## RD Sharma Solutions Class 10 Maths Chapter 14 – Free PDF Download

The** RD Sharma Solutions for Class 10 Maths Chapter 14 – Co-ordinate Geometry** explain each concept in a comprehensive manner to help students obtain a thorough knowledge of concepts. Coordinates are ordered pairs of numbers, and the study of Geometry using coordinates is called Co-ordinate Geometry. Class 10 Mathematics explains planes and graphical interpretations. The faculty at BYJU’S have created the RD Sharma Solutions to boost problem-solving skills among students.

**Chapter 14 Co-ordinate Geometry** consists of five exercises. RD Sharma Solutions for Class 10 Maths provide descriptive answers for the problems of these exercises. The main purpose of preparing these solutions is to boost confidence in solving any type of problem with ease among students. Students can access the solutions PDF updated for 2023-24 from the links given here. Important concepts discussed in this chapter are as follows:

- To find the distance between two points whose coordinates are given.
- Find the coordinates of the point which divides the line segment joining two given points in a given ratio.
- The method of finding the area of a triangle in terms of the coordinates of its vertices.

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### Access the RD Sharma Solutions for Class 10 Maths Chapter 14 – Co-ordinate Geometry

### RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.1 Page No: 14.4

**1. On which axis do the following points lie?**

**(i) P (5, 0)**

**(ii) Q (0, -2)**

**(iii) R (-4, 0)**

**(iv) S (0, 5)**

**Solution:**

(i) P (5, 0) lies on the *x-axis*

(ii) Q (0, -2) lies on the *y-axis (negation half)*

(iii) R (-4, 0) lies on the *x-axis (negative half)*

(iv) S (0, 5) lies on the *y-axis*

### RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.2 Page No: 14.15

**1. Find the distance between the following pair of points.**

**(i) (- 6, 7) and (-1, -5)**

**(ii) (a + b, b + c) and (a -b, c – b)**

**(iii) (a sin α, – b cos α) and (- a cos α, b sin α)**

**(iv) (a, 0) and (0, b)**

**Solution:**

(i) Let the given points be P (- 6, 7) and Q (- 1, – 5).

Here,

x_{1} = – 6, y_{1} = 7 and

x_{2} = -1, y_{2} = – 5

(ii) Let the given points be P (a + b, b + c) and Q (a – b, c – b).

Here,

x_{1} = a + b, y_{1 }= b + c and

x_{2} = a – b, y_{2} = c – b

(iii) Let the given points be P(a sinα, – b cos α) and Q(-a cos α, b sin α).

x_{1 }= a sin α, y_{1} = – b cos α and

x_{2} – a cos α, y_{2} = b sin α

(iv) Let the given points be P(a, 0) and Q (0, b).

Here,

x_{1} = a, y_{1} = 0, x_{2} = 0, y_{2} = b,

**2. Find the value of a when the distance between the points (3, a) and (4, 1) is √10.**

**Solution:**

Let the given points be P (3, a) and Q(4, 1).

Here,

On squaring on both sides, we have

⇒ 10 = 2 + a^{2} – 2a

⇒ a^{2} – 2a + 2 – 10 = 0

⇒ a^{2} – 2a – 8 = 0

By splitting the middle team,

⇒ a^{2} – 4a + 2a – 8 = 0

⇒ a(a – 4) + 2(a – 4) = 0

⇒ (a – 4) (a + 2) = 0

⇒ a = 4, a = – 2

*Thus, there are 2 possible values for a which are 4 and -2.*

**3. If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.**

**Solution:**

Let the given points be P(2, 1) and Q(1,- 2) and R(x, y).

Also, PR = QR (given)

But, PR = QR

⇒ x^{2 }+ 5 – 4x + y^{2} –2y = x^{2 }+ 5 – 2x + y^{2 }+ 4y

⇒ x^{2 }+ 5 – 4x + y^{2} – 2y = x^{2 }+ 5 – 2x + y^{2 }+ 4y

⇒ – 4x + 2x – 2y – 4y = 0

⇒ – 2x – 6y = 0

⇒ – 2(x + 3y) = 0

⇒ -2(x + 3y) = 0

⇒ x + 3y = 0/-2

*⇒ x + 3y = 0*

Hence, proved.

