# Class 10 RD Sharma Solutions – Chapter 14 Coordinate Geometry – Exercise 14.2 | Set 3

### Question 40. Prove that the points (3, 0), (4, 5), (-1, 4), and (-2, -1) taken in order, form a rhombus. Also, find its area.

**Solution:**

Let us considered the given points are A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1)

Now we find the length of the sides and diagonals,

By using distance formula

So, AB =

AB

^{2 }= (4 + 3)^{2 }+ (5 – 0)^{2 }= (1)

^{2 }+ (5)^{2}= 1 + 25 = 26

Similarly, BC

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

^{2 }+ (-1)^{2 }= 25 + 1 = 26CD

^{2 }= (-2 + 1)^{2 }+ (-1 – 4)^{2}= (-1)

^{2 }+ (-5)^{2 }= 1 + 25 = 26and DA

^{2 }= (3 + 2)^{2 }+ (0 + 1)^{2 }= (5)

^{2 }+ (1)^{2 }= 25 + 1 = 26Diagonal AC

^{2 }= (-1 – 3)^{2 }+ (4 – 0)^{2}= (-4)

^{2 }+ (4)^{2 }= 16 + 16 = 32and BD

^{2 }= (-2 – 4)^{2 }+ (-1 – 5)^{2}= (-6)

^{2 }+ (-6)^{2 }= 36 + 36 = 72So, we conclude that the sides AB = BC = CD = DA and diagonal AC is not equal to BD

Hence, ABCD is a rhombus

Now we find the area of rhombus ABCD = Product of diagonals/2

= (√32 × √72)/2

= (√16 × 2 × 2 × 36)/2

= 4 × 2 × 6/2

= 24 sq. units

### Question 41. In the seating arrangement of desks in a classroom three students Rohini, Sandhya, and Bina are seated at A (3, 1), B (6, 4), and C (8, 6). Do you think they are seated in a line?

**Solution:**

Given that A (3, 1), B (6, 4) and C (8, 6)

Now we find the length of the sides and diagonals,

By using distance formula

AB =

AB

^{2 }= (6 – 3)^{2 }+ (4 – 1)^{2}= (3)

^{2 }+ (3)^{2 }= 9 + 9 = 18Similarly, BC

^{2 }= (8 – 6)^{2 }+ (6 – 4)^{2}= (2)

^{2 }+ (2)^{2 }= 4 + 4 = 8and BC

^{2 }= (8 – 6)^{2 }+ (6 – 4)^{2}= (2)

^{2 }+ (2)^{2 }= 4 + 4 = 8and CA

^{2 }= (3 – 8)^{2 }+ (1 – 6)^{2}= (-5)

^{2 }+ (-5)^{2 }= 25 + 25 = 50AB = √18 = √9 * 2 = 3√2

BC = √8 = √4 * 2 = 2√2

and CA = √50 = √25 * 2 = 5√2

AB + BC = 3√2 + 2√2 = 5√2 = CA

Hence, A, B and C are collinear points. Hence, they are seated in a line.

### Question 42. Find a point on y-axis which is equidistant from the points (5, -2) and (-3, 2).

**Solution:**

Let us assume P be the point lies on y-axis. So, its x = 0, so the coordinates of P is (0, y)

It is given that the point P (0, y) is equidistant from the points A(5, -2) and B(-3, 2).

So, PA = PB

Also, PA

^{2 }= PB^{2}Now by using distance formula, we get

(5 – 0)

^{2 }+ (-2 – y)^{2 }= (-3 – 0)^{2 }+ (2 – y)^{2}25 + 4 + y

^{2 }+ 4y = 9 + 4 – 4y + y^{2}y

^{2 }+ 4y + 4y – y^{2 }= 13 – 298y = -16

y = -16/8 = 2

Hence, the required point P is (0,-2)

### Question 43. Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).

**Solution:**

Let us considered P (x, y) is equidistant from A(3, 6) and B(-3, 4)

So, PA = PB

Also, PA

^{2 }= PB^{2}Now by using distance formula, we get

On squaring both sides, we get

(x – 3)

^{2 }+ (y – 6)^{2 }= (x + 3)^{2 }+ (y – 4)^{2}x

^{2 }– 6x + 9 + y^{2 }– 12y + 36 = x^{2 }+ 6x + 9 = y^{2 }– 8y + 16-6x – 12y + 45 = 6x – 8y + 25

