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

In this section, we delve into Chapter 14 of the Class 10 RD Sharma textbook, which focuses on Coordinate Geometry. Exercise 14.2, Set 3, is designed to help students apply the principles of coordinate geometry to solve various problems involving points, lines, and distances in the Cartesian plane.

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

This section provides detailed solutions for Exercise 14.2, Set 3, from Chapter 14 of the Class 10 RD Sharma textbook. These solutions are aimed at helping students develop a solid understanding of coordinate geometry, ensuring they can confidently solve problems related to points and lines on the Cartesian plane.

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 = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(4-3)^2+(5-0)^2}

AB² = (4 + 3)² + (5 - 0)² 

= (1)² + (5)² 

= 1 + 25 = 26 

Similarly, BC² = (-1 - 4)² + (4 - 5)² 

= (-5)² + (-1)² = 25 + 1 = 26

CD² = (-2 + 1)² + (-1 - 4)²

= (-1)² + (-5)² = 1 + 25 = 26

and DA² = (3 + 2)² + (0 + 1)² 

= (5)² + (1)² = 25 + 1 = 26

Diagonal AC² = (-1 - 3)² + (4 - 0)²

= (-4)² + (4)² = 16 + 16 = 32

and BD² = (-2 - 4)² + (-1 - 5)² 

= (-6)² + (-6)² = 36 + 36 = 72

So, 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 = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(6-3)^2+(4-1)^2} 

AB² = (6 - 3)² + (4 - 1)²

= (3)² + (3)² = 9 + 9 = 18

Similarly, BC² = (8 - 6)² + (6 - 4)²

= (2)² + (2)² = 4 + 4 = 8

and BC² = (8 - 6)² + (6 - 4)² 

= (2)² + (2)² = 4 + 4 = 8 

and CA² = (3 - 8)² + (1 - 6)²

= (-5)² + (-5)² = 25 + 25 = 50

AB = √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, PA² = PB²

Now by using distance formula, we get

(5 - 0)² + (-2 - y)² = (-3 - 0)² + (2 - y)²     

25 + 4 + y² + 4y = 9 + 4 - 4y + y²

y² + 4y + 4y - y² = 13 - 29

8y = -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, PA² = PB²

Now by using distance formula, we get

\sqrt{(x-3)^2+(y-6)^2}=\sqrt{[x-(-3)]^2+(y-4)^2}\\

On squaring both sides, we get

(x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²

x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 = y² - 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

 AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(3-0)^2+(p-2)^2}=\sqrt{(3)^2+p^2-4p+4}\\ =\sqrt{9+p^2-4p+4}=\sqrt{p^2-4p+13}\\ AC =\sqrt{(p-0)^2+(5-2)^2}\\ =\sqrt{p^2+(3)^2}=\sqrt{p^2+9}

It is given that AB = AC

√p² - 4p + 13 = √p² + 9

So, on squaring both side, we get

= p² - 4p + 13 = p² + 9

p² - 4p - p² = 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 = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(-2-7)^2+(5-10)^2}=\sqrt{(-9)^2+(-5)^2}\\ =\sqrt{81+25}=\sqrt{106}

Similarly, BC = \sqrt{(3+2)^2+(-4-5)^2}=\sqrt{106}

and AC = \sqrt{(3-7)^2+(-4-10)^2}=\sqrt{212}

So, we conclude that AB = BC = √106 and AB² + BC² = AC²

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

\sqrt{(x-7)^2+(3+1)^2}=\sqrt{(x-6)^2+(3-8)^2}

On squaring both sides, we get

(x - 7)² + (4)² = (x - 6)² + (-5)²

x² - 14x + 49 + 16 = x² - 12x + 36 + 25

x² - 14x + 65 = x² - 12x + 61

x² - 14x + 12x - x² = 61 - 65

-2x = -4

x = -4/-2 = 2

x = 2

Now we find the distance AP=\sqrt{(2-7)^2+(4)^2}=\sqrt{(-5)^2+(4)^2} = \sqrt{41}

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

\sqrt{(3-8)^2+(y+3)^2} = \sqrt{(3-7)^2+(y-6)^2}

On squaring both sides, we get

(3 - 8)² + (y + 3)² = (-4)² + (y - 6)²

(-5)² + y² + 6y + 9 = 16 + y² - 12y + 36

25 + y² + 6y + 9 = 16 + y² - 12y + 36

y² + 6y - y² + 12y = 36 - 9 - 25 + 16

18y = 18

y = 18/18 = 1

y = 1

Now we find the distance AQ=\sqrt{(3-8)^2+(1+3)^2}\\ =\sqrt{(-5)^2+(4)^2}=\sqrt{25+16}=\sqrt{41}