** **

**4. Find the value of x, y if the distances of the point (x, y) from (- 3, 0) as well as from (3, 0) are 4.**

**Solution:**

Let the given points be P(x, y), Q( -3, 0) and R(3, 0).

On squaring on both sides, we get

⇒ 16 = x^{2} + 9 + 6x + y^{2}

⇒ x^{2} + y^{2} = 7 – 6x …… (1)

On squaring on both sides,

⇒16 = x^{2} + 9 – 6x + y^{2}

⇒ x^{2} + y^{2} = 16 – 9 + 6x

⇒ x^{2 }+ y^{2 }= 7 + 6x …. (2)

Equating (1) and (2), we have

7 – 6x = 7 + 6x

⇒ 7 – 7 = 6x + 6x

⇒ 0 = 12x

⇒ x = 12

Then, substituting the value of x = 0 in (2),

x^{2} + y^{2} = 7+ 6x

0 + y^{2} = 7 + 6 × 0

y^{2} = 7

y = + √7

*As y can have two values, the points are (12, √7) and (12, -√7).*

**5. The length of a line segment is of 10 units, and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.**

**Solution: **

Given,

The length of the line segment is 10 units.

The coordinates of one end-point are (2, -3), and the abscissa of the other end is 10.

So, let the ordinate of the other end be k.

*Therefore, the ordinates of the other end can be 3 or -9.*

**6. Show that the points A(- 4, -1), B(-2, – 4), C(4, 0) and D(2, 3) are the vertices points of a rectangle.**

**Solution: **

** v**

Given: Points A(- 4, -1), B(-2, – 4), C(4, 0) and D(2, 3)

Required to prove: the points are the vertices points of a rectangle.

Vertices of rectangle ABCD are A(- 4, -1), B(-2, – 4), C(4, 0) and D(2, 3)

We know that

As the opposite sides are equal and also the diagonals are equal.

*Therefore, the given points are the vertices of a rectangle.*

- Hence, proved.

**7. Show that the points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.**

**Solution:**

Given: Points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2)

Required to prove: the points are the vertices points of a parallelogram.

Vertices of a parallelogram ABCD are A (1, -2), B (3, 6), C (5, 10) and D (3, 2)

We know that

Finding the diagonals,

It’s seen that the opposite sides of the quadrilateral formed by the given four points are equal

i.e., (AB = CD) & (DA = BC)

Also, the diagonals BD and AC are found to be unequal.

*Hence, the given points form a parallelogram.*

- Hence, proved.

**8. Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square.**

**Solution: **

Given: Points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4)

Required to prove: the points are the vertices points of a square.

Vertices of a square ABCD are A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4)

We know that

*As the opposite sides are equal and also the diagonals are equal, the given vertices are, therefore, the vertices of a square.*

- Hence, proved.

**9. Prove that the points (3, 0), (6, 4) and (- 1, 3) are vertices of a right-angled isosceles triangle.**

**Solution:**

Let the vertices of the triangle ABC be A(3, 0), B(6, 4) and C (- 1, 3)

We know that,

It’s seen that AB = AC. Thus, it’s an isosceles triangle.

Verifying the Pythagoras theorem, we have

BC^{2} = AB^{2} + AC^{2}

(√50)^{2} = (√25)^{2} + (√25)^{2}

50 = 25 + 25

50 = 50

As BC^{2} = AB^{2} + AC^{2}

*Therefore, the given vertices are of a right-angled isosceles triangle.*

- Hence, proved.

**10.** **Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right-angled triangle. Find the area of the triangle and the length of the hypotenuse.**

**Solution:**

From the given question,

Let consider the vertices of a triangle ABC as A(2, -2), B(-2, 1) and C(5, 2)

We know that

It’s seen that AB = AC, thus, the triangle is an isosceles triangle.

Verifying the Pythagoras theorem, we have

BC^{2} = AB^{2} + AC^{2}

(√50)^{2} = (√25)^{2} + (√25)^{2}

50 = 25 + 25

50 = 50

As BC^{2} = AB^{2} + AC^{2}

Therefore, the given triangle is right angled triangle.

Now,

- Hence, proved.