-6x – 6x – 12y + 8y + 45 – 25 = 0

-12 – 4y + 20 = 0

3x + y – 5 = 0

3x + y = 5

### Question 44. If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p.

**Solution:**

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

Now by using distance formula, we get

It is given that AB = AC

√p

^{2 }– 4p + 13 = √p^{2 }+ 9So, on squaring both side, we get

= p

^{2 }– 4p + 13 = p^{2 }+ 9p

^{2 }– 4p – p^{2 }= 9 – 13-4p = -4

p = 1

Hence, the value of p is 1

### Question 45. Prove that the points (7, 10), (-2, 5), and (3, -4) are the vertices of an isosceles right triangle.

**Solution:**

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

Now we find the length of the sides

By using distance formula

Now AB =

Similarly, BC =

and AC =

So, we conclude that AB = BC = √106 and AB

^{2 }+ BC^{2 }= AC^{2}Hence, ABC is an isosceles right triangle

### Question 46. If the point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8), find the value of x and the distance AP.

**Solution:**

It is given that Point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8)

So, PA = PB

On squaring both sides, we get

(x – 7)

^{2 }+ (4)^{2 }= (x – 6)^{2 }+ (-5)^{2}x

^{2 }– 14x + 49 + 16 = x^{2 }– 12x + 36 + 25x

^{2 }– 14x + 65 = x^{2 }– 12x + 61x

^{2 }– 14x + 12x – x^{2 }= 61 – 65-2x = -4

x = -4/-2 = 2

x = 2

Now we find the distance

### Question 47. If A (3, y) is equidistant from points P (8, -3) and Q (7, 6), find the value of y and find the distance AQ.

**Solution:**

It is given that point A (3, y) is equidistant from P (8, -3) and Q (7, 6)

So, AP = AQ

On squaring both sides, we get

(3 – 8)

^{2 }+ (y + 3)^{2 }= (-4)^{2 }+ (y – 6)^{2}(-5)

^{2 }+ y^{2 }+ 6y + 9 = 16 + y^{2 }– 12y + 3625 + y

^{2 }+ 6y + 9 = 16 + y^{2 }– 12y + 36y

^{2 }+ 6y – y^{2 }+ 12y = 36 – 9 – 25 + 1618y = 18

y = 18/18 = 1

y = 1

Now we find the distance

### Question 48. If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

**Solution:**

Given that the A (0, -3) and B (0, 3) are the two vertices of an equilateral triangle ABC

Let us assume that the coordinates of the third vertex be C (x, y)

In equilateral triangle, AC = AB

So,

(x – 0)

^{2 }+ (y + 3)^{2 }= (0 – 0)^{2 }+ (3 + 3)^{2}x

^{2 }+ (y + 3)^{2 }= 0 + (6)^{2 }= 36x

^{2 }+ y^{2 }+ 6y + 9 = 36x

^{2 }+ y^{2 }+ 6y = 36 – 9 = 27 …….(i)Also, BC = AB

(x – 0)

^{2 }+ (y – 3)^{2 }= 36x

^{2 }+ y^{2 }+ 9 – 6y = 36x

^{2 }+ y^{2 }– 6y = 36 – 9 = 27 ……..(ii)So, from eq (i) and (ii), we get

x

^{2 }+ y^{2 }+ 6y = x^{2 }+ y^{2 }– 6yx

^{2 }+ y^{2 }+ 6y – x^{2 }– y^{2 }+ 6y = 012y = 0

y = 0

Now put the value of y in eq(i)

x

^{2 }+ y^{2 }+ 6y = 27x

^{2 }+ 0 + 0 = 27x = ±√27 = ±3√3

So, the coordinates of third point is(3√3, 0) or (-3√3, 0)

### Question 49. If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3), find k. Also, find the length of AP.

**Solution:**

Given that the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3)

So, AP = BP

(2 + 2)

^{2 }+ (2 – k)^{2 }= (2 + 2k)^{2 }+ (2 + 3)^{2}(4)

^{2 }+ (2 – k)^{2 }= (2 + 2k)^{2 }+ (5)^{2}16 + 4 + k

^{2 }– 4k = 4 + 4k^{2 }+ 8k + 254k

^{2 }+ 8k + 29 – 16 – 4 – k^{2 }+ 4k = 03k

^{2 }+ 12k + 9 = 0k

^{2 }+ 4k + 3 = 0k

^{2 }+ k + 3k + 3 = 0k(k + 1) + 3(k + 1) = 0

(k + 1)(k + 3) = 0

So, the value of k either k + 1 = 0, then k = -1

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

Therefore, k = -1, -3

Now we find the distance

### Question 50. Show that ∆ABC, where A (-2, 0), B (2, 0), C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.

**Solution:**

Given that In ∆ABC, the vertices are A (-2, 0), B (2, 0), C (0, 2)

