ML Aggarwal Solution Class 9 Chapter 19 Coordinate Geometry Exercise 19.4

 Exercise 19.4

Question 1

Find the distance between the following pairs of points :

(i) (2, 3), (4, 1)

(ii) (0, 0), (36, 15)

(iii) (a, b), (-a, -b)

Sol :

(i) Distance between (2,3) and (4,1)

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

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

$=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

(ii) Distance (0,0) and (36,15)

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

$=\sqrt{(36-0)^{2}+(15-0)^{2}}=\sqrt{36^{2}+15^{2}}$

$=\sqrt{1296+225}=\sqrt{1521}=39 \mathrm{~cm}$

(iii) Distance between (a,b) and (-a,-b)

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

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

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

$=\sqrt{4\left(a^{2}+b^{2}\right)}=2 \sqrt{a^{2}+b^{2}}$

Question 2

A is a point on y-axis whose ordinate is 4 and B is a point on x-axis whose abscissa is -3. Find the length of the line segment AB.

Sol :

∵A lies on y-axis

∴Abscissa=0, and ordinate=4

i.e. A(0,4)

∵B lies on x-axis

∴Ordinate=0 and abscissa=-3

i.e. B(-3,0)

∴AB$=\sqrt{(-3-0)^{2}+(0-4)^{2}}$

$=\sqrt{(-3)^{2}+(-4)^{2}}$

$=\sqrt{9+16}=\sqrt{25}$

=5 units

Question 3

Find the value of a, if the distance between the points A (-3, -14) and B (a, -5) is 9 units.

Sol :

Distance A(-3,-14) and B(a,-5)

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

$=\sqrt{(a+3)^{2}+(-5+14)^{2}}$

$=\sqrt{(a+3)^{2}+(9)^{2}}$

$=\sqrt{a^{2}+9+6 a+81}$

$\therefore \sqrt{\left(a^{2}+6 a+90\right)}=9$

Squaring both sides

$a^{2}+6 a+90=81$

$\Rightarrow a^{2}+6 a+90-81=0$

$\Rightarrow a^{2}+6 a+9=0$

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

$\therefore a+3=0 \Rightarrow a=-3$

Question 4

(i) Find points on the x-axis which are at a distance of 5 units from the point (5, -4).

(ii) Find points on the y-axis are at a distance of 10 units from the point (8, 8) ?

(iii) Find points (or points) which are at a distance of √10 from the point (4, 3) given that the ordinate of the point or points is twice the abscissa.

Sol :

(i) Let the points on x-axis be (x,0) then

Distance between (x,0) and (5,-4)=5 units

$\Rightarrow \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=5$

$\Rightarrow \quad \sqrt{(5-x)^{2}+(-4-0)^{2}}=5$

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

Squaring both sides

$(5-x)^{2}+16=25$

$\Rightarrow 25-10 x+x^{2}+16-25=0$

$\Rightarrow x^{2}-10 x+16=0$

$\Rightarrow x^{2}-2 x-8 x+16=0$

$\Rightarrow x(x-2)-8(x-2)=0 $

$\Rightarrow(x-2)(x-8)=0$

Either x-2=0, then x=2

or x-8=0, then x=8

∴The points are (2,0) and (8,0)

(ii) Let the co-ordinates of points or points be (x,y) which are at a distance of 10 units from the points (8,8)

$\therefore \sqrt{(8-x)^{2}+(8-y)^{2}}=10$

Squaring both sides

$(8-x)^{2}+(8-y)^{2}=100$

$\Rightarrow 64+x^{2}-16 x+64+y^{2}-16 y=100$

$x^{2}+y^{2}-16 x-16 y+128=100$

∵Points are on y-axis

∴x=0

Hence $(0)^{2}+y^{2}-16 \times 0-16 y+128=100$

$\Rightarrow y^{2}-16 y+128-100=0$

$\Rightarrow y^{2}-16 y+28=0$

$\Rightarrow y^{2}-14 y-2 y+28=0$

$\Rightarrow y(y-14)-2(y-14)=0$

$\Rightarrow(y-14)(y-2)=0$

Either y-14=0, then y=14

or y-2=0, then y=2

∴:Points will be (0,14) and (0,2)