**11. Prove that the points (2a, 4a), (2a, 6a) and (2a + **√3a, 5a**) are the vertices of an equilateral triangle.**

**Solution: **

From given,

Let’s consider the vertices of a triangle ABC as A(2 a, 4 a), B(2 a, 6 a) and C(2a + √3a, 5a)

We know that

As all the sides are equal, the triangle is an equilateral triangle.

*Thus, the given vertices are of an equilateral triangle.*

- Hence, proved.

**12. Prove that the points (2, 3), (-4, -6) and (1, 3/2) do not form a triangle.**

**Solution: **

From given,

Let’s consider the vertices of a triangle ABC as A(2, 3), B(-4, -6) and C(1, 3/2)

We know that

*Thus, the given vertices do not form a triangle as the sum of two sides of a triangle is not greater than the third side.*

- Hence, proved.

**13. The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC, right-angled at B. Find the values of a and hence the area of triangle ABC. **

**Solution: **

Given,

A right triangle ABC, right-angled at B.

Points A (2, 9), B (a, 5) and C (5, 5)

So, AC is the hypotenuse.

Thus, from Pythagoras’ theorem, we have

AC^{2} = AB^{2} + BC^{2}

^{2}+ (5 – 9)

^{2}] = [(a – 2)

^{2}+ (5 – 9)

^{2}] + [(5 – a)

^{2}+ (5 – 5)

^{2}] [3

^{2 }+ (-4)

^{2}] = [(a – 2)

^{2}+ (-4)

^{2}] + [(5 – a)

^{2}+ 0]

9 + 16 = a^{2} – 4a + 4 + 16 + 25 – 10a + a^{2}

2a^{2} – 14a + 20 = 0

a^{2} – 7a + 10 = 0

(a – 5)(a – 2) = 0 [By factorisation method]

So,

a = 5 or 2

Here, a = 5 is not possible as it coincides with point C. So, for a triangle to form, the value of a = 2 is correct.

Thus, the coordinates of point B are (2, 5).

Now, the area of triangle ABC.

*Therefore, the area of triangle ABC is 6 sq. units.*

**14. Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus. **

**Solution: **

Let A(2, -1), B(3 ,4), C(-2, 3) and D(-3, -2)

Then, we have,

Length of AB = √[(3 – 2)^{2} + (4 – (-1))^{2}] = √[(1)^{2} + (5)^{2}] = √[1 + 25] = √26 units

Length of BC = √[(3 – (-2))^{2} + (4 – 3)^{2}] = √[(5)^{2} + (1)^{2}] = √[25 + 1] = √26 units

Length of CD = √[(-2 – (-3))^{2} + (3 – 2)^{2}] = √[(-5)^{2} + (1)^{2}] = √[25 + 1] = √26 units

Length of AD = √[(-3 – 2)^{2} + (-2 – (-1))^{2}] = √[(-5)^{2} + (-1)^{2}] = √[25 + 1] = √26 units

As AB = BC = CD = AD

We can say that

*Quadrilateral ABCD is a rhombus.*

**15. Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3. **

**Solution: **

Let the third vertex be C (x, y).

And given A (2, 0) and B (2, 5).

We have

Length of AB = √[(2 – 2)^{2} + (5 – 0)^{2}] = √[(0)^{2} + (5)^{2}] = √[0 + 25] = 5 units

Length of BC = √[(x – 2)^{2} + (y – 5)^{2}] = √[x^{2} – 4x + 4 + y^{2} – 10y + 25]

= √[ x^{2} – 4x + y^{2} – 10y + 29] units

Length of AC = √[(x – 2)^{2} + (y – 0)^{2}] = √[x^{2} – 4x + 4 + y^{2}] units

Given that

AC = BC = 3

So, AC^{2} = BC^{2} = 9

x^{2} – 4x + 4 + y^{2} = x^{2} – 4x + y^{2} – 10y + 29

10y = 25

y = 25/10 = 2.5

And,

AC^{2} = 9

x^{2} – 4x + 4 + y^{2} = 9

x^{2} – 4x + 4 + (2.5)^{2} = 9

x^{2} – 4x + 4 + 6.25 = 9

x^{2} – 4x + 1.25 = 0

D = (-4)^{2} – 4 x 1 x 1.25 = 16 – 5 = 11

So, the roots are

x = -(-4) + √11/ 2 = (4 + 3.31)/ 2 = 3.65

And,

x = -(-4) – √11/ 2 = (4 – 3.31)/ 2 = 0.35

*Therefore, the third vertex can be C (3.65, 2.5) or (0.35, 2.5).*

**16. Which point on the x-axis is equidistant from (5, 9) and (-4, 6)?**

**Solution: **

Let A (5, 9) and B (-4, 6) be the given points.

