ML Aggarwal Coordinate Geometry Exe-19.4 Class 9 ICSE Maths Solutions

ML Aggarwal Coordinate Geometry Exe-19.4 Class 9 ICSE Maths APC Understanding Solutions. Solutions of  Exe-19.4. This post is the Solutions of  ML Aggarwal Chapter 19 – Coordinate Geometry for ICSE Maths Class-9.  APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-19 Coordinate Geometry for ICSE Board Class-9Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Coordinate Geometry Exe-19.4 Class 9 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 9th
Chapter-19 Coordinate Geometry
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Exe-19.4 Questions
Edition 2021-2022

Exe-19.4 Solutions of ML Aggarwal for ICSE Class-9 Ch-19, Coordinate Geometry

Note:- Before viewing Solutions of Chapter – 19 Coordinate Geometry Class-9 of ML AggarwaSolutions .  Read the Chapter Carefully. Then solve all example given in Exercise-19.1, Exercise-19.2, Exercise-19.3, Exercise-19.4, MCQs, Chapter Test.


Coordinate Geometry Exercise-19.4

ML Aggarwal Class 9 ICSE Maths Solutions

Page 465

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)

Answer :

(i) Let P(x1, y1) and Q(x2 , y2) be the

Co-ordinates of P = (2,3)

Co-ordinates of Q = (4,1)

Here x1 = 2, y1 = 3 , x2 = 4, y2 = 1

By distance formula d(P,Q) = √[(x2-x1)2+(y2-y1)2]

d(P,Q) = √[(4-2)2+(1-3)2]

= √[(2)2+(-2)2]

= √(4+4)

= √8

= √(4×2)

= 2√2

distance between P and Q is 2√2 units.

(ii) Let P(x1, y1) and Q(x2 , y2) be the given points

Co-ordinates of P = (0,0)

Co-ordinates of Q = (36,15)

Here x1 = 0, y1 = 0 , x2 = 36, y2 = 15

By distance formula d(P,Q) = √[(x2-x1)2+(y2-y1)2]

d(P,Q) = √[(36-0)2+(15-0)2]

= √[(36)2+(15)2]

= √(1296+225)

= √1521

= 39

distance between P and Q is 39 units.

(iii) Let P(x1, y1) and Q(x2 , y2) be the given points

Co-ordinates of P = (a,b)

Co-ordinates of Q = (-a,-b)

Here x1 = a, y1 = b , x2 = -a, y2 = -b

By distance formula d(P,Q) = √[(x2-x1)2+(y2-y1)2]

d(P,Q) = √[(-a-a)2+(-b-b)2]

= √[(-2a)2+(-2b)2]

= √(4a2+4b2)

= √4(a2+b2)

= 2√(a2+b2)

distance between P and Q is 2√(a2+b2) units.

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.

Answer :

A is a point on Y axis and ordinate is 4.

So the x-coordinate is 0.

coordinates of A are (0,4)

Given B is a point on X axis and abscissa is -3.

So the y-coordinate is 0.

coordinates of B are (-3,0)

By distance formula , Length of AB, d(AB) = √[(x2-x1)2+(y2-y1)2]

d(AB) = √[(-3-0)2+(0-4)2]

= √(-32+-42)

= √(9+16)

= √25

= 5

Hence,

the length of line segment AB is 5 units.


Coordinate Geometry Exercise-19.4

ML Aggarwal Class 9 ICSE Maths Solutions

Page 466

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

Answer :

Given distance between A(-3,-14) and B(a,-5) is 9 units.

By distance formula , Length of AB, d(AB) = √[(x2-x1)2+(y2-y1)2]

9 = √[(a-(-3))2+(-5-(-14))2]

⇒ 9 = √[(a+3)2+(-5+14)2)]

⇒ 9 = √[(a+3)2+92]

⇒ 9 = √[(a2+6a+9+81)]

⇒ 9 = √[(a2+6a+90)]

81 = a2+6a+90

⇒ a2+6a+90-81 = 0

⇒ a2+6a+9 = 0

⇒ (a+3)(a+3) = 0

⇒ a+3 = 0

⇒ a = -3

Hence,

the value of a is -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.