(iii) Let the abscissa of point=x

the ordinate=2x

∵Points (x,2x) is at a distance of √10 from the point (4,3) then

$\sqrt{(x-4)^{2}+(2 x-3)^{2}}=\sqrt{10}$

Squaring both sides

$(x-4)^{2}+(2 x-3)^{2}=10$

$\Rightarrow x^{2}-8 x+16+4 x^{2}-12 x+9=10$

$\Rightarrow 5 x^{2}-20 x+25-10=0$

$\Rightarrow 5 x^{2}-20 x+15=0$

$\Rightarrow x^{2}-4 x+3=0$ (Dividing by 5)

$\Rightarrow x^{2}-x-3 x+3=0$

$\Rightarrow x(x-1)-3(x-1)=0$

$\Rightarrow(x-1)(x-3)=0$

Either x-1=0, then x=1

or x-3=0, then x=3

∴Points will be (1,2) and (3,6)

Question 5

Find the point on the x-axis which, is equidistant from the points (2, -5) and (-2, 9).

Sol :

Using distance formula 

Let the required point on x-axis be (x,0)

Then distance between (x,0) and (2,-5) is equal to the distance between (x,0) and (-2,9)

$\therefore \sqrt{(2-x)^{2}+(-5-0)^{2}}$

$=\sqrt{(-2-x)^{2}+(9-0)^{2}}$

Squaring both sides

$(2-x)^{2}+(-5)^{2}=(-2-x)^{2}+9^{2}$

$4-4 x+x^{2}+25=4+4 x+x^{2}+81$

-4 x+29=85+4x 

$\Rightarrow 4 x+4 x=-85+29$

$\Rightarrow 8 x=-56 \Rightarrow x=\frac{-56}{5}=-7$

$\therefore x=-7$

$\therefore$ Point $=(-7,0)$

Question 6

Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.

Sol :

Using distance formula

Points are P(6,-1) Q(1,3) and R(x, 8) and PQ=QR

∴$(1-6)^{2}+(3+1)^{2}=(x-1)^{2}+(8-3)^{2}$

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

$25+16=(x-1)^{2}+25$

$(x-1)^{2}=16 $

$\Rightarrow x^{2}-2 x+1=16^{\prime}$

$\Rightarrow x^{2}-2 x+1-16=0$

$\Rightarrow x^{2}-2 x-15=0$

$\Rightarrow x^{2}-5 x+3 x-15=0$

$\Rightarrow x(x-5)+3(x-5)=0$

$\Rightarrow(x-5)(x+3)=0$

Either x-5=0, then x=5

or x+3=0, then x=-3

∴x=5 ,-3

Question 7

If Q (0, 1) is equidistant from P (5, -3) and R (x, 6) find the values of x.

Sol :

$\because \mathrm{Q}(0,1)$ is equidistant from $\mathrm{P}(5,-3)$ and $\mathrm{R}(x, 6)$ find the value of x

$\therefore \mathrm{QP}=\mathrm{QR}$

$\Rightarrow(5-0)^{2}+(-3-1)^{2}=(x-0)^{2}+(6-1)^{2}$

$\Rightarrow(5)^{2}+(-4)^{2}=x^{2}+5^{2}$

$ 25+16=x^{2}+25 \Rightarrow x^{2}=16=(+4)^{2}$

$\therefore x=\pm 4 $

$\therefore x=4,-4$

Question 8

Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

Sol :

$\because$ Points (x, y) is equidistant from the points (7,1) and (3,5)

$=(x-7)^{2}+(y-1)^{2}=(x-3)^{2}+(y-5)^{2}$

$=x^{2}-14 x+49+y^{2}-2 y+1=x^{2}-6 x+9+ y^{2}-10 y+25$

$49+1-9-25=-6 x-10 y+14 x+2 y$

50-34=8x-8y 

$\Rightarrow 8 x-8 y=16$

$\Rightarrow x-y=2$ (Dividing by 8)

Question 9

The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from the points Q (2, -5) and U (-3, 6), then find the coordinates of P.