Let the point on the x-axis equidistant from the above points be C(x, 0).

Now, we have

AC = √[(x – 5)^{2} + (0 – 9)^{2}] = √[x^{2} – 10x + 25 + 81] = √[x^{2} – 10x + 106]

And,

BC = √[(x – (-4))^{2} + (0 – 6)^{2}] = √[x^{2} + 8x + 16 + 36] = √[x^{2} + 8x + 52]

As AC = BC (given condition)

So, AC^{2} = BC^{2}

x^{2} – 10x + 106 = x^{2} + 8x + 52

18x = 54

x = 3

*Therefore, the point on the x-axis is (3, 0).*

**17. Prove that the points (-2, 5), (0, 1) and (2, -3) are collinear.**

**Solution: **

Let A (-2, 5), B(0, 1) and C (2, -3) be the given points.

So, we have

AB = √[(0 – (-2))^{2} + (1 – 5)^{2}] = √[(2)^{2} + (-4)^{2}] = √[4 + 16] = √20 = 2√5 units

BC = √[(2 – 0)^{2} + (-3 – 1)^{2}] = √[(2)^{2} + (-4)^{2}] = √[4 + 16] = √20 = 2√5 units

AC = √[(2 – (-2))^{2} + (-3 – 5)^{2}] = √[(4)^{2} + (-8)^{2}] = √[16 + 64] = √80 = 4√5 units

Now, it’s seen that

AB + BC = AC

2√5 + 2√5 = 4√5

4√5 = 4√5

*Therefore, we can conclude that the given points (-2, 5), (0, 1) and (2, -3) are collinear.*

**18. The coordinates of point P are (-3, 2). Find the coordinates of the point Q, which lies on the line joining P and origin such that OP = OQ. **

**Solution: **

Let the coordinates of Q be taken as (x, y).

As Q lies on the line joining P and O(origin) with OP = OQ.

Then, by mid-point theorem,

(x – 3)/2 = 0

And,

(y + 2)/ 2 = 0

∴ x = 3, y = -2

*Therefore, the coordinates of point Q are (3, -2).*

**19. Which point on the y-axis is equidistant from (2, 3) and (-4, 1)?**

**Solution: **

Let A (2, 3) and B (-4, 1) be the given points.

Let the point on the y-axis equidistant from the above points be C (0, y).

Now, we have

AC = √[(0 – 2)^{2} + (y – 3)^{2}] = √[y^{2} – 6y + 9 + 4] = √[y^{2} – 6y + 13]

And,

BC = √[(0 – (-4))^{2} + (y – 1)^{2}] = √y^{2} – 2y + 1 + 16] = √[y^{2} – 2y + 17]

As AC = BC (given condition)

So, AC^{2} = BC^{2}

y^{2} – 6y + 13 = y^{2} – 2y + 17

-4y = 4

y = -1

*Therefore, the point on the y-axis is (0, -1).*

**20. The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex. **

**Solution: **

Let A (3, 4), B (3, 8) and C (9, 8) be the given points.

And let the fourth vertex be D(x, y).

We know that

In a parallelogram, the diagonals bisect each other.

So, the mid-point of AC should be the same as the mid-point of BC.

By mid-point theorem,

Mid-point of AC = (3 + 9/ 2), (4 + 8/ 2) = (6, 6)

Now,

The mid-point of BD = (3 + x/ 2, 8 + y/ 2)

And this point must be equal to (6, 6).

So, we have

(3 + x)/ 2 = 6 (8 + y)/ 2 = 6

3 + x = 12 8 + y = 12

x = 9 y = 4

*Therefore, the fourth vertex is D (9, 4).*

**21. Find a point which is equidistant from points A (-5, 4) and B (-1, 6). How many such points are there?**

**Solution: **

Let P(x, y) be the equidistant point from points A (-5, 4) and B (-1, 6).