Answer :

(i) Given the point is on x axis. So y-coordinate is 0.

Let the points on X-axis be A(x,0) which is at a distance of 5 units from B(5,-4).

By distance formula, distance between AB = √[(x2-x1)2+(y2-y1)2]

5 = √[(5-x)2+(-4-0)2]

⇒ 5 = √[(5-x)2+-42]

⇒ 5 = √[(25+x2-10x+16)]

⇒ 5 = √[(x2-10x+41)]

25 = x2-10x+41

⇒ x2-10x+41-25 =0

⇒ x2-10x+16 =0

⇒ (x-2)(x-8) = 0

⇒ x-2 = 0 or x-8 = 0

⇒ x = 2 or x = 8

Hence

the points are (2,0) and (8,0).

(ii) Given the point is on Y axis. So x-coordinate is 0.

Let the points on Y-axis be A(0,y) which is at a distance of 10 units from B(8,8).

By distance formula , distance between AB = √[(x2-x1)2+(y2-y1)2]

10 = √[(8-0)2+(8-y)2]

⇒ 10= √[(8)2+(8-y2]

⇒ 10 = √[(64+64+y2-16y)]

⇒ 10 = √[(y2-16y+128)]

100 = y2-16y+128

⇒ y2-16y+128-100 =0

⇒ y2-16y+28 = 0

⇒ (y-14)(y-2) = 0

⇒ y-14 = 0 or y-2 = 0

⇒ y = 14 or y = 2

Hence the points are (0,14) and (0,2).

(iii) Let the abscissa of the point be x.

Then ordinate = 2x

So the coordinates of the point are (x,2x).

Since the point is at a distance of √10 from the point (4,3),

√[(4-x)2+(3-2x)2] = √10

(4-x)2+(3-2x)2 = 10

⇒ x2+16-8x+4x2 -12x+9-10 = 0

⇒ 5x2-20x+15 = 0

Divide by 5

x2-4x+3 = 0

⇒ (x-3)(x-1) = 0

⇒ x-3 = 0 or x-1 = 0

⇒ x = 3 or x = 1

So 2x = 2×3 = 6 or 2x = 2×1 = 2

Hence,

the points are (3,6) and (1,2).

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

Answer :

Let the point on X axis be (x,0) which is equidistant from (2,-5) and (-2,9).

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

√[(2-x)2+(-5-0)2] = √[(-2-x)2+(9-0)2[By distance formula]

⇒ √(4-4x+x2+25) = √(4+4x+x2+81)

⇒ √(x2-4x+29) = √(x2+4x+85)

x2-4x+29 = x2+4x+85

⇒ -4x-4x = 85-29

⇒ -8x = 56

⇒ x = 56/-8

⇒ x = -7

Hence,

the point is (-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.

Answer :

Coordinates of P are (6,-1).

Coordinates of Q are (1,3).

Coordinates of R are (x,8).

PQ = QR

By distance formula, √[(1-6)2+(3-(-1))2] = √[(x-1)2+(8-3)2]

⇒ √[(-5)2+(42] = √[(x-1)2+(5)2]

⇒ √[(25+16)] = √[x2-2x+1+25]

⇒ √(41) = √[x2-2x+26]

41= x2-2x+26

⇒ x2-2x+26-41 = 0

⇒ x2-2x+15 = 0

⇒ (x+3)(x-5) = 0

⇒ (x+3)= 0 or (x-5) = 0

x = -3 or x = 5

Hence,

the value of x is -3 or 5.

Question 7. If Q (0, 1) is equidistant from P (5, -3) and R (x, 6) find the values of x.

Answer :

Q(0,1) is equidistant from P(5,-3) and R(x,6).