Sol :

$\because x$ -coordinates of a point P= twice its y coordinate 

Let coordinates of point P be (2x, x) 

$\because \mathrm{P}$ is equidistant from points Q(2,-5) and R(-3,6)

$\therefore \mathrm{PQ}=\mathrm{PR}$

Now, $(2 x-2)^{2}+(x+5)^{2}=(2 x+3)^{2}+(x-6)^{2}$

$\Rightarrow 4 x^{2}-8 x+4+x^{2}+10 x+25$

$=4 x^{2}+12 x+9+x^{2}-12 x+36$

$2 x+29=45$

$2 x=45-29=16$

$x=\frac{16}{2}=8$

$\therefore$ Coordinates of points $\mathrm{P}$ will be $(2 \times 8,8)$

i.e., $(16,8)$

Question 10

If the points A (4,3) and B (x, 5) are on a circle with centre C (2, 3), find the value of x.

Sol :

Points A(4,3) and B(x, 5) are on the circle whose centre C(2,3) 

$\therefore \mathrm{AC}=\mathrm{BC}$

(radii of the same circle)

$\Rightarrow(4-2)^{2}+(3-3)^{2}=(x-2)^{2}+(5-3)^{2}$

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

$4=(x-2)^{2}+4$

$\Rightarrow(x-2)^{2}=4-4=0$

$\therefore x-2=0 \Rightarrow x=2$

$\therefore x=2$

Question 11

If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p.

Sol :

Points A(0,2) is equidistant from B(3, p) and C(p, 5)

$\therefore \mathrm{AB}=\mathrm{AC}$

$(3-0)^{2}+(p-2)^{2}=(p-0)^{2}+(5-2)^{2}$

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

$9+p^{2}-4 p+4=p^{2}+9$

-4p+4=0 

$\Rightarrow 4 p=4 \Rightarrow p=\frac{4}{4}=1$

Question 12

Using distance formula, show that (3, 3) is the centre of the circle passing through the points (6, 2), (0, 4) and (4, 6).

Sol :

To show O(3,3) is the centre of a circle passing through the points A(6,2), B(0,4) and C(4,6)

$\therefore \mathrm{OA}=\mathrm{OB}=\mathrm{OC}$

Now OA $=\sqrt{(6-3)^{2}+(2-3)^{2}}$

$=\sqrt{3^{2}+(-1)^{2}}=\sqrt{9+1}=\sqrt{10}$

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

$=\sqrt{9+1}+\sqrt{10}$

and $\mathrm{OC}=\sqrt{(4-3)^{2}+(6-3)^{2}}=\sqrt{1^{2}+3^{2}}$

$=1+9=\sqrt{10}$

$\because \mathrm{OA}=\mathrm{OB}=\mathrm{OC}$

$\therefore \mathrm{O}$ is the centre of the circle passing through the points A ,B and C.

Question 13

The centre of a circle is C (2α – 1, 3α + 1) and it passes through the point A (-3, -1). If a diameter of the circle is of length 20 units, find the value(s) of α.

Sol :

Centre of a circle is C[(2 a-1),(3 a+1)] and it passes through the points A(-3,-1) and length of diameter =20 units

i.e., length of radius $=\frac{20}{2}=10$ units

⇒AC=10 

Now $\mathrm{AC}=\sqrt{(2 a-1+3)^{2}}$ $+\sqrt{(2 a-1+3)^{2}+(3 a+1+1)^{2}}$

$=\sqrt{(2 a+2)^{2}+(3 a+2)^{2}}$

$\therefore \sqrt{(2 a+2)^{2}+(3 a+2)^{2}}=10$

Squaring

$(2 a+2)^{2}+(3 a+2)^{2}=10^{2}$

$4 a^{2}+8 a+4+9 a^{2}+12 a+4=100$

$13 a^{2}+20 a+8-100=0$

$\Rightarrow 13 a^{2}+20 a-92=0$

$\Rightarrow 13 a^{2}-26 a+46 a-92=0$

$\Rightarrow 13 a(a-2)+46(a-2)=0$

$\Rightarrow(a-2)(13 a+46)=0$

Either a-2=0, then a=2

or 13a+46=0, then $13=-46 \Rightarrow a=\frac{-46}{13}$

Hence a=2 , $\frac{-46}{13}$

Question 14

Using distance formula, show that the points A (3, 1), B (6, 4) and C (8, 6) are coliinear.