So, the mid-point can be the required point.

(x, y) = ( (-5 – 1/ 2), (4 + 6)/2 )

(x , y) = (-6/2 , 10/2) = (-3, 5)

Thus, the required point is (-3, 5)

Now,

We also know that AP = BP

So, AP^{2} = BP^{2}

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

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

10x + 41 – 8y = 2x + 37 – 12y

8x + 4y + 4 = 0

2x + y + 1 = 0

*Therefore, all the points which lie on line 2x + y + 1 = 0 are equidistant from A and B.*

**22. The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units.**

**Solution: **

Given,

Diameter of the circle = 10√2 units

So, the radius = 5√2 units

Let the centre of a circle be 0(2a, a-7), and the circle passes through point P (11, -9).

Then, OP is the radius of the circle.

OP = 5√2

OP^{2} = (5√2) = 50

(11- 2a)^{2} + (-9 – a + 7)^{2} = 50

121 – 44a + 4a^{2} + 4 + a^{2} + 4a = 50

5a^{2} – 40a + 75 = 0

a^{2} – 8a + 15 = 0

(a – 5)(a – 3) = 0 [Factorisation method]

*So, a = 5 or a = 3*

**23. Ayush starts walking from his house to the office. Instead of going to the office directly, he goes to the bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in kilometers.**

**Solution: **

The position of Ayush’s house is (2, 4), and the position of the bank is (5, 8).

So, the distance between the house and the bank,

d_{1} = √[(5 – 2)^{2} + (8 – 4)^{2}] = √[(3)^{2} + (4)^{2}] = √[9 + 16] = √25 = 5 km

The position of the bank is (5, 8), and the position of the school is (13, 14).

So, the distance between the bank and the school

d_{2} = √[(13 – 5)^{2} + (14 – 8)^{2}] = √[(8)^{2} + (6)^{2}] = √[64 + 36] = √100 = 10 km

The position of the school is (13, 14), and the position of the office is (13, 26).

So, the distance between the school and the office is

d_{3} = √[(13 – 13)^{2} + (26 – 14)^{2}] = √[(0)^{2} + (12)^{2}] = √144 = 12 km

Let d be the total distance covered by Ayush.

d = d_{1} + d_{2} + d_{3} = 5 + 10 + 12 = 27 km

Let the D be the shortest distance from Ayush’s house to the office.

D = √[(13 – 2)^{2} + (26 – 4)^{2}] = √[(11)^{2} + (22)^{2}] = √[121+ 484] = √605 = 24.6 km

*Thus, the extra distance covered by Ayush = d – D = 27 – 24.6 = 2.4 km.*

**24. Find the value of k, if the point P(0, 2) is equidistant from (3, k) and (k, 5).**

**Solution: **

Let the point P (0, 2) be equidistant from A (3, k) and B (k, 5).

So, PA = PB

PA^{2} = PB^{2}

(3 -0)^{2} + (k -2)^{2} = (k – 0)^{2} + (5 – 2)^{2}

9 + k^{2} + 4 – 4k – k^{2} – 9 = 0

4 – 4k = 0

-4k = -4

*Therefore, the value of k = 1.*

**25. If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the (i) interior (ii) exterior of the triangle.**

**Solution: **

Let B (-4, 3) and C (4, 3) be the given two vertices of the equilateral triangle.

Let A (x, y) be the third vertex.

Then, we have

AB = BC = AC

Let us consider the part AB = BC.

AB^{2} = BC^{2}

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

16 + x^{2} + 8x + 9 + y^{2} – 6y = 64

x^{2} + y^{2} + 8x – 6y = 39

Now, let us consider AB = AC.

AB^{2} = AC^{2}

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

16 + x^{2} + 8x + 9 + y^{2} – 18y = 16 + x^{2} – 8x + 9 + y^{2} – 6y

16x = 0

x = 0

Now, BC = AC

BC^{2} = AC^{2}

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

64 + 0 = 16 + 9 + y^{2 }– 6y

64 = 16 + (3 – y)^{2}

(3 – y)^{2} = 48

3 – y = ± 4√3

y = 3 ± 4√3

Therefore, the coordinates of the third vertex.