So PQ = QR

By distance formula, √[(5-0)2+(-3-1))2] = √[(x-0)2+(6-1)2]

√[(5)2+(-42] = √[(x2+(5)2]

⇒ √[(25+16)] = √[x2+25]

⇒ √(41) = √[x2+25]

41 = x2+25

⇒ x2+25-41 = 0

⇒ x2-16= 0

⇒ (x-4)(x+4) = 0

⇒ (x-4) = 0 or (x+4) = 0

⇒ x = 4 or x = -4

Hence,

the value of x is 4 or -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).

Answer :

point (7,1) be Q and the point (3,5) be R.

Let P(x,y) be the point equidistant from Q(7,1) and R(3,5).

So PQ = PR

By distance formula, √[(7-x)2+(1-y))2] = √[(3-x)2+(5-y)2]

⇒ √[x2-14x+49+y2-2y+1] = √[x2-6x+9+y2-10y+25]

⇒ √[x2-14x+y2-2y+50] = √[x2-6x+y2-10y+34]

x2-14x+y2-2y+50 = x2-6x+y2-10y+34

⇒ -14x+6x-2y+10y+50-34 = 0

⇒ -8x+8y+16 = 0

Divide by 8

-x+y+2 = 0

⇒ y = x-2

Hence,

the required relation is y = x-2.

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.

Answer :

Let the y co-ordinate be x.

Then x coordinate is 2x.

So coordinates of P are (2x,x).

P is equidistant from the points Q (2, -5) and U (-3, 6).

PQ = PU

By distance formula, √[(2-2x)2+(-5-x))2] = √[(-3-2x)2+(6-x)2]

⇒ √[(4-8x+4x2+25+10x+x2] = √[9+12x+4x2+36-12x+x2]

⇒ √[29+2x+5x2] = √[45+5x2]

29+2x+5x2 = 45+5x2

⇒ 2x+29-45 = 0

⇒ 2x-16 = 0

⇒ 2x = 16

⇒ x = 16/2

⇒ x = 8

So 2x = 2×8 = 16

P(2x,x) = P(16,8)

Hence,

the coordinates of P are (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.

Answer :

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

AC = BC [Radii of same circle]

By distance formula, √[(2-4)2+(3-3)2] = √[(2-x)2+(3-5)2]

⇒ √[(-2)2+0] = √[4-4x+x2+(-2)2]

⇒ √4 = √[4-4x+x2+4]

⇒ √4 = √[8-4x+x2]

⇒ 4 = 8-4x+x2

⇒ x2-4x+4 = 0

⇒ (x-2)(x-2) = 0

⇒ x = 2

Hence,

the value of x is 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.

Answer :

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

AB = AC

By distance formula, √[(3-0)2+(p-2)2] = √[(p-0)2+(5-2)2]

⇒ √[(3)2+(p-2)2] = √[(p)2+(3)2]

⇒ √[9+p2-4p+4] = √[p2+9]

⇒ √[p2-4p+13] = √[p2+9]

p2-4p+13 = p2+9

⇒ -4p+13-9 = 0

⇒ -4p+4 = 0

⇒ -4p = -4

p = -4/-4 = 1

Hence,

the value of p is 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).

Answer :

Let C(3, 3) is the centre of the circle passing through the points P(6, 2), Q(0, 4) and R(4, 6).

CP = CQ = CR

By distance formula, CP = √[(x2-x1)2+(y2-y1)2]

⇒ CP = √[(6-3)2+(2-3)2]

⇒ CP = √[(3)2+(-1)2]

⇒ CP = √[9+1]

⇒ CP = √10

By distance formula, CQ = √[(x2-x1)2+(y2-y1)2]

⇒ CQ = √[(0-3)2+(4-3)2]

⇒ CQ = √[(3)2+(1)2]

⇒ CQ = √[9+1]

⇒ CQ = √10

By distance formula, CR = √[(x2-x1)2+(y2-y1)2]

⇒ CR = √[(4-3)2+(6-3)2]

⇒ CR = √[(1)2+(3)2]

⇒ CR = √[1+9]

⇒ CR = √10

Since CP = CQ = CR,

C(3,3) is the centre of the circle passing through the points P(6, 2), Q(0, 4) and R(4, 6).