Sol :

To show that the points A(3,1), B(6,4) and C(8,6) are collinear, if sum of any two lines is equal to the third line

Now, $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(6-3)^{2}+(4-1)^{2}}=\sqrt{3^{2}+3^{2}}$

$=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$\mathrm{BC}=\sqrt{(8-6)^{2}+(6-4)^{2}}=\sqrt{2^{2}+2^{2}}$

$=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

$A C=\sqrt{(8-3)^{2}+(6-1)^{2}}=\sqrt{5^{2}+5^{2}}$

$=\sqrt{25+25}=\sqrt{25 \times 2}=5 \sqrt{2}$

$\therefore \mathrm{AB}+\mathrm{BC}=3 \sqrt{2}+2 \sqrt{2}=5 \sqrt{2}=\mathrm{AC}$

$\therefore$ Points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.

Question 15

Check whether the points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

Sol :

To check that points A(5,-2), B(6,4), C(7,-2) are the vertices of an isosceles triangle ABC

Now ,$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(6-5)^{2}+(4+2)^{2}}=\sqrt{1^{2}+6^{2}}$

$=\sqrt{1+36}=\sqrt{37}$

$\mathrm{BC}=\sqrt{(7-6)^{2}+(-2-4)^{2}}$

$=\sqrt{1^{2}+(-6)^{2}}=\sqrt{1+36}=\sqrt{37}$

$A C=\sqrt{(7-5)^{2}+(-2+2)^{2}}$

$=\sqrt{(2)^{2}+0^{2}}=\sqrt{4}=2$

$\therefore$ Two sides AB=BC

$\therefore \triangle \mathrm{ABC}$ is an isosceles triangle

$\therefore$ Whose vertices are A, B, C

Question 16

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

Sol :

Three points of a triangle are

A(-5,6), B(-4,-2) and (7,5)

Now, $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(-4+5)^{2}+(-2-6)^{2}}=\sqrt{(1)^{2}+(-8)^{2}}$

$=\sqrt{1+64}=\sqrt{65}$

$\mathrm{BC}=\sqrt{(7+4)^{2}+(5+2)^{2}}=\sqrt{11^{2}+7^{2}}$

$=\sqrt{121+49}=\sqrt{170}$

$\mathrm{CA}=\sqrt{(7+5)^{2}+(5-6)^{2}}$

$=\sqrt{12^{2}+(-1)^{2}}=\sqrt{144+1}=\sqrt{145}$

$\because$ All the sides are different

$\therefore \Delta \mathrm{ABC}$ is a scalene.

Question 17

Show that the points (1, 1), (- 1, – 1) and (-√3,√3) form an equilateral triangle.

Sol :

Let the vertices of a $\Delta \mathrm{ABC}$ be A(1,1) , B(-1,-1) and C$(-\sqrt{3}, \sqrt{3})$

then $\mathrm{AB}=\sqrt{\left[1-(-1)^{2}\right]+[1-(-1)]^{2}}$

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

$=\sqrt{4+4}=\sqrt{8}=\sqrt{4 \times 2}=2 \sqrt{2}$ units

$\mathrm{BC}=\sqrt{[-\sqrt{3}-(-1)]^{2}+(\sqrt{3}-(-1))^{2}}$

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

$=\sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}}=\sqrt{8}$

$=\sqrt{4 \times 2}=2 \sqrt{2}$ units.

$A C=\sqrt{[-\sqrt{3}-1]^{2}+(\sqrt{3}-1)^{2}}$

$=\sqrt{3+1+2 \sqrt{3}+3+1-2 \sqrt{3}}$

$=\sqrt{8}=\sqrt{4 \times 2}=2 \sqrt{2}$ units

$\because \mathrm{AB}=\mathrm{BC}=\mathrm{AC}=2 \sqrt{2}$ units

$\therefore \Delta \mathrm{ABC}$ is an equilateral triangle.