(i) When the origin lies in the interior of the triangle is *(0, 3 – 4√3).*

(ii) When the origin lies in the exterior of the triangle is *(0, 3 + 4√3).*

**26. Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus. **

**Solution: **

Let A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) be the given points.

Then, we have

AB = √[(-5 + 3)^{2} + (-5 – 2)^{2}] = √[(2)^{2} + (7)^{2}] = √[4 + 49] = √53 units

BC = √[(2 + 5)^{2} + (-3 + 5)^{2}] = √[(7)^{2} + (2)^{2}] = √[49 + 4] = √53 units

CD = √[(4 – 2)^{2} + (4 + 3)^{2}] = √[(2)^{2} + (7)^{2}] = √[4 + 49] = √53 units

AD = √[(4 + 3)^{2} + (4 – 2)^{2}] = √[(7)^{2} + (2)^{2}] = √[49 + 4] = √53 units

And the diagonals

AC = √[(2 + 3)^{2} + (-3 – 2)^{2}] = √[(5)^{2} + (-5)^{2}] = √[25 + 25] = 5√2 units

BD = √[(4 + 5)^{2} + (4 + 5)^{2}] = √[(9)^{2} + (9)^{2}] = √[81 + 81] = 9√2 units

It’s seen that

As AB = BC = CD = AD and the diagonals AC ≠ BD

ABCD is a rhombus.

Now,

*Area of rhombus ABCD = ½ x AC x BD = ½ x 5√2 x 9√2 = 45 sq. units*

**27. Find the coordinates of the circumcenter of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find the circumradius. **

**Solution: **

Let A(3, 0), B(-1, -6) and C(4 , -1) be the given points.

Let O(x, y) be the circumcenter of the triangle.

Then, OA = OB = OC

OA^{2} = OB^{2}

(x – 3)^{2} + (y – 0)^{2} = (x + 1)^{2} + (y + 6)^{2}

x^{2 }+ 9 – 6x + y^{2} = x^{2} + 1 + 2x + y^{2} + 36 + 12y

-8x -12y = 28

2x + 3y = -7 …..(i) [After simplification]

Again,

OB^{2} = OC^{2}

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

x^{2} + 2x + 1 + y^{2} + 36 + 12y = x^{2} + 16 – 8x + y^{2} + 1 + 2y

10x + 10y = -20

x + y = -2 …..(ii) [After simplification]

Hence, the circumcenter of the triangle is (1, -3)

Circumradius = Distance from any of the given points (say B)

=√[(1 + 1)^{2} + (-3 + 6)^{2}] = √(4 + 9)

*= √13 units*

**28. Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4).**

**Solution:**

Let A(7, 6) and B(-3, 4) be the given points.

Let P(x, 0) be the point on the x-axis such that PA = PB

So, PA^{2} = PB^{2}

(x – 7)^{2} + (0 – 6)^{2} = (x + 3)^{2} + (0 – 4)^{2}

x^{2} + 49 – 14x + 36 = x^{2} + 9 + 6x + 16

-20x = -60

x = 3

*Therefore, the point on the x-axis is (3, 0).*

### RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.3 Page No: 14.28

**1. Find the coordinates of the point which divides the line segment joining (-1, 3) and (4, – 7) internally in the ratio 3: 4.**

**Solution:**

Let P(x, y) be the required point.

By section formula, we know that the coordinates are

Here,

x_{1} = – 1 y_{1} = 3

x_{2 }= 4 y_{2} = -7

m: n = 3: 4

Then,

*Therefore, the coordinates of P are (8/7, – 9/7).*

**2. Find the points of trisection of the line segment joining the points.**

**(i) (5, – 6) and (-7, 5)**

**(ii) (3, – 2) and (-3, – 4)**

**(iii) (2, – 2) and (-7, 4)**

**Solution:**

(i) Let P and Q be the point of trisection of AB, such that AP = PQ = QB.