Hence proved.

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.

Answer :

Centre of a circle is C(2α-1, 3α+1) and it passes through the point A (-3, -1).

Diameter of the circle = 20

radius = 20/2= 10

AC = 10 [radius]

AC = √[(x2-x1)2+(y2-y1)2]

⇒ 10 = √[(2α-1-(-3))2+(3α+1-(-1))2]

⇒ 10 = √[(2α-1+3)2+(3α+1+1)2]

⇒ 10 = √[(2α+2)2+(3α+2)2]

100 = [(2α+2)2+(3α+2)2]

⇒ 100 = 4α2+8α+4+9α2+12α+4

⇒ 100 = 13α2+20α+8

⇒ 13α2+20α+8-100 = 0

⇒ 13α2+20α-92 = 0

⇒ 13α2-26α+46 α -92 = 0

⇒ 13α(α-2)+46(α-2) = 0

⇒ (α-2)( 13α+46) = 0

⇒ α-2 = 0 or 13α+46 = 0

⇒ α = 2 or 13α = -46

⇒ α = 2 or α = -46/13

Hence,

the value is α = 2 or α = -46/13 .

Question 14. Using distance formula, show that the points A (3, 1), B (6, 4) and C (8, 6) are collinear.

Answer :

points are A (3, 1), B (6, 4) and C (8, 6).

If AB+BC = AC, then the three points are collinear.

AB = √[(x2-x1)2+(y2-y1)2]

⇒ AB = √[(6-3)2+(4-1)2]

⇒ AB = √[(3)2+(3)2]

⇒ AB = √[9+9]

⇒ AB = √18

⇒ AB = √(9×2)

⇒ AB = 3√2

BC = √[(x2-x1)2+(y2-y1)2]

⇒ BC = √[(8-6)2+(6-4)2]

⇒ BC = √[(2)2+(2)2]

⇒ BC= √[4+4]

⇒ BC = √8

⇒ BC = √(4×2)

⇒ BC = 2√2

AC = √[(x2-x1)2+(y2-y1)2]

⇒ AC = √[(8-3)2+(6-1)2]

⇒ AC = √[(5)2+(5)2]

⇒ AC= √[25+25]

⇒ AC = √50

⇒ AC = √(25×2)

⇒ AC = 5√2

AB+BC = 3√2+ 2√2 = 5√2 = AC

Hence proved.

So A, B, C are collinear.

Question 15. Check whether the points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

Answer :

Let A( 5, -2), B(6, 4) and C(7, -2) are the vertices of an isosceles triangle.

AB = √[(x2-x1)2+(y2-y1)2]

⇒ AB = √[(6-5)2+(4-(-2))2]

⇒ AB = √[(1)2+(4+2)2]

⇒ AB = √[1+62]

⇒ AB = √(1+36)

⇒ AB = √37

AC = √[(x2-x1)2+(y2-y1)2]

⇒ AC = √[(7-5)2+(-2-(-2))2]

⇒ AC = √[(2)2+(-2+2)2]

⇒ AC= √[4+0]

⇒ AC = √4

⇒ AC = 2

BC = √[(x2-x1)2+(y2-y1)2]

⇒ BC = √[(7-6)2+(-2-4)2]

⇒ BC = √[(1)2+(-6)2]

⇒ BC= √[1+36]

⇒ BC = √37

⇒ BC = √37

Here, AB = BC.

Hence,

ABC is an isosceles triangle.

Question 16. Name the type of triangle formed by the points A (-5, 6), B (-4, -2) and (7, 5).

Answer :

The three vertices of the triangle are A (-5, 6), B (-4, -2) and (7, 5).

