NCERT Solutions Class 9 Maths Chapter 3 Coordinate Geometry 

4.0NCERT Questions with Solutions for Class 9 Maths Chapter 3 - Detailed Solutions

Exercise : 3.1

1. Find the distance between the following pairs of points (a) $(2,3),(4,1)$ (b) $(-5,7),(-1,3)$ (c) $(a,b),(-a,-b)$ Sol. (a) The given points are : $\mathrm{A}(2,3),\mathrm{B}(4,1)$. Required distance $$\begin{gathered} =AB=BA=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ AB=\sqrt{(4-2{)}^2+(1-3{)}^2}=\sqrt{(2{)}^2+(-2{)}^2} \\ =\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\text{ units } \end{gathered}$$ (b) Here $x_1=-5,y_1=7$ and $x_2=-1,y_2=3$ $\therefore$ The required distance $=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$ $=\sqrt{[-1-(-5){]}^2+(3-7{)}^2}$ $=\sqrt{(-1+5{)}^2+(-4{)}^2}$ $=\sqrt{16+16}=\sqrt{32}=\sqrt{2\times 16}$ $=4\sqrt{2}$ units (c) Here ${\mathrm{x}}_1=\mathrm{a},{\mathrm{y}}_1=\mathrm{b}$ and ${\mathrm{x}}_2=-\mathrm{a},{\mathrm{y}}_2=-\mathrm{b}$ $\therefore$ The required distance $=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$ $=\sqrt{(-a-a{)}^2+(-b-b{)}^2}$ $=\sqrt{(-2a{)}^2+(-2b{)}^2}=\sqrt{4{\mathrm{a}}^2+4{\mathrm{b}}^2}$ $=\sqrt{4\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}=2\sqrt{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}$ units
2. Find the distance between the points $(0,0)$ and $(36,15)$. Sol. Let the points be $A(0,0)$ and $B(36,15)$ $\therefore AB=\sqrt{(36-0{)}^2+(15-0{)}^2}$ $=\sqrt{(36{)}^2+(15{)}^2}=\sqrt{1296+225}$ $=\sqrt{1521}=\sqrt{39^2}=39$
3. Determine if the points $(1,5),(2,3)$ and $(-2,-11)$ are collinear. Sol. The given points are : $A(1,5),B(2,3)$ and $C(-2,-11)$. Let us calculate the distance : $\mathrm{AB},\mathrm{BC}$ and CA by using distance formula. $\mathrm{AB}=\sqrt{(2-1{)}^2+(3-5{)}^2}=\sqrt{(1{)}^2+(-2{)}^2}$ $=\sqrt{1+4}=\sqrt{5}$ units $\mathrm{BC}=\sqrt{(-2-2{)}^2+(-11-3{)}^2}$ $=\sqrt{(-4{)}^2+(-14{)}^2}=\sqrt{16+196}=\sqrt{212}$ $=2\sqrt{53}$ units $\mathrm{CA}=\sqrt{(-2-1{)}^2+(-11-5{)}^2}$ $=\sqrt{(-3{)}^2+(-16{)}^2}=\sqrt{9+256}=\sqrt{265}$ $=\sqrt{5}\times \sqrt{53}$ units From the above we see that: $\mathrm{AB}+\mathrm{BC}\ne \mathrm{CA}$ Hence the above stated points $A(1,5)$, $B(2,3)$ and $C(-2,-11)$ are not collinear.
4. Check whether $(5,-2),(6,4)$ and $(7,-2)$ are the vertices of an isosceles triangle. Sol. Let the points be $\mathrm{A}(5,-2),\mathrm{B}(6,4)$ and $\mathrm{C}(7,-2)$. $\therefore \mathrm{AB}=\sqrt{(6-5{)}^2+[4-(-2){]}^2}$ $=\sqrt{(1{)}^2+(6{)}^2}=\sqrt{1+36}=\sqrt{37}$ $BC=\sqrt{(7-6{)}^2+(-2-4{)}^2}$ $=\sqrt{(1{)}^2+(-6{)}^2}=\sqrt{1+36}=\sqrt{37}$ $\mathrm{AC}=\sqrt{(7-5{)}^2+(-2-(-2){)}^2}$ $=\sqrt{(+2{)}^2+(0{)}^2}=\sqrt{4+0}=2$ We have $\mathrm{AB}=\mathrm{BC}\ne \mathrm{AC}$. $\therefore △\mathrm{ABC}$ is an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think ABCD is a square?" Chameli disagrees. Using distance formula, find which of them is correct.