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

x² + (y + 3)² = 0 + (6)² = 36

x² + y² + 6y + 9 = 36

x² + y² + 6y = 36 - 9 = 27              .......(i)

Also, BC = AB 

(x - 0)² + (y - 3)² = 36

x² + y² + 9 - 6y = 36

x² + y² - 6y = 36 - 9 = 27              ........(ii)

So, from eq (i) and (ii), we get

x² + y² + 6y = x² + y² - 6y

x² + y² + 6y - x² - y² + 6y = 0

12y = 0

y = 0

Now put the value of y in eq(i)

x² + y² + 6y = 27

x² + 0 + 0 = 27

x = ±√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 + k² - 4k = 4 + 4k² + 8k + 25

4k² + 8k + 29 - 16 - 4 - k² + 4k = 0

3k² + 12k + 9 = 0

k² + 4k + 3 = 0

k² + k + 3k + 3 = 0

k(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 AP=\sqrt{(4)^2+(2-k)^2}\\ =\sqrt{16+(2+1)^2}\\ =\sqrt{16+9}=\sqrt{25}=5

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, 

AB=\sqrt{(2+2)^2+(0-0)^2}\\ =\sqrt{16}=4\\ BC=\sqrt{(0-2)^2+(2-0)^2}\\ =\sqrt{4+4}=\sqrt{8}=2\sqrt{2}

Now, PQ=\sqrt{(4+4)^2+(0-0)^2}\\ =\sqrt{64}=8\\ QR=\sqrt{(0-4)^2+(4-0)^2}\\ =\sqrt{32}=4\sqrt{2}\\ RP=\sqrt{(0+4)^2+(4-0)^2}\\ =\sqrt{16+16}=\sqrt{32}=4\sqrt{2} 

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 AB=\sqrt{(-2-3)^2+(3-4)^2}=\sqrt{(-5)^2+(-1)^2}\\ \sqrt{25+1}=\sqrt{26}\\ BC=\sqrt{(x+2)^2+(y-3)^2}\\ CA=\sqrt{(3-x)^2+(4-y)^2}

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

So, BC = AB

\sqrt{(x+2)^2+(y-3)^2}=\sqrt{26}

On squaring both side we get

(x + 2)² + (y - 3)² = 26

x² + 4x + 4 + y² - 6y + 9 = 26

x² + y² + 4x - 6y + 13 = 26

x² + y² + 4x - 6y = 26 - 13 = 12     ...........(i) 

Similarly, CA = AB 

\sqrt{(3-x)^2+(4-y)^2}=\sqrt{26}

On squaring both side we get

(3 - x)² + (4 - y)² = 26

9 + x² - 6x + 16 + y² - 8y = 26

x² + y² - 6x - 8y + 25 = 26

x² + y² - 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

x² + (6 - 5x)² + 4x - 6(6 - 5x) = 13

x² + 36 + 25x² - 60x + 4x - 36 +  30x - 13 = 0

26x² - 26x - 13 = 0

2x² - 2x - 1 = 0

Here, a = 2, b = -2, c = -1

x=\frac{-b±\sqrt{b^2-4ac}}{2a}\\ =\frac{-(-2)±\sqrt{(-2)^2-4\times 2\times (-1)}}{2\times 2}\\ =\frac{2±\sqrt{4+8}}{4}\\ =\frac{2±\sqrt{12}}{4}\\=\frac{2±2\sqrt{3}}{4}=\frac{1±\sqrt{3}}{2}

 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 OA² = OB² = OC²

Now OA = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

OA² = (x + 2)² + (y + 3)²

= x² + 4x + 4 + y² + 6y + 9

= x² + y² + 4x + 6y + 13

OB² = (x + 1)² + (y + 0)²

= x² + 2x + 1 + y²

= x² + y² + 2x + 1

OC² = (x - 7)² + (y + 6)²

= x² - 14x + 49 + y² + 12y + 36

= x² + y² - 14x + 12y + 85

OA² = OB²

x² + y² + 4x + 6y + 13 = x² + y² + 2x + 1

4x + 6y -2y = 1 - 13

2x + 6y = -12

x + 3y = -6       .........(i)