AB = √[(x2-x1)2+(y2-y1)2]

⇒ AB = √[(-4-(-5))2+(4-(-2-6))2]

⇒ AB = √[(1)2+(-8)2]

⇒ AB = √[1+64]

⇒ AB = √65

AC = √[(x2-x1)2+(y2-y1)2]

⇒ AC = √[(7-(-5))2+(5-6)2]

⇒ AC = √[(12)2+(-1)2]

⇒ AC= √[144+1]

⇒ AC = √145

BC = √[(x2-x1)2+(y2-y1)2]

⇒ BC = √[(7-(-4))2+(5-(-2))2]

⇒ BC = √[(11)2+(7)2]

⇒ BC= √[121+49]

⇒ BC = √170

Length of all sides of the triangle are different.

ABC is a scalene triangle.

Question 17. Show that the points (1, 1), (- 1, – 1) and (-√3,√3) form an equilateral triangle.

Answer :

Let A(1,1), B(-1,-1) and C(-√3, √3) be the vertices of ABC.

AB = √[(x2-x1)2+(y2-y1)2]

⇒ AB = √[(-1-1)2+(-1-1)2]

⇒ AB = √[(-2)2+(-2)2]

⇒ AB = √[4+4]

⇒ AB = √8

BC = √[(x2-x1)2+(y2-y1)2]

⇒ BC = √[(-√3-(-1))2+(√3-(-1))2]

⇒ BC = √[(-√3+1)2+(√3+1)2]

⇒ BC = √[3-2√3+1+3+2√3+1]

⇒ BC = √8

AC = √[(x2-x1)2+(y2-y1)2]

⇒ AC = √[(-√3-1)2+(√3-1)2]

⇒ AC = √[3+2√3+1+3-2√3+1]

⇒ AC= √8

Here AB = BC = AC.

the points form an equilateral triangle.

Question 18. Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.

Answer :

Let A(7,10), B(-2,5) and C(3,-4) be the vertices of ABC.

AB = √[(x2-x1)2+(y2-y1)2]

⇒ AB = √[(-2-7)2+(5-10)2]

⇒ AB = √[(-9)2+(-5)2]

⇒ AB = √[81+25]

⇒ AB = √106

BC = √[(x2-x1)2+(y2-y1)2]

⇒ BC = √[(3-(-2))2+(-4-5)2]

⇒ BC = √[(5)2+(-9)2]

⇒ BC = √(25+81)

⇒ BC = √106

AC = √[(x2-x1)2+(y2-y1)2]

⇒ AC = √[(3-7)2+(-4-10)2]

⇒ AC = √[(-4)2+(-14)2]

⇒ AC= √(16+196)

⇒ AC√212

Here AB = BC.

ABC is an isosceles triangle.

AB2+BC2 = 106+106

AB2+BC2 = 212 = AC2

So, ABC is a right triangle.

Hence,

ABC 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.

Answer :

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

AB = √[(x2-x1)2+(y2-y1)2]

⇒ AB = √[(-2-0)2+(a-3)2]

⇒ AB = √(4+a2-6a+9)

⇒ AB = √( a2-6a+13)

BC = √[(x2-x1)2+(y2-y1)2]

⇒ BC = √[(-1-(-2))2+(4-a)2]

⇒ BC = √[(1)2+16-8a+a2]

⇒ BC = √(17-8a+a2)

⇒ BC = √(a2-8a+17)

AC = √[(x2-x1)2+(y2-y1)2]

⇒ AC = √[(-1-0)2+(4-3)2]

⇒ AC = √[(-1)2+(1)2]

⇒ AC= √(1+1)

⇒ AC= √2

BC= AC2 +AB2

⇒ a2-8a+17 = 2+a2-6a+13

⇒ -8a+6a+17-2-13 = 0

⇒ -2a+2 = 0

⇒ 2a = 2

⇒ a = 2/2 = 1

Hence,

the value of a is 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.