• Sol. Let the number of horizontal columns represent the x-coordinates whereas the vertical rows represent the $y$ coordinates. $\therefore$ The points are : $\mathrm{A}(3,4),\mathrm{B}(6,7),\mathrm{C}(9,4)$ and $D(6,1)$ $\therefore \mathrm{AB}=\sqrt{(6-3{)}^2+(7-4{)}^2}$ $=\sqrt{(3{)}^2+(3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$ $BC=\sqrt{(9-6{)}^2+(4-7{)}^2}$ $=\sqrt{3^2+(-3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$ $\mathrm{CD}=\sqrt{(6-9{)}^2+(1-4{)}^2}$ $=\sqrt{(-3{)}^2+(-3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$ $\mathrm{AD}=\sqrt{(6-3{)}^2+(1-4{)}^2}$ $=\sqrt{(3{)}^2+(-3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

Since, $AB=BC=CD=AD$ i.e., All the four sides are equal Also AC $=\sqrt{(9-3{)}^2+(4-4{)}^2}$ $=\sqrt{(+6{)}^2+(0{)}^2}=6$ and $\mathrm{BD}=\sqrt{(6-6{)}^2+(1-7{)}^2}=\sqrt{(0{)}^2+(-6{)}^2}=$ 6 i.e., $\mathrm{BD}=\mathrm{AC}$ $\Rightarrow$ Both the diagonals are also equal. $\therefore \mathrm{ABCD}$ is a square. Thus, Champa is correct as ABCD is square.

