_1 \quad \Rightarrow \quad my+ny=my_2+ny_1$

$\Rightarrow \quad y=\frac{my_2+ny_1}{m+n}$

Thus, the coordinates of $P$ are $(\frac{mx_2+nx_1}{m+n}, \frac{\mathrm{my_{2 } + \mathrm { ny } _ { 1 }}}{m+n})$

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REMARKS

If $\mathbf{P}$ is the mid-point of $\mathbf{A B}$, then it divides $\mathbf{A B}$ 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{aligned} x & =\frac{3 x-4+2 \times 6}{3+2} \\ \text { and } \quad y & =\frac{3 \times 5+2 \times 3}{3+2} \\ \Rightarrow \quad x & =0 \text { and } y=\frac{21}{5} \end{aligned} $

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

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(ii) For external division, we have

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

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 $(\frac{7 k+4}{k+1}, \frac{7 k+4}{k+1})$

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

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.

$\therefore \quad \frac{6 \lambda-3}{k+1}=0 \quad \Rightarrow \quad k=\frac{1}{2} \quad$ 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 $AD=3 AB$. Find the coordinates of $D$.

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

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

$=(4,-11)$.

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

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Sol. Suppose the line $3 x+y-9=0$ 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 $(\frac{2 k+1}{k+1}, \frac{7 k+3}{k+1})$ But, $C$ lies on $3 x+y-9=0$, therefore

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

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 $\triangle ABC$ 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 $A D$ in the ratio $\mathbf{2}: \mathbf{1}$ are

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

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 $\mathbf{A B C}$ 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 $C N$ perpendicular from $B, A$ and $C$ respectively, to the $X$-axis. ABLM, AMNC and BLNC are all trapeziums.

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

Area of $\triangle A B C=\frac{1}{2}(y_2+y_1) x_1-x_2)+\frac{1}{2}(y_1+y_3)(x_3-x_1)-\frac{1}{2}(y_2+y_3)(x_3-x_2)$

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

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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 $\triangle A B C=0$.

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7.6 AREA OF QUADRILATERAL :

Let the vertices of Quadrilateral $A B C D$ 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 $\triangle ABC+$ Area of $\triangle ACD$

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Ex. 14 The vertices of $\triangle ABC$ 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, $x_1=-2, y_1=1 ; x_2=5, y_2=4 ; x_3=2 y_3=-3$

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

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

Taking positive sign

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

Taking negative sign

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

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

So, $\quad-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 $\triangle ABC+$ Area of $\triangle ACD$

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

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

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

Area of $\triangle ACD$

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

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

So, $\quad$ 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 $(a^{2}, a b)$ 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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CO-ORDINATE GEOMETRY

3. Length of the median from $B$ on $A C$ 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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CO-ORDINATE GEOMETRY

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) $A(at_1{ }^{2}, 2 at_1)$ and $B(at_2{ }^{2}, 2 at_2)$.

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 $b x=a y$.

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. $\quad x=7$ or $-1$

CO-ORDINATE GEOMETRY

4. Show that the points $(a, a),(-a,-a)$ and $-\sqrt{3 a}, \sqrt{3 a})$ 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. $\quad \frac{25}{2}$ 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 $A B C$, find the length of the median passing through the vertex $A$.

CO-ORDINATE GEOMETRY

Sol. 7. $\quad 5$ 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. $\quad (-4,-2)$

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. $\quad (1,-3)$

CO-ORDINATE GEOMETRY

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

CO-ORDINATE GEOMETRY

Sol. 11. $\quad K=-15$

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 $\triangle APB=10$.

CO-ORDINATE GEOMETRY

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

CO-ORDINATE GEOMETRY

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

CO-ORDINATE GEOMETRY

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

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. $\quad 3: 2, a=5$

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. $\quad (3,0)$

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. $\quad p=7 / 3, q=0$

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. $\quad a=1, b=3$

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. $\quad (-2,-5)$

CO-ORDINATE GEOMETRY

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

[CBSE - 2008]

CO-ORDINATE GEOMETRY

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

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. $\quad k=-2$

CO-ORDINATE GEOMETRY

23. Find the area of the $\triangle A B C$ with vertices $A(-5,7), B(-4,-5)$ and $C(4,5)$.

[CBSE - 2008]

CO-ORDINATE GEOMETRY

Sol. 23. $\quad 53$ 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 $3 x+y-5=0$.

[CBSE - 2008]

CO-ORDINATE GEOMETRY

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

[CBSE - 2008]

CO-ORDINATE GEOMETRY

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