Answer :

Let the points A(0, – 1), B(- 2, 3), C(6, 7) and D(8, 3) be the vertices of a rectangle.

AB = √[(x2-x1)2+(y2-y1)2]

⇒ AB = √[(-2-0)2+(3-(-1))2]

⇒ AB = √(-2)2+(4)2

⇒ AB = √(4+16)

⇒ AB = √20

⇒ AB = √(4×5)

⇒ AB = 2√5

BC = √[(x2-x1)2+(y2-y1)2]

⇒ BC = √[(6-(-2))2+(7-3)2]

⇒ BC = √[(8)2+42]

⇒ BC = √(64+16)

⇒ BC = √(80)

⇒ BC = √(5×16)

⇒ BC = 4√5

CD = √[(x2-x1)2+(y2-y1)2]

⇒ CD = √[(8-6)2+(3-7)2]

⇒ CD = √[(2)2+(-4)2]

⇒ CD = √(4+16)

⇒ CD = √(20)

⇒ CD = √(4×5)

⇒ CD = 2√5

AD = √[(x2-x1)2+(y2-y1)2]

⇒ AD = √[(8-0)2+(3-(-1))2]

⇒ AD = √[(8)2+(4)2]

⇒ AD = √(64+16)

⇒ AD = √(80)

⇒ AD = √(5×16)

⇒ AD = 4√5

Here AB = CD and BC = AD.

Hence,

these are the vertices of a rectangle.

Area of ▭ABCD = AB × BC

= 2√5 × 4√5

= 40 sq. units.

Hence the area of ▭ABCD is 40 sq. units.


Coordinate Geometry Exercise-19.4

ML Aggarwal Class 9 ICSE Maths Solutions

Page 467

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.

Answer :

P (2, -1), Q (3, 4), R (-2, 3) and S (-3, -2) be four points in a plane.

PQ = √[(x2-x1)2+(y2-y1)2]

⇒ PQ = √[(3-2)2+(4-(-1))2]

⇒ PQ = √(1)2+(5)2

⇒ PQ = √(1+25)

⇒ PQ = √26

QR = √[(x2-x1)2+(y2-y1)2]

⇒ QR = √[(-2-3)2+(3-4)2]

⇒ QR = √[(-5)2+(-1)2]

⇒ QR = √[25+1]

⇒ QR = √26

RS = √[(x2-x1)2+(y2-y1)2]

⇒ RS = √[(-3-(-2))2+(-2-3)2]

⇒ RS = √[(-1)2+(-5)2]

⇒ RS = √[1+25]

⇒ RS = √26

PS = √[(x2-x1)2+(y2-y1)2]

⇒ PS = √[(-3-2)2+(-2-(-1))2]

⇒ PS = √[(-5)2+(-1)2]

⇒ PS = √[25+1]

⇒ PS = √26

Here PQ = QR = RS = PS.

So, it can be a rhombus or a square.

PR = √[(-2-2)2+(3-(-1))2[Distance formula]

⇒ PR = √[(-4))2+(4)2]

⇒ PR = √(16+16) = √32 = √(16×2) = 4√2

Diagonal, QS = √[(-3-3)2+(-2-4)2[Distance formula]

⇒ QS = √[(-6))2+(-6)2]

⇒ QS = √[36+36] = √(2×36) = 6√2

Here, diagonals are not equal. So PQRS is not a square. It is a rhombus.

Area of rhombus PQRS = ½ ×PR×QS

= ½ × 4√2×6√2

= 24 sq units.

Hence,

the area of the rhombus PQRS is 24 sq. units.

Question 22. Prove that the points A (2, 3), B {-2, 2), C (-1, -2) aqd D (3, -1) are the vertices of a square ABCD.

Answer :

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

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.