5. Name the quadrilateral formed, if any, by the following points, and give reasons for your answer. (i) $(-1,-2),(1,0),(-1,2),(-3,0)$ (ii) $(-3,5),(3,1),(0,3),(-1,-4)$ (iii) $(4,5),(7,6),(4,3),(1,2)$ Sol. (i) $\mathrm{A}(-1,-2),\mathrm{B}(1,0),\mathrm{C}(-1,2),\mathrm{D}(-3,0)$ Determine distances : AB, BC, CD, DA, AC and BD. $\mathrm{AB}=\sqrt{(1+1{)}^2+(0+2{)}^2}=\sqrt{4+4}$ $=\sqrt{8}=2\sqrt{2}$ $\mathrm{BC}=\sqrt{(-1-1{)}^2+(2-0{)}^2}=\sqrt{4+4}$ $=\sqrt{8}=2\sqrt{2}$ $\mathrm{CD}=\sqrt{(-3+1{)}^2+(0-2{)}^2}=\sqrt{4+4}$ $=\sqrt{8}=2\sqrt{2}$ DA $=\sqrt{(-1+3{)}^2+(-2-0{)}^2}=\sqrt{4+4}$ $=\sqrt{8}=2\sqrt{2}$ $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ The sides of the quadrilateral are equal ...(1) $\mathrm{AC}=\sqrt{(-1+1{)}^2+(2+2{)}^2}=\sqrt{0+16}=4$ $\mathrm{BD}=\sqrt{(-3-1{)}^2+(0-0{)}^2}=\sqrt{16+0}=4$ Diagonal AC $=$ Diagonal BD. From (1) and (2) we conclude that ABCD is a square. (ii) Let the points be $\mathrm{A}(-3,5),\mathrm{B}(3,1),\mathrm{C}(0,3)$ and $D(-1,-4)$. $\therefore \mathrm{AB}=\sqrt{[3-(-3){]}^2+(1-5{)}^2}$ $=\sqrt{6^2+(-4{)}^2}=\sqrt{36+16}$ $=\sqrt{52}=2\sqrt{13}$ $BC=\sqrt{(0-3{)}^2+(3-1{)}^2}=\sqrt{9+4}=\sqrt{13}$ $\mathrm{CD}=\sqrt{(-1-0{)}^2+(-4-3{)}^2}=\sqrt{(-1{)}^2+(-7{)}^2}$ $=\sqrt{1+49}=\sqrt{50}$ DA $=\sqrt{[-3-(-1){]}^2+[5-(-4){]}^2}$ $=\sqrt{(-2{)}^2+(9{)}^2}$ $=\sqrt{4+81}=\sqrt{85}$ $\mathrm{AC}=\sqrt{[0-(-3){]}^2+(3-5{)}^2}=\sqrt{(3{)}^2+(-2{)}^2}$ $=\sqrt{9+4}=\sqrt{13}$ $\mathrm{BD}=\sqrt{(-1-3{)}^2+(-4-1{)}^2}=\sqrt{(-4{)}^2+(-5{)}^2}$ $=\sqrt{16+25}=\sqrt{41}$ We see that $\sqrt{13}+\sqrt{13}=2\sqrt{13}$ i.e., $AC+BC=AB$ $\Rightarrow A,B$ and $C$ are collinear. Thus, $ABCD$ is not a quadrilateral. (iii) Let the points be $\mathrm{A}(4,5),\mathrm{B}(7,6),\mathrm{C}(4,3)$ and $D(1,2)$. $\therefore \mathrm{AB}=\sqrt{(7-4{)}^2+(6-5{)}^2}=\sqrt{3^2+1^2}=\sqrt{10}$ $BC=\sqrt{(4-7{)}^2+(3-6{)}^2}$ $=\sqrt{(-3{)}^2+(-3{)}^2}=\sqrt{18}$ $\mathrm{CD}=\sqrt{(1-4{)}^2+(2-3{)}^2}$ $=\sqrt{(-3{)}^2+(-1{)}^2}=\sqrt{10}$ DA $=\sqrt{(1-4{)}^2+(2-5{)}^2}=\sqrt{9+9}=\sqrt{18}$ $\mathrm{AC}=\sqrt{(4-4{)}^2+(3-5{)}^2}=\sqrt{0+(-2{)}^2}=2$ $\mathrm{BD}=\sqrt{(1-7{)}^2+(2-6{)}^2}=\sqrt{36+16}=\sqrt{52}$ Since, $\mathrm{AB}=\mathrm{CD},\mathrm{BC}=\mathrm{DA}$ [opposite sides of the quadrilateral are equal] And $\mathrm{AC}\ne \mathrm{BD}\Rightarrow$ Diagonals are unequal. $\therefore \mathrm{ABCD}$ is a parallelogram.