So, P divides AB internally in the ratio of 1: 2, thereby applying the section formula, the coordinates of P will be

Now, Q also divides AB internally in the ratio of 2:1, so their coordinates will be

(ii) Let P and Q be the points of trisection of AB such that AP = PQ = QB

P divides AB internally in the ratio of 1: 2. Hence, by applying the section formula, the coordinates of P are

Now, Q also divides internally in the ratio of 2: 1

So, the coordinates of Q are given by

(iii) Let P and Q be the points of trisection of AB such that AP = PQ = OQ

P divides AB internally in the ratio 1:2. So, the coordinates of P, by applying the section formula, are given by

Now, Q also divides AB internally in the ration 2: 1. And the coordinates of Q are given by

**3. Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (-2, -1), (1, 0), (4, 3) and (1, 2) meet.**

**Solution:**

Let A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) be the given points.

Let P(x, y) be the point of intersection of the diagonals of the parallelogram formed by the given points.

We know that diagonals of a parallelogram bisect each other.

*Therefore, the coordinates of P are (1, 1).*

**4. Prove that the points (3, 2), (4, 0), (6, -3) and (5, -5) are the vertices of a parallelogram.**

**Solution:**

Let A(3, -2), B(4, 0), C(6, -3) and D(5, -5).

Let P(x, y) be the point of intersection of diagonals AC and BD of ABCD.

The mid-point of AC is given by

Again, the mid-point of BD is given by

Thus, we can conclude that diagonals AC and BD bisect each other.

And we know that diagonals of a parallelogram bisect each other.

*Therefore, ABCD is a parallelogram.*

**5. If P(9a – 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.**

**Solution: **

Given that, P(9a – 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3:1

Then, by THE section formula

Coordinates of P are

And,

Solving for a, we have

(9a – 2) x 4 = 24a + 3a + 1

36a – 8 = 27a + 1

9a = 9

a = 1

Now, solving for b, we have

4 x –b = 15 – 3

-4b = 12

b = -3

*Therefore, the values of a and b are 1 and -3, respectively.*

**6. If (a, b) is the mid-point of the line segment joining the points A (10, -6), B(k, 4) and a – 2b = 18, find the value of k and the distance AB.**

**Solution:**

As (a, b) is the mid-point of the line segment A(10, -6) and B(k, 4).

So,

(a, b) = (10 + k / 2, -6 + 4/ 2)

a = (10 + k)/ 2 and b = -1

2a = 10 + k

k = 2a – 10

Given a – 2b = 18

Using b = -1 in the above relation, we get

a – 2(-1) = 18

a = 18 – 2 = 16

So,

k = 2(16) – 10 = 32 – 10 = 22

Thus,

*AB = √[(22 – 10) ^{2} + (4 + 6)^{2}] = √[(12)^{2} + (10)^{2}] = √[144 + 100] = 2√61 units*

**7. Find the ratio in which the point (2, y) divides the line segment joining the points A(-2, 2) and B(3, 7). Also, find the value of y. **

**Solution: **

Let the point P(2, y) divide the line segment joining the points A(-2, 2) and B(3, 7) in the ratio k: 1

Then, the coordinates of P are given by

And, given the coordinates of P are (2, y).

So,

2 = (3k – 2)/ (k + 1) and y = (7k + 2)/ (k + 1)

Solving for k, we get

2(k + 1) = (3k – 2)

2k + 2 = 3k – 2

k = 4

Using k to find y, we have

y = (7(4) + 2)/ (4 + 1)

= (28 + 2)/5

= 30/5

y = 6

*Therefore, the ratio is 4: 1 and y = 6*

**8. If A(-1, 3), B(1, -1) and C(5, 1) are the vertices of a triangle ABC, find the length of the median through A. **

**Solution:**

Let AD be the median through A.

AD is the median, AND D is the mid-point of BC.

So, the coordinates of D are (1 + 5/ 2, -1 + 1/ 2 ) = (3 , 0)

Therefore,

*Length of median AD = √[(3 + 1) ^{2} + (0 – 3)^{2}] = √[(4)^{2} + (-3)^{2}] = √[16 + 9] = √25 = 5 units*

**9. If the points P, Q(x, 7), R, S(6, y) in this order divide the line segment joining A(2, p) and B (7, 10) in 5 equal parts, find x, y and p.**

**Solution: **

From the question, we have

AP = PQ = QR = RS = SB

So, Q is the mid-point of A and S.