In ∆PQR, the vertices are P (-4, 0), Q (4, 0), R (0, 4)

Show that ∆ABC ~ ∆PQR

So,

Now,

So, AB/PQ = 4/8 = 1/2

BC/QR = 2√2/4√2 = 1/2

CA/PQ = 2√2/4√2 = 1/2

So, AB/PQ = BC/QR = CA/RP

By using SSS

∆ABC ~ ∆PQR

### Question 51. An equilateral triangle has two vertices at the points (3, 4), and (-2, 3). Find the coordinates of the third vertex.

**Solution:**

Given that the A (3, 4) and B (-2, 3) are the two vertices of an equilateral triangle ABC

Let us assume that the coordinates of the third vertex be C (x, y)

Now

As we know that in equilateral triangle, AB = BC = CA

So, BC = AB

On squaring both side we get

(x + 2)

^{2 }+ (y – 3)^{2 }= 26x

^{2 }+ 4x + 4 + y^{2 }– 6y + 9 = 26x

^{2 }+ y^{2 }+ 4x – 6y + 13 = 26x

^{2 }+ y^{2 }+ 4x – 6y = 26 – 13 = 12 ………..(i)Similarly, CA = AB

On squaring both side we get

(3 – x)

^{2 }+ (4 – y)^{2 }= 269 + x

^{2 }– 6x + 16 + y^{2 }– 8y = 26x

^{2 }+ y^{2 }– 6x – 8y + 25 = 26x

^{2 }+ y^{2 }– 6x – 8y = 26 – 25 = 1 …….(ii)Now on subtracting eq(ii) from (i), we get

10x + 2y = 12

5x + y = 6 ……..(iii)

y = 6 – 5x

Now substituting the value of y in eq(i), we get

x

^{2 }+ (6 – 5x)^{2 }+ 4x – 6(6 – 5x) = 13x

^{2 }+ 36 + 25x^{2 }– 60x + 4x – 36 + 30x – 13 = 026x

^{2 }– 26x – 13 = 02x

^{2 }– 2x – 1 = 0Here, a = 2, b = -2, c = -1

When x = (1 + √3)/2, then y = 6 – 5x = 6 – 5((1 + √3)/2) = (7 – 5√3)/2

Or when x = (1 – √3)/2, then y = 6 – 5x = 6 – 5((1 – √3)/2) = (7 + 5√3)/2

Hence, the co-ordinates of the point C is ((1 + √3)/2, (7 – 5√3)/2) or ((1 – √3)/2, (7 + 5√3)/2)

### Question 52. Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).

**Solution:**

Given that the vertices of ∆ABC are A(-2, -3), B(-1, 0), and C(7, -6), and

let us assume that O is the circumcenter ∆ABC. So, the coordinates of O will be (x, y)

So, OA = OB = OC

Or OA

^{2 }= OB^{2 }= OC^{2}Now

OA

^{2 }= (x + 2)^{2 }+ (y + 3)^{2}= x

^{2 }+ 4x + 4 + y^{2 }+ 6y + 9= x

^{2 }+ y^{2 }+ 4x + 6y + 13OB

^{2 }= (x + 1)^{2}+ (y + 0)^{2}= x

^{2 }+ 2x + 1 + y^{2}= x

^{2 }+ y^{2 }+ 2x + 1OC

^{2 }= (x – 7)^{2 }+ (y + 6)^{2}= x

^{2 }– 14x + 49 + y^{2 }+ 12y + 36= x

^{2 }+ y^{2 }– 14x + 12y + 85OA

^{2 }= OB^{2}x

^{2 }+ y^{2 }+ 4x + 6y + 13 = x^{2 }+ y^{2 }+ 2x + 14x + 6y -2y = 1 – 13

2x + 6y = -12

x + 3y = -6 ………(i)

OB

^{2 }= OC^{2}x

^{2 }+ y^{2 }+ 2x + 1 = x^{2 }+ y^{2 }– 14x + 12y + 852x + 14x – 2y = 85 – 1

16x – 12y = 84

4x – 3y = 21 ………(ii)