6. Find the point on the $x$-axis which is equidistant from $(2,-5)$ and $(-2,9)$. Sol. We know that any point on x -axis has its ordinate $=0$ Let the required point be $\mathrm{P}(\mathrm{x},0)$. Let the given points be $A(2,-5)$ and $B(-2,9)$ $\therefore AP=\sqrt{(x-2{)}^2+5^2}=\sqrt{x^2-4x+4+25}$ $=\sqrt{x^2-4x+29}$ $BP=\sqrt{[x-(-2){]}^2+(-9{)}^2}$ $=\sqrt{(x+2{)}^2+(-9{)}^2}$ $=\sqrt{x^2+4x+4+81}=\sqrt{x^2+4x+85}$ Since, $A$ and $B$ are equidistant from $P$, $\therefore \mathrm{AP}=\mathrm{BP}$ $\Rightarrow \sqrt{x^2-4x+29}=\sqrt{x^2+4x+85}$ $\Rightarrow {\mathrm{x}}^2-4\mathrm{x}+29={\mathrm{x}}^2+4\mathrm{x}+85$ $\Rightarrow {\mathrm{x}}^2-4\mathrm{x}-{\mathrm{x}}^2-4\mathrm{x}=85-29$ $\Rightarrow -8x=56\Rightarrow x=\frac{56}{-8}=-7$ $\therefore$ The required point is $(-7,0)$
7. Find the values of $y$ for which the distance between the points $P(2,-3)$ and $Q(10,y)$ is 10 units. Sol. Distance between $\mathrm{P}(2,-3)$ and $\mathrm{Q}(10,\mathrm{y})=$ 10 units $\Rightarrow \sqrt{(10-2{)}^2+(y+3{)}^2}=10$ $\Rightarrow 64+(y+3{)}^2=100$ $\Rightarrow (y+3{)}^2=36$ $\Rightarrow y^2+6y+9=36$ $\Rightarrow {\mathrm{y}}^2+6\mathrm{y}-27=0$ $\Rightarrow y^2+9y-3y-27=0$ $\Rightarrow \mathrm{y}(\mathrm{y}+9)-3(\mathrm{y}+9)=0$ $\Rightarrow (y+9)(y-3)=0$ $\Rightarrow y+9=0$ or $\mathrm{y}-3=0$ $\Rightarrow y=-9$ or 3 Hence, there can be two values of $y$ which are -9 and 3 .
8. If $\mathrm{Q}(0,1)$ is equidistant from $\mathrm{P}(5,-3)$ and $R(x,6)$, find the values of $x$. Also find the distances QR and PR. Sol. Here, $\mathrm{QP}=\sqrt{(5-0{)}^2+[(-3)-1{]}^2}=\sqrt{5^2+(-4{)}^2}$ $=\sqrt{25+16}=\sqrt{41}$ $\mathrm{QR}=\sqrt{(\mathrm{x}-0{)}^2+(6-1{)}^2}$ $=\sqrt{{\mathrm{x}}^2+5^2}=\sqrt{{\mathrm{x}}^2+25}$ $\mathrm{QP}=\mathrm{QR}$ $\sqrt{41}=\sqrt{{\mathrm{x}}^2+25}$ Squaring both sides, we have ${\mathrm{x}}^2+25=41$ $\Rightarrow {\mathrm{x}}^2+25-41=0$ $\Rightarrow x^2-16=0\Rightarrow x=\pm \sqrt{16}=\pm 4$ Thus, the point $R$ is $(4,6)$ or $(-4,6)$ Now, $\mathrm{QR}=\sqrt{[(\pm 4)-(0){]}^2+(6-1{)}^2}$ $=\sqrt{16+25}=\sqrt{41}$ and PR $=\sqrt{(\pm 4-5{)}^2+(6+3{)}^2}$ $\Rightarrow \mathrm{PR}=\sqrt{(-4-5{)}^2+(6+3{)}^2}$ or $\sqrt{(4-5{)}^2+(6+3{)}^2}$ $\Rightarrow \mathrm{PR}=\sqrt{(-9{)}^2+9^2}$ or $\sqrt{1+81}$ $\Rightarrow \mathrm{PR}=\sqrt{2\times 9^2}$ or $\sqrt{82}$ $\Rightarrow \mathrm{PR}=9\sqrt{2}$ or $\sqrt{82}$
9. Find a relation between $x$ and $y$ such that the point $(x,y)$ is equidistant from the point $(3,6)$ and $(-3,4)$. Sol. $\mathrm{A}(3,6)$ and $\mathrm{B}(-3,4)$ are the given points. Point $P(x,y)$ is equidistant from the points $A$ and $B$. $\Rightarrow \mathrm{PA}=\mathrm{PB}$ $\Rightarrow \sqrt{(x-3{)}^2+(y-6{)}^2}=\sqrt{(x+3{)}^2+(y-4{)}^2}$ $\Rightarrow (x-3{)}^2+(y-6{)}^2=(x+3{)}^2+(y-4{)}^2$ $\Rightarrow \left({\mathrm{x}}^2-6\mathrm{x}+9\right)+\left({\mathrm{y}}^2-12\mathrm{y}+36\right)$ $\Rightarrow \left({\mathrm{x}}^2+6\mathrm{x}+9\right)+\left({\mathrm{y}}^2-8\mathrm{y}+16\right)$ $\Rightarrow -6x-12y+45=6x-8y+25$ $\Rightarrow 12\mathrm{x}+4\mathrm{y}-20=0\Rightarrow 3\mathrm{x}+\mathrm{y}-5=0$