Then,

x = (2 + 6)/ 2 = 8/2 = 4

7 = (y + p)/ 2

y + p = 14 ….. (1)

Now, since S divides QB in the ratio 2: 1

So, p = 14 – 9 = 5

*Therefore, x = 4, y = 9 and p = 5*

**10. If a vertex of a triangle is (1, 1) and the middle points of the sides through it be (-2, 3) and (5, 2), find the other vertices. **

**Solution: **

Let A(1, 1) be the given vertex and D(-2, 3) and E(5, 2) be the mid-points of AB and AC.

Now, D and E are the mid-points of AB and AC.

So, the coordinates of B are (-5, 5).

Again,

So, the coordinates of C are (9, 3).

*Therefore, the other vertices of the triangle are (-5, 5) and (9, 3).*

**11. (i) In what ratio is the segment joining the points, (-2, -3) and (3, 7), divided by the y-axis? Also, find the coordinates of the point of division. **

**Solution: **

Let P(-2, -3) and Q(9, 3) be the given points.

Suppose the y-axis divides PQ in the ratio k: 1 at R(0, y).

So, the coordinates of R are given by

Now, equating,

= 0

3k – 2 = 0

k = 2/3

Therefore, the ratio is 2: 3

Putting k = 2/3 in the coordinates of R, we get

*R (0, 1)*

**(ii) In what ratio is the line segment joining (-3, -1) and (-8, -9) divided at the point (-5, -21/5)?**

**Solution: **

Let A(-3, -1) and B(-8, -9) be the given points.

And, let P be the point that divides AB in the ratio k: 1

So, the coordinates of P are given by

But, given the coordinates of P,

On equating, we get

(-8k – 3)/ (k + 1) = -5

-8k – 3 = -5k – 5

3k = 2

k = 2/3

*Thus, point P divides AB in the ratio 2: 3*

**12. If the mid-point of the line joining (3, 4) and (k, 7) is (x, y) and 2x + 2y + 1 = 0, find the value of k. **

**Solution: **

As (x, y) is the mid-point,

x = (3 + k)/ 2 and y = (4 + 7)/ 2 = 11/2

Also,

Given that the mid-point lies on line 2x + 2y + 1 = 0

2[(3 + k)/ 2] + 2(11/2) + 1 = 0

3 + k + 11 + 1 = 0

*Thus, k = -15*

**13. Find the ratio in which the point P(3/4, 5/12) divides the line segments joining the point A(1/2, 3/2) and B(2, -5). **

**Solution: **

Given,

Points A(1/2, 3/2) and B(2, -5)

Let the point P(3/4, 5/12) divide the line segment AB in the ratio k: 1

Then, we know that

P(3/4, 5/12) = (2k + ½)/ (k +1) , (2k + 3/2)/ (k + 1)

Now, equating the abscissa, we get

¾ = (2k + ½)/ (k +1)

3(k + 1) = 4(2k + 1/2)

3k + 3 = 8k + 2

5k = 1

k = 1/5

*Therefore, the ratio in which the point P(3/4, 5/12) divides is 1: 5*

**14. Find the ratio in which the line joining (-2, -3) and (5, 6) is divided by (i) x-axis and (ii) y-axis. Also, find the coordinates of the point of division in each case. **

**Solution: **

Let A(-2, -3) and B(5, 6) be the given points.

(i) Suppose the x-axis divides AB in the ratio k: 1 at the point P.

Then, the coordinates of the point of division are

P lies on the x-axis, and the y-coordinate is zero.

So,

6k – 3/ k + 1 = 0

6k – 3 = 0

k = ½

Thus, the required ratio is 1: 2

Using k in the coordinates of P,

We get P (1/3, 0)

(ii) Suppose the y-axis divides AB in the ratio k: 1 at point Q.

Then, the coordinates of the point od division are given by

Q lies on the y-axis, and the x-ordinate is zero.

So,

5k – 2/ k + 1 = 0

5k – 2 = 0

k = 2/5

Thus, the required ratio is 2: 5

Using k in the coordinates of Q,

*We get Q (0, -3/7)*

**15. Prove that the points (4, 5), (7, 6), (6, 3), and (3, 2) are the vertices of a parallelogram. Is it a rectangle?**

**Solution: **

Let A (4, 5), B(7, 6), C(6, 3) and D(3, 2) be the given points.

And P be the point of intersection of AC and BD.

Coordinates of the mid-point of AC are (4+6/2 , 5+3/2