Question 18

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

Sol :

Let points are A(7,10) , B(-2,5) and C(3,-4)

Now $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(-2-7)^{2}+(5-10)^{2}}=\sqrt{(-9)^{2}+(-5)^{2}}$

$=\sqrt{81+25}=\sqrt{106}$

Similarly, $\mathrm{BC}=\sqrt{(3+2)^{2}+(-4-5)^{2}}$

$=(5)^{2}+(-9)^{2}$

$=\sqrt{25+81}=\sqrt{106}$

and $A C=\sqrt{(3-7)^{2}+(-4-10)^{2}}$

$=\sqrt{(-4)^{2}+(-14)^{2}}=\sqrt{16+196}=\sqrt{212}$

We see that AB=BC$=\sqrt{106}$

$\therefore$ It is an isosceles triangle

and $\mathrm{AB}^{2}+\mathrm{BC}^{2}=(\sqrt{106})^{2}+(\sqrt{106})^{2}$

=106+106=212

and $\mathrm{AC}^{2}=(\sqrt{212})^{2}=212$

$\therefore \mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}$

$\therefore$ It is an isosceles right triangle

Question 19

The points A (0, 3), B (- 2, a) and C (- 1, 4) are the vertices of a right angled triangle at A, find the value of a.

Sol :

∵ A,B and C are the vertices of a right angled $\Delta \mathrm{ABC}$, right angle at A

$\therefore \mathrm{AB}^{2}=(-2-0)^{2}+(a-3)^{2}$

$=(-2)^{2}+(a-3)^{2}=4+(a-3)^{2}$

$A C^{2}=(-1-0)^{2}+(4-3)^{2}=(-1)^{2}+(1)^{2}$

=1+1+2

$\mathrm{BC}^{2}=(-1+2)^{2}+(4-a)^{2}$

$=(1)^{2}+(4-a)^{2}$

$\therefore \mathrm{AB}^{2}+\mathrm{AC}^{2}=\mathrm{BC}^{2}$  (By Pythagoras theorem)

$\Rightarrow 4+(a-3)^{2}+2=1+(4-a)^{2}$

$=4+a^{2}-6 a+9+2=1+16-8 a+a^{2}$

$\Rightarrow a^{2}-6 a+15=a^{2}-8 a+17$

$\Rightarrow 8 a-6 a=17-15$

$\Rightarrow 2 a=2 \Rightarrow a=1$

Question 20

Show that the points (0, – 1), (- 2, 3), (6, 7) and (8, 3), taken in order, are the vertices of a rectangle. Also find its area.

Sol :

Let A(0,-1), B(-2,3), C(6,7) and D(8,3) are the vertices of a quadrilateral ABCD

Now $\mathrm{AB}=\sqrt{(-2-0)^{2}+[3-(-1)]^{2}}$

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

$=\sqrt{20}=\sqrt{4 \times 5}=2 \sqrt{5}$ units

$\mathrm{BC}=\sqrt{[6-(-2)]^{2}+(7-3)^{2}}$

$=\sqrt{(6+2)^{2}+(4)^{2}}$

$=\sqrt{(8)^{2}+(4)^{2}}=\sqrt{64+16}$

$=\sqrt{80}=\sqrt{16 \times 5}=4 \sqrt{5}$ units

$C D=\sqrt{(8-6)^{2}+(3-7)^{2}}$

$=\sqrt{(2)^{2}+(-4)^{2}}=\sqrt{4+16}$

$=\sqrt{20}=\sqrt{4 \times 5}=2 \sqrt{5}$ units.

$\mathrm{AD}=\sqrt{(8-0)^{2}+[3-(-1)]^{2}}$

$=\sqrt{(8)^{2}+(3+1)^{2}}=\sqrt{64+16}=\sqrt{80}$

$=\sqrt{16 \times 5}=4 \sqrt{5}$ units

$\because \mathrm{AB}=\mathrm{CD}$ and $\mathrm{BC}=\mathrm{AD}$

$\therefore \mathrm{ABCD}$ is a rectangle.

Question 21

If P (2, -1), Q (3, 4), R (-2, 3) and S (-3, -2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.

Sol :

Four points are P(2,-1), Q(3,4), R(-2,3) and S(-3,-2) are the vertices of a quad.