Exercise: 3.2

1. Find the co-ordinates of the point which divides the line joining of $(-1,7)$ and $(4,-$ $3)$ in the ratio $2:3$. Sol. Let the required point be $\mathrm{P}(\mathrm{x},\mathrm{y})$. Here the end points are $(-1,7)$ and $(4,-3)$ Ratio $=2:3=m_1:m_2$ $\mathrm{x}=\frac{{\mathrm{m}}_1{\mathrm{x}}_2+{\mathrm{m}}_2{\mathrm{x}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}=\frac{(2\times 4)+3(-1)}{2+3}$ $=\frac{8-3}{5}=\frac{5}{5}=1$ and $\mathrm{y}=\frac{{\mathrm{m}}_1{\mathrm{y}}_2+{\mathrm{m}}_2{\mathrm{y}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}$ $=\frac{2\times (-3)+(3\times 7)}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3$ Thus, the required point is $P(1,3)$.
2. Find the coordinates of the points of trisection of the line segment joining (4, $1)$ and $(-2,-3)$. Sol.

Points $P$ and $Q$ trisect the line segment joining the points $\mathrm{A}(4,-1)$ and $\mathrm{B}(-2,-3)$, i.e., $\mathrm{AP}=\mathrm{PQ}=\mathrm{QB}$. Here, P divides AB in the ratio $1:2$ and Q divides AB in the ratio $2:1$. x -coordinate of $\mathrm{P}=\frac{1\times (-2)+2\times (4)}{1+2}=\frac{6}{3}=2$; y -coordinate of $\mathrm{P}=\frac{1\times (-3)+2\times (-1)}{1+2}=\frac{-5}{3}$ Thus, the coordinates of P are $\left(2,\frac{-5}{3}\right)$. Now, x coordinate of $\mathrm{Q}=\frac{2\times (-2)+1\times (4)}{2+1}=0$; $y$-coordinate of $\mathrm{Q}=\frac{2\times (-3)+1\times (-1)}{2+1}=-\frac{7}{3}$ Thus, the coordinates of Q are $\left(0,-\frac{7}{3}\right)$. Hence, the points of trisection are $\mathrm{P}\left(2,\frac{-5}{3}\right)$ and $\mathrm{Q}\left(0,-\frac{7}{3}\right)$.

3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along $AD$, as shown in fig. Niharika runs $\frac{1}{4}$ th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$ th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? Sol. Let us consider 'A' as origin, then

$AB$ is the $x$-axis. $AD$ is the $y$-axis. Now, the position of green flag-post is $\left(2,\frac{100}{4}\right)$ or $(2,25)$ And, the position of red flag-post is $\left(8,\frac{100}{5}\right)$ or $(8,20)$ $\Rightarrow$ Distance between both the flags $=\sqrt{(8-2{)}^2+(20-25{)}^2}$ $=\sqrt{6^2+(-5{)}^2}=\sqrt{36+25}=\sqrt{61}\mathrm{m}$ Let the mid-point of the line segment joining the two flags be $\mathrm{M}(\mathrm{x},\mathrm{y})$.

$\therefore \mathrm{x}=\frac{2+8}{2}$ and $\mathrm{y}=\frac{25+20}{2}$ or $x=5$ and $y=22.5$ Thus, the blue flag is on the $5^{\text{th }}$ line at a distance 22.5 m above AB .

3. Find the ratio in which the line segment joining the points $(-3,10)$ and $(6,-8)$ is divided by $(-1,6)$. Sol. Let the required ratio be K : 1
   