From eq (i), we get

x = -3y – 6

On substituting the value of x in eq (ii)

4(-3y – 6) – 3y = 21

-12 – 24 – 3y = 21

-15y = 21 + 24

-15y = 45

y = -45/15 = -3

x = -3y – 6 = -3 × (-3) – 6

= + 9 – 6 = 3

Hence, the co-ordinates of O are (3,-3)

### Question 53. Find the angle subtended at the origin by the line segment whose endpoints are (0, 100) and (10, 0).

**Solution:**

Let us considered the co-ordinates of the end points of a line segment are

A (0, 100), B (10, 0) and origin is O (0, 0)

So, the angle subtended by the line PQ at the origin is 90° or π/2

### Question 54. Find the centre of the circle passing through (5, -8), (2, -9), and (2, 1).

**Solution:**

Given that O is the centre of the circle and A(5, -8), B(2, -9), and C (2, 1) are the points on the circle.

So, let us considered the co-ordinates of O be (x, y)

Therefore, OA = OB = OC

OA

^{2 }= OB^{2 }= OC^{2}Now

OA

^{2 }= (x – 5)^{2 }+ (y + 8)^{2}= x

^{2 }– 10x + 25 + y^{2 }+ 16y + 64= x

^{2 }+ y^{2 }– 10x + 16y + 89Similarly, OB

^{2 }= (x – 5)^{2 }+ (y + 9)^{2}= x

^{2 }+ 4 – 4x + y^{2 }+ 81 + 18y= x

^{2 }+ y^{2 }– 4x + 18y + 85and OC

^{2 }= x^{2 }– 4x + 4 + y^{2 }– 2y + 1= x

^{2 }+ y^{2 }– 4x – 2y + 5OA

^{2 }= OB^{2}x

^{2 }+ y^{2 }– 10x + 16y + 89 = x^{2 }+ y^{2 }– 4x + 18y + 85-10x + 4x + 16y – 18y = 85 – 89

-6x – 2y = -4

3x + y = 2 …….(i)

OB

^{2 }= OC^{2}x

^{2 }+ y^{2 }– 4x + 18y + 85 = x^{2 }+ y^{2 }– 4x – 2y + 518y + 2y = 5 – 85

20y = -80

y = -80/10 = -4

Now substitute the value of y in eq(i), we get

3x + y = 2

3x – 4 = 2

3x = 2 + 4 = 6

x = 6/3 = 2

Hence, the co-ordinates of O are(2,-4)

### Question 55. If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices.

**Solution:**

Given that the two opposite points of a ABCD square are A(5, 4) and C(1, -6)

Let us considered that the co-ordinates of B be (x, y).

So, join AC

As we know that the sides of a square are equal, so,

AB = BC

AB

^{2 }= BC^{2}(x – 5)

^{2 }+ (y – 4)^{2 }= (x – 1)^{2 }+ (y + 6)^{2}x

^{2 }– 10x + 25 + y^{2 }– 8y + 16 = x^{2 }– 2x + 1 +y^{2 }+ 12y + 36-10x + 2x – 8y – 12y = 37 – 41

-8x – 20y = -4

2x + 5y = 1

2x = 1 – 5y

x = (1 – 5y)/2

So, ABC is a right-angled triangle

Now by using Pythagoras theorem, we get

AC

^{2 }= AB^{2 }+ BC^{2}(5 – 1)

^{2 }+ (4 + 6)^{2 }= x^{2 }– 10x + 25 + y^{2 }– 8y + 16 + x^{2 }– 2x + 1 + y^{2 }+ 12y + 36(4)

^{2 }+ (10)^{2 }= 2x^{2 }+ 2y^{2 }– 12x + 4y + 7816 + 100 = 2x

^{2 }+ 2y^{2 }– 12x + 4y + 782x

^{2 }+ 2y^{2 }– 12x + 4y + 78 – 16 – 100 = 02x

^{2 }+ 2y^{2 }– 12x + 4y – 38 = 0x

^{2 }+ y^{2 }– 6x + 2y – 19 = 0 …..(i)Now substituting x = (1 – 5y)/2 in eq(i), we get