Now, $\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(3-2)^{2}+(4+1)^{2}}=\sqrt{1^{2}+5^{2}}$

$=\sqrt{1+25}=\sqrt{26}$

$\mathrm{QR}=\sqrt{(-2-3)^{2}+(3-4)^{2}}$

$=\sqrt{(-5)^{2}+(1)^{2}}=\sqrt{25+1}=\sqrt{26}$

$\mathrm{RS}=\sqrt{(-3+2)^{2}+(-2-3)^{2}}$

$=\sqrt{(-1)^{2}+(-5)^{2}}=\sqrt{1+25}=\sqrt{26}$

and $\mathrm{SP}=\sqrt{(-3-2)^{2}+(-2+1)^{2}}$

$=\sqrt{(-5)^{2}+(-1)^{2}}=\sqrt{25+1}=\sqrt{26}$

$\because \mathrm{PQ}=\mathrm{QR}=\mathrm{RS}=\mathrm{SP}$

$\therefore$ PQRS is a square or a rhombus

Now, diagonal $\mathrm{PR}=\sqrt{(-2-2)^{2}+(3+1)^{2}}$

$=\sqrt{(-4)^{2}+(4)^{2}}=\sqrt{16+16}$

$=\sqrt{32}=4 \sqrt{2} \mathrm{~cm}$

and $\mathrm{QS}=\sqrt{(-3-3)^{2}+(-2-4)^{2}}$

$=\sqrt{(-6)^{2}+(-6)^{2}}=\sqrt{36+36}$

$=\sqrt{72}=6 \sqrt{2}$

$\therefore \mathrm{PQ}=\mathrm{QS}$

$\therefore$ PQRS is a rhombus not a square 

Now, area of rhombus PQRS

$=\frac{1}{2} \times \mathrm{PR} \times \mathrm{QS}$

$=\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}$

$=\frac{1}{2} \times 4 \times 6 \times 2=24$ sq. units

Question 22

Prove that the points A (2, 3), B {-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD.

Sol :

Points A(2,3), B(-2,2) , C(-1,-2) and D(3,-1)

$\therefore \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

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

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

$=\sqrt{16+1}=\sqrt{17}$

Similarly,

$\mathrm{BC}=\sqrt{(-1+2)^{2}+(-2-2)^{2}}$

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

$=\sqrt{1+16}=\sqrt{17}$

$\mathrm{CD}=\sqrt{(3+1)^{2}+(-1+2)^{2}}$

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

$=\sqrt{16+1}=\sqrt{17}$

and $\mathrm{DA}=\sqrt{(2-3)^{2}+(3+1)^{2}}$

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

$=\sqrt{1+16}=\sqrt{17}$

$A C=\sqrt{(-1-2)^{2}+(-2-3)^{2}}$

$=\sqrt{(-3)^{2}+(-5)^{2}}$

$=\sqrt{9+25}=\sqrt{34}$

$\mathrm{BD}=\sqrt{(3+2)^{2}+(-1-2)^{2}}$

$=\sqrt{(5)^{2}+(-3)^{2}}$

$=\sqrt{25+9}=\sqrt{34}$

$\because$ Sides AB, BC, CD and DA are equal and diagonals AC and BD are also equal

$\therefore \mathrm{ABCD}$ is a square

Question 23

Name the type of quadrilateral formed by the following points and give reasons for your answer :

(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)

(ii) (4, 5), (7, 6), (4, 3), (1, 2)

Sol :

(i) A(-1,-2) ,B(1,0) , C(-1,2) and D(-3,0)

Now $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$\mathrm{AB}^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$

$=(1+1)^{2}+(0+2)^{2}=(2)^{2}+(2)^{2}$

=4+4=8

Similarly,

$\mathrm{BC}^{2}=(-1-1)^{2}+(2-0)^{2}=(-2)^{2}+(2)^{2}$

=4+4=8

$C D^{2}=(-3+1)^{2}+(0-2)^{2}=(-2)^{2}+(-2)^{2}$

=4+4=8

$\mathrm{DA}^{2}=(-1+3)^{2}+(-2+0)^{2}=(2)^{2}+(-2)^{2}$

=4+4=8

Diagonal $\mathrm{AC}^{2}=(-1+1)^{2}+(2+2)^{2}$

$=(0)^{2}+(4)^{2}$

=0+16=16

and $\mathrm{BD}^{2}=(-3-1)^{2}+(0-0)^{2}=(-4)^{2}+0=16$

$\because$ The sides are equal and diagonal are also equal

$\therefore$ The quadrilateral ABCD is a square

(ii) Points are A(4,5), B(7,6), C(4,3), D(1,2)

Now $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$\mathrm{AB}^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$