   Comapring $x$ - coordinate Comparingy - coordinate $$\begin{gathered} \frac{\mathrm{k}(6)+1(-3)}{\mathrm{k}+1}=-1 \\ \frac{\mathrm{k}\times (-8)+1\times (10)}{\mathrm{k}+1}=6 \\ \Rightarrow 6\mathrm{k}-3=-\mathrm{k}-1 \\ \Rightarrow -8\mathrm{k}+10=6\mathrm{k}+6 \\ \Rightarrow 7\mathrm{k}=2 \\ \Rightarrow -8\mathrm{k}-6\mathrm{k}=6-10 \\ \Rightarrow \mathrm{k}=\frac{2}{7} \\ \Rightarrow -14\mathrm{k}=-4 \\ \Rightarrow \mathrm{k}=\frac{2}{7} \end{gathered}$$ $\therefore$ The required ratio is $2:7$.
4. Find the ratio in which the line segment joining $A(1,-5)$ and $B(-4,5)$ is divided by the x -axis. Also find the coordinates of the point of division. Sol. The given points are : $\mathrm{A}(1,-5)$ and $\mathrm{B}(-4$, 5). Let the required ratio $=\mathrm{k}:1$ and the required point be $\mathrm{P}(\mathrm{x},0)$. Since, the point P lies on x -axis. To find the ratio $\mathrm{x}=\frac{{\mathrm{m}}_1{\mathrm{x}}_2+{\mathrm{m}}_2{\mathrm{x}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}$ and $0=\frac{{\mathrm{m}}_1{\mathrm{y}}_2+{\mathrm{m}}_2{\mathrm{y}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}$ $\Rightarrow \mathrm{x}=\frac{-4\mathrm{k}+1}{\mathrm{k}+1}$ and $0=\frac{5\mathrm{k}-5}{\mathrm{k}+1}$ $\Rightarrow \mathrm{x}(\mathrm{k}+1)=-4\mathrm{k}+1$ and $5\mathrm{k}-5=0\Rightarrow \mathrm{k}=1$ $\Rightarrow \mathrm{x}(\mathrm{k}+1)=-4\mathrm{k}+1$ $\Rightarrow \mathrm{x}(1+1)=-4+1$ $[\because \mathrm{k}=1]$ $\Rightarrow 2\mathrm{x}=-3$ $\Rightarrow \mathrm{x}=-\frac{3}{2}$ $\therefore$ The required ratio $\mathrm{k}:1=1:1$ Coordinates of P are $(\mathrm{x},0)=\left(\frac{-3}{2},0\right)$
5. If $(1,2),(4,y),(x,6)$ and $(3,5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$. Sol. Mid-point of the diagonal AC has xcoordinate $=\frac{x+1}{2}$ and $y$-coordinate $=\frac{6+2}{2}=4$ i.e., $\left(\frac{\mathrm{x}+1}{2},4\right)$ is the mid-point of AC .

Similarly, mid-point of the diagonal BD is $\left(\frac{4+3}{2},\frac{y+5}{2}\right)$,i.e., $\left(\frac{7}{2},\frac{y+5}{2}\right)$ We know that the two diagonals AC and $BD$ bisect each other at $M$. Therefore, $\left(\frac{\mathrm{x}+1}{2},4\right)$ and $\left(\frac{7}{2},\frac{\mathrm{y}+5}{2}\right)$ coincide $\Rightarrow \frac{\mathrm{x}+1}{2}=\frac{7}{2}$ and $\frac{\mathrm{y}+5}{2}=4$ $\Rightarrow \mathrm{x}=6$ and $\mathrm{y}=3$

7. Find the coordinates of a point A, where $AB$ is the diameter of a circle whose centre is $(2,-3)$ and $B$ is $(1,4)$. Sol. Here, centre of the circle is $O(2,-3)$ Let the end points of the diameter be $A(x$, y) and $\mathrm{B}(1,4)$

The centre of a circle bisects the diameter. $\therefore 2=\frac{\mathrm{x}+1}{2}\Rightarrow \mathrm{x}+1=4$ or $\mathrm{x}=3$ And $-3=\frac{\mathrm{y}+4}{2}\Rightarrow \mathrm{y}+4=-6$ or $\mathrm{y}=-10$ Here, the coordinates of A are $(3,-10)$

If $A$ and $B$ are $(-2,-2)$ and $(2,-4)$, respectively, find the coordinates of P such that $AP=\frac{3}{7}AB$ and $P$ lies on the line segment AB. $AP=\frac{3}{7}AB$, $\mathrm{BP}=\mathrm{AB}-\mathrm{AP}=\mathrm{AB}-\frac{3}{7}\mathrm{AB}=\frac{4}{7}\mathrm{AB}$ $\frac{\mathrm{AP}}{\mathrm{BP}}=\frac{\frac{3}{7}\mathrm{AB}}{\frac{4}{7}\mathrm{AB}}=\frac{3}{4}$ Thus, P divides AB in the ratio $3:4$. x -coordinate of $\mathrm{P}=\frac{3\times (2)+4\times (-2)}{3+4}=-\frac{2}{7}$ $y$-coordinate of $\mathrm{P}=\frac{3\times (-4)+4\times (-2)}{3+4}=-\frac{20}{7}$ Hence, the coordiantes of P are $\left(-\frac{2}{7},-\frac{20}{7}\right)$.

8. Find the coordinates of the points which divide the line segment joining $\mathrm{A}(-2,2)$ and $B(2,8)$ into four equal parts. Sol. Here, the given points are $\mathrm{A}(-2,2)$ and $\mathrm{B}(2,8)$ Let $P_1,P_2$ and $P_3$ divide $AB$ in four equal parts.