1 + 25y

^{2 }– 10y + 4y^{2 }– 12 + 60y + 8y – 76 = 029y

^{2 }+ 58y – 87 = 0y

^{2 }+ 2y – 3 = 0y

^{2 }+ 3y – y – 3 = 0y(y + 3) – 1(y + 3) = 0

(y + 3)(y – 1) = 0

The value of y can be either y + 3 = 0, then y = -3

or y – 1 = 0, then y = 1

When y = 1, then

x = (1 – 5y)/2

= (1 – 5(1))/2

= -2

When y = -3, then

x = (1 – 5(-3))/2

= 8

So, the other points of ABCD square are(-2,1) and (8,-3)

### Question 56. Find the centre of the circle passing through (6, -6), (3, -7), and (3, 3).

**Solution:**

Let us considered O be the centre of the circle is (x, y)

It is given that centre of the circle passing through (6, -6), (3, -7), and (3, 3)

Join OA, OB and OC

So, OA = OB = OC

OA

^{2 }= (x -6 )^{2 }+ (y + 6)^{2 }OB

^{2 }= (x – 3)^{2 }+ (y + 7)^{2}and OC

^{2 }= (x – 3)^{2 }+ (y-3)^{2}As we know that OA

^{2 }= OB^{2}So, (x – 6)

^{2 }+ (y + 6)^{2 }= (x – 3)^{2 }+ (y + 7)^{2}x

^{2 }– 12x + 36 + y^{2}+12y + 36 = x^{2 }– 6x + 9 + y^{2 }+ 14y + 49x

^{2 }– 12x + 36 + y^{2}+12y + 36 – x^{2 }+ 6x – 9 – y^{2 }– 14y – 49 = 0-12x + 12y + 72 + 6x – 14y – 58 = 0

-6x – 2y + 14 = 0

-6x – 2y = -14

3x + y = 7 …..(i)

Also, OB

^{2 }= OC^{2}(x – 3)

^{2 }+ (y + 7)^{2 }= (x – 3)^{2 }+ (y – 3)^{2}x

^{2 }– 6x + 9 + y^{2}+ 14y + 49 = x^{2 }– 6x + 9 + y^{2 }– 6y + 9x

^{2 }+ y^{2 }– 6x + 58 + 14y – x^{2 }– y^{2 }+ 6x + 6y – 18 = 020y + 40 = 0

20y = -40

y = -40/20 = -2

3x + (-2) = 7

3x = 7 + 2 = 9

x = 9/3 = 3

Hence, the co-ordinates of the centre are(3,-2)

### Question 57. Two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

**Solution:**

Let us considered ABCD is square, in which the co-ordinates are A (-1, 2) and C (3, 2).

Let us assume the coordinates of B are (x, y)

Now, join AC

As we know that the sides of a square are equal, so,

AB = BC

AB

^{2 }= BC^{2}Now

AB

^{2 }= (x + 1)^{2 }+ (y – 2)^{2}Similarly, BC

^{2 }= (x – 3)^{2 }+ (y – 2)^{2}As we know that AB = BC

So,

(x + 1)

^{2 }+ (y – 2)^{2 }= (x – 3)^{2 }+ (y – 2)^{2}(x + 1)

^{2 }= (x – 3)^{2}x

^{2 }+ 2x + 1 = x^{2 }– 6x + 9x

^{2 }+ 2x + 6x – x^{2 }= 9 – 1 = 88x = 8

x = 8/8 = 1

Now in right triangle ABC

AC

^{2 }= AB^{2 }+ BC^{2}(3 + 1)

^{2 }+ (2 – 2)^{2 }= (x + 1)^{2 }+ (y – 2)^{2 }+ (x – 3)^{2 }+ (y – 2)^{2}(4)

^{2 }+ (0)^{2 }= x^{2 }+ 2x + 1 + y^{2 }– 4y + 4 + x^{2 }– 6x + 9 + y^{2 }– 4y + 416 = 2x

^{2 }+ 2y^{2 }– 4x – 8y + 182x

^{2 }+ 2y^{2 }– 4x – 8y = 16 – 182x

^{2 }+ 2y^{2 }– 4x – 8y = -2x

^{2 }+ y^{2 }– 2x – 4y = -1 …….(i)Now substitute the value of x, in eq(i), we get

(1)

^{2 }+ y^{2 }– 2 × 1 – 4y = -11 + y

^{2 }– 2 – 4y = -1y

^{2 }– 4y = -1 – 1 + 2 = 0y(y – 4) = 0

So the value of the y can be either y = 0

or y – 4 = 0, then y = 4

Hence, the coordinates of other points will be (1, 0) and (1, 4)

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