$=(7-4)^{2}+(6-5)^{2}$

$=(3)^{2}+(1)^{2}$

=9+1=10

Similarly $\mathrm{BC}^{2}=(4-7)^{2}+(3-6)^{2}$

$=(3)^{2}+(-3)^{2}=9+9=18$

$C D^{2}=(1-4)^{2}+(2-3)^{2}$

$=(-3)^{2}+(-1)^{2}$

=9+1=10

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

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

=9+9=18

∴Here AB=CD and BC=DA

Diagonal $\mathrm{AC}^{2}=(4-4)^{2}+(3-5)^{2}$

$=(0)^{2}+(-2)^{2}$

=0+4=4

and $B D^{2}=(1-7)^{2}+(2-6)^{2}$

$=(-6)^{2}+(-4)^{2}$

=36+16=52

$\because$ Opposite sides are equal and diagonals are not equal

$\therefore$ It is a parallelogram

Question 24

Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, -2) and (2, -2). Also, find its circumradius.

Sol :

O is the circumference of $\triangle \mathrm{ABC}$ 

Whose vertices are A(8,6), B(8,-2), C(2,-2)

Let coordinates of centre O(x,y)

$\because \mathrm{OA}=\mathrm{OB}=\mathrm{OC} \Rightarrow \mathrm{OA}^{2}=\mathrm{OB}^{2}=\mathrm{OC}^{2}$

$\therefore \mathrm{OA}^{2}=\mathrm{OB}^{2}$

$\Rightarrow(x-8)^{2}+(y-6)^{2}=(x-8)^{2}+(y+2)^{2}$

$x^{2}-16 x+64+y^{2}-12 y+36$

$=x^{2}-16 x+64+y^{2}+4 y+4$

$36-4=4 y+12 y \Rightarrow 16 y=32$

$\Rightarrow y=\frac{32}{16}=2$

Similarly $\mathrm{OB}^{2}=\mathrm{OC}^{2}$

$(x-8)^{2}+(y+2)^{2}=(x-2)^{2}+(y+2)^{2}$

$x^{2}-16 x+64=x^{2}-4 x+4$

16x-4x=64-4

$\Rightarrow 12 x=60 \Rightarrow x=\frac{60}{12}=5$

$\therefore$ Coordinates of $\mathrm{O}$ are $(5,2)$

$\therefore \mathrm{OA}=\sqrt{(x-8)^{2}+(y-6)^{2}}$

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

$=\sqrt{9+16}=\sqrt{25}=5$ units

Question 25

If two opposite vertices of a square are (3, 4) and (1, -1), find the coordinates of the other two vertices.

Sol :

Length of hypotenuse of square =√2 ×side of square

∴AC=√2 AB

⇒$\mathrm{AC}^{2}=2 \mathrm{AB}^{2}$  [squaring both sides]

⇒$(3-1)^{2}+[4-(-1)]^{2}=2\left[(x-3)^{2}+(y-4)^{2}\right]$

⇒$29=2\left[(x-3)^{2}+\left(\frac{23-4 x}{10}-4\right)^{2}\right]$  [From (i)]

$\Rightarrow 29=2\left[(x-3)^{2}+\frac{(-4 x-17)^{2}}{100}\right]$

$\Rightarrow 2900=2\left[100\left(x^{2}-6 x+9\right)+\left(16 x^{2}+289+\right.\right.136 x)]$

$\Rightarrow 116 x^{2}-464 x+1189=1450$

$\Rightarrow 116 x^{2}-464 x-261=0$

$\Rightarrow x=\frac{464 \pm \sqrt{(-464)^{2}-4 \times 116 \times(-261)}}{2 \times 116}$ $\left(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right)$

$\Rightarrow x=\frac{464 \pm \sqrt{336400}}{2 \times 116} \Rightarrow x=\frac{464 \pm 580}{232}$

$\Rightarrow x=\frac{464+580}{232}$ or $x=\frac{464-580}{232}$

$x=\frac{1044}{232}=\frac{9}{2}$ or $x=\frac{-116}{232}=\frac{-1}{2}$

When $x=\frac{9}{2}$

$y=\frac{23-4 \times\left(\frac{-1}{2}\right)}{10}=\frac{25}{10}=\frac{5}{2}$

Thus , the coordinates of the remaining vertices of square are $\left(\frac{9}{2}, \frac{1}{2}\right)$ and $\left(\frac{-1}{2}, \frac{5}{2}\right)$