$\because {\mathrm{AP}}_1={\mathrm{P}}_1{\mathrm{P}}_2={\mathrm{P}}_2{\mathrm{P}}_3={\mathrm{P}}_3\mathrm{B}$ Obviously, ${\mathrm{P}}_2$ is the mid-point of AB $\therefore$ Coordinates of ${\mathrm{P}}_2$ are $\left(\frac{-2+2}{2},\frac{2+8}{2}\right)$ or $(0,5)$ Again, ${\mathrm{P}}_1$ is the mid-point of ${\mathrm{AP}}_2$. $\therefore$ Coordinates of ${\mathrm{P}}_1$ are $\left(\frac{-2+0}{2},\frac{2+5}{2}\right)$ or $\left(-1,\frac{7}{2}\right)$ Also $P_3$ is the mid-point of $P_{2}B$. $\therefore$ Coordinates of ${\mathrm{P}}_3$ are $\left(\frac{0+2}{2},\frac{5+8}{2}\right)$ or $\left(1,\frac{13}{2}\right)$ Thus, the coordinates of $P_1,P_2$ and $P_3$ are $\left(-1,\frac{7}{2}\right),(0,5)$ and $\left(1,\frac{13}{2}\right)$ respectively.

• Find the area of a rhombus if its vertices are $(3,0),(4,5),(-1,4)$ and $(-2,-1)$ taken in order. Sol. As diagonals AC and BD bisect each other at right angle to each other at 0 .
  
  $\mathrm{AC}=\sqrt{(-1-3{)}^2+(4-0{)}^2}$ $=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$ $\mathrm{BD}=\sqrt{(4+2{)}^2+(5+1{)}^2}$ $=\sqrt{36+36}=6\sqrt{2}$ Then $\mathrm{OA}=\frac{1}{2}\mathrm{AC}=\frac{1}{2}\times 4\sqrt{2}=2\sqrt{2}$ $\mathrm{OB}=\frac{1}{2}\mathrm{BD}=\frac{1}{2}\times 6\sqrt{2}=3\sqrt{2}$ Area of $△\mathrm{AOB}=\frac{1}{2}(\mathrm{OA})\times (\mathrm{OB})=\frac{1}{2}\times 2\sqrt{2}$ $\times 3\sqrt{2}=6$ sq. units Hence, the area of the rhombus ABCD $=4\times$ area of $△AOB=4\times 6=24$ sq. units.