OB² = OC²

x² + y² + 2x + 1 = x² + y² - 14x + 12y + 85

2x + 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               

OA² = OB² = OC²

Now OA=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

OA² = (x - 5)² + (y + 8)²

= x² - 10x + 25 + y² + 16y + 64

= x² + y² - 10x + 16y + 89

Similarly, OB² = (x - 5)² + (y + 9)²

= x² + 4 - 4x + y² + 81 + 18y

= x² + y² - 4x + 18y + 85

and OC² = x² - 4x + 4 + y² - 2y + 1

= x² + y² - 4x - 2y + 5

OA² = OB²

x² + y² - 10x + 16y + 89 = x² + y² - 4x + 18y + 85

-10x + 4x + 16y - 18y = 85 - 89

-6x - 2y = -4               

3x + y = 2                         .......(i)

OB² = OC²

x² + y² - 4x + 18y + 85 = x² + y² - 4x - 2y + 5

18y + 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                   

AB² = BC²

(x - 5)² + (y - 4)² = (x - 1)² + (y + 6)²

x² - 10x + 25 + y² - 8y + 16 = x² - 2x + 1 +y² + 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

AC² = AB² + BC²

(5 - 1)² + (4 + 6)² = x² - 10x + 25 + y² - 8y + 16 + x² - 2x + 1 + y² + 12y + 36

(4)² + (10)² = 2x² + 2y² - 12x + 4y + 78

16 + 100 = 2x² + 2y² - 12x + 4y + 78

2x² + 2y² - 12x + 4y + 78 - 16 - 100 = 0

2x² + 2y² - 12x + 4y - 38 = 0

x² + y² - 6x + 2y - 19 = 0     .....(i)            

Now substituting x = (1 - 5y)/2 in eq(i), we get

(\frac{1-5y}{2})^2+y^2-6(\frac{1-5y}{2})+2y-19=0\\ \frac{1+25y^2-10y}{4}+y^2-3(1-5y)+2y-19=0

1 + 25y² - 10y + 4y² - 12 + 60y + 8y - 76 = 0

29y² + 58y - 87 = 0

y² + 2y - 3 = 0

y² + 3y - y - 3 = 0

y(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         

OA² = (x -6 )² + (y + 6)² 

OB² = (x - 3)² + (y + 7)²        

and OC² = (x - 3)² + (y-3)²

As we know that OA² = OB²

So, (x - 6)² + (y + 6)² = (x - 3)² + (y + 7)²

x² - 12x + 36 + y²+12y + 36 = x² - 6x + 9 + y² + 14y + 49

x² - 12x + 36 + y²+12y + 36 - x² + 6x - 9 - y² - 14y - 49 = 0

-12x + 12y + 72 + 6x - 14y - 58 = 0

-6x - 2y + 14 = 0

-6x - 2y = -14

3x + y = 7                 .....(i)

Also, OB² = OC²

(x - 3)² + (y + 7)² = (x - 3)² + (y - 3)²

x² - 6x + 9 + y²+ 14y + 49 = x² - 6x + 9 + y² - 6y + 9

x² + y² - 6x + 58 + 14y - x² - y² + 6x + 6y - 18 = 0

20y + 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                        

AB² = BC²

Now AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(x+1)^2+(y-2)^2}

AB² = (x + 1)² + (y - 2)²

Similarly, BC² = (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

x² + 2x + 1 = x² - 6x + 9

x² + 2x + 6x - x² = 9 - 1 = 8

8x = 8

x = 8/8 = 1

Now in right triangle ABC

AC² = AB² + BC²

(3 + 1)² + (2 - 2)² = (x + 1)² + (y - 2)² + (x - 3)² + (y - 2)²

(4)² + (0)² = x² + 2x + 1 + y² - 4y + 4 + x² - 6x + 9 + y² - 4y + 4 

16 = 2x² + 2y² - 4x - 8y + 18

2x² + 2y² - 4x - 8y = 16 - 18

2x² + 2y² - 4x - 8y = -2

x² + y² - 2x - 4y = -1    .......(i)

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

(1)² + y² - 2 × 1 - 4y = -1

1 + y² - 2 - 4y = -1

y² - 4y = -1 - 1 + 2 = 0

y(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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