AB = √[(-2-2)2+(2-3)2]

= √[(-4)2+(-1)2]

= √[(16+1)]

= √17

BC = √[(-2-(-1))2+(2-(-2))2]

= √[(-1)2+(4)2]

= √[(1+16)]

= √17

CD = √[(3-(-1))2+(-1-(-2))2]

= √[(4)2+(1)2]

= √[(16+1)]

= √17

AD = √[(3-2)2+(-1-3)2]

= √[(1)2+(-4)2]

= √[(1+16)]

= √17

Here AB = BC = CD = AD.

All the sides are equal .

Diagonal AC = √[(-1-2)2+(-2-3)2]

= √[(-3)2+(-5)2]

= √[9+25]

= √34

Diagonal BD = √[(3-(-2))2+(-1-2)2]

√[(5)2+(-3)2]

= √[(25+9)]

= √34

AC = BD

So,

diagonals are also equal.

Hence,

the points are the vertices of 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)

Answer :

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)

(i) Let A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0)are the given points.

AB = √[(1-(-1))2+(0-(-2))2]

= √[(2)2+(2)2]

= √(4+4)

= √8

BC = √[(-1-1))2+(2-0)2]

= √[(-2)2+(2)2]

= √(4+4)

= √8

CD = √[(-3-(-1))2+(0-2)2]

= √[(-2)2+(-2)2]

= √(4+4)

= √8

AD = √[(-3-(-1))2+(0-(-2))2]

= √[(-2)2+(2)2]

= √(4+4)

= √8

Diagonal AC = √[(-1-(-1))2+(2-(-2))2]

= √[(0)2+(4)2]

= √[16]

= 4

Diagonal BD = √[(-3-1)2+(0-0)2]

√(-4)2+0

= √16

= 4

AC = BD

So, diagonals are also equal.

Also, AB = BC = CD = AD.

All the sides are equal .

Hence,

quadrilateral ABCD is a square.

(ii) Let A(4, 5), B(7, 6), C(4, 3), D(1, 2)are the given points.

AB = √[(7-4)2+(6-5)2]

= √[(3)2+(1)2]

= √(9+1)

= √10

BC = √[(4-7)2+(3-6)2]

= √[(-3)2+(-3)2]

= √(9+9)

= √18

CD = √[(1-4)2+(2-3)2]

= √[(-3)2+(-1)2]

= √(9+1)

= √10

AD = √[(1-4)2+(2-5)2]

= √[(-3)2+(-3)2]

= √(9+9)

= √18

Diagonal AC = √[(4-4)2+(3-5)2]

= √[(0)2+(-2)2]

= √4

= 2

Diagonal BD = √[(1-7)2+(2-6)2]

√(-6)2+-42

= √(36+16)

= √52

AC ≠ BD

AB = CD

BC = AD.

Since, opposite sides are equal and diagonals are not equal, ABCD 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.

Answer 24:

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

Let O(x,y) be the circumcentre of the circle.

Let A(8, 6), B(8, -2) and C(2, -2) be the vertices of the triangle.

OB = OC [Radii of same circle]

√[(8-x)2+(-2-y)2] = √[(2-x)2+(-2-y)2]

(8-x)2+(-2-y)2 = (2-x)2+(-2-y)2

⇒ 64+x2-16x+4+4y+y2 = 4-4x+x2+4+4y+y2

⇒ 64-16x = 4-4x

⇒ 12x = 60

⇒ x = 60/12 = 5

OA = OB

√[(8-x)2+(6-y)2] = √[(8-x)2+(-2-y)2]

(8-x)2+(6-y)2= (8-x)2+(-2-y)2

⇒ 36-12y+y2 = 4+4y+y2

⇒ -12y-4y = 4-36

⇒ -16y = -32

⇒ y = 32/16 = 2

Hence,

the coordinates of O are (5,2).

OA = √[(8-5)2+(6-2)2]

= √[(3)2+(4)2]

= √[9+16]

= √25 = 5

Hence the circumradius is 5 units.

—  : End of ML Aggarwal Coordinate Geometry Exe-19.4 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

Thanks

Please Share with Your Friends

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.