Exercise : 3.3

1. Find the area of the triangle whose vertices are : (i) $(2,3),(-1,0),(2,-4)$ (ii) $(-5,-1),(3,-5),(5,2)$ Sol. (i) Let the vertices of the triangles be $\mathrm{A}(2,3),\mathrm{B}(-1,0)$ and $\mathrm{C}(2,-4)$ Here ${\mathrm{x}}_1=2,{\mathrm{y}}_1=3$, ${\mathrm{x}}_2=-1,{\mathrm{y}}_2=0$ $x_3=2,y_3=-4$ Area of a $\Delta$ $=\frac{1}{2}\left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|$ Area of a $\Delta$ $=\frac{1}{2}|2\{0-(-4)\}+(-1)\{-4-(3)\}+2\{3-0\}|$ $=\frac{1}{2}|2(0+4)+(-1)(-4-3)+2(3)|$ $=\frac{1}{2}|8+7+6|=\frac{1}{2}|21|=\frac{21}{2}$ sq.units (ii) $\mathrm{A}(-5,-1),\mathrm{B}(3,-5),\mathrm{C}(5,2)$ are the vertices of the given triangle. ${\mathrm{x}}_1=-5,{\mathrm{x}}_2=3,{\mathrm{x}}_3=5;{\mathrm{y}}_1=-1,{\mathrm{y}}_2=-5,{\mathrm{y}}_3=2$. Area of the $△ABC$ $=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$ $=\frac{1}{2}|-5\times (-5-2)+3\times (2+1)+5\times (-1+5)|$ $=\frac{1}{2}|35+9+20|=\frac{1}{2}|64|=32$ sq. units
2. In each of the following find the value of ' $k$ ', for which the points are collinear. (i) $(7,-2),(5,1),(3,k)$ (ii) $(8,1),(\mathrm{k}-4),(2,-5)$. Sol. The given three points will be collinear if the area of triangle formed by them is equal to zero. (i) Let $\mathrm{A}(7,-2),\mathrm{B}(5,1)$ and $\mathrm{C}(3,\mathrm{k})$ be the vertices of a triangle. $\therefore$ The given points will be collinear, if ar $(△ABC)=0$ or $7(1-k)+5(k+2)+3(-2-1)=0$ $\Rightarrow 7-7\mathrm{k}+5\mathrm{k}+10+(-6)-3=0$ $\Rightarrow 17-9+5\mathrm{k}-7\mathrm{k}=0$ $\Rightarrow 8-2\mathrm{k}=0\Rightarrow 2\mathrm{k}=8$ $\Rightarrow \mathrm{k}=\frac{8}{2}=4$ The required value of $\mathrm{k}=4$. (ii) $\mathrm{A}(8,1),\mathrm{B}(\mathrm{k},-4),\mathrm{C}(2,-5)$ are the given points. ${\mathrm{x}}_1=8,{\mathrm{x}}_2=\mathrm{k},{\mathrm{x}}_3=2$ $y_1=1,y_2=-4,y_3=-5$ the condition for the three points to be collinear is ${\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)=0$ $8\times (-4+5)+\mathrm{k}\times (-5-1)+2\times (1+4)=0$ i.e. $8-6k+10=0$, i.e., $6k=18$, i.e., $k=3$
3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $(0,-1),(2,1)$ and $(0,3)$. Find the ratio of this area of the area of the given triangle. Sol. Let the vertices of the triangle be $\mathrm{A}(0,-1)$, $B(2,1)$ and $C(0,3)$. Let $D,E$ and $F$ be the mid-points of the sides $\mathrm{BC},\mathrm{CA}$ and AB respectively. Then : Coordinates of $D$ are $\left(\frac{2+0}{2},\frac{1+3}{2}\right)\text{ i.e., }\left(\frac{2}{2},\frac{4}{2}\right)\text{ or }(1,2)$ Coordinates of Eare $\left(\frac{0+0}{2},\frac{3+(-1)}{2}\right)$ i.e., $(0,1)$ Coordinates of F are $\left(\frac{2+0}{2},\frac{1+(-1)}{2}\right)$ i.e., $(1,0)$ Now, $\mathrm{ar}(△\mathrm{ABC})$

$$\begin{gathered} =\frac{1}{2}|0(1-3)+2\{3-(-1)\}+0(-1-1)| \\ =\frac{1}{2}|0(-2)+8+0(-2)| \end{gathered}$$ $=\frac{1}{2}|0+8+0|=\frac{1}{2}\times 8=4$ sq. units Now, ar ( $△\mathrm{DEF}$ ) $=\frac{1}{2}|1(1-0)+0(0-2)+1(2-1)|$ $=\frac{1}{2}|1(1)+0+1(1)|$ $=\frac{1}{2}|1+0+1|=\frac{1}{2}\times 2=1$ sq. unit $\therefore \frac{\mathrm{ar}(△\mathrm{DEF})}{\mathrm{ar}(△\mathrm{ABC})}=\frac{1}{4}$ $\therefore \mathrm{ar}(△\mathrm{DEF}):\mathrm{ar}(△\mathrm{ABC})=1:4$.

4. Find the area of the quadrilateral whose vertices taken in order are $(-4,-2),(-3,-$ 5), $(3,-2)$ and $(2,3)$. Sol. Join A and C. The given points are

$\mathrm{A}(-4,-2),\mathrm{B}(-3,-5),\mathrm{C}(3,-2)$ and $\mathrm{D}(2,3)$ Area of $△ABC$ $=\frac{1}{2}|(-4)(-5+2)-3(-2+2)+3(-2+5)|$ $=\frac{1}{2}|12+0+9|=\frac{21}{2}=10.5$ sq. units Area of $△ACD$ $=\frac{1}{2}|(-4)(-2-3)+3(3+2)+2(-2+2)|$ $=\frac{1}{2}|20+15|=\frac{35}{2}=17.5$ sq. units. Area of quadrilateral ABCD $=\mathrm{ar}(△\mathrm{ABC})+\mathrm{ar}(△\mathrm{ACD})$ $=(10.5+17.5)$ sq. units $=28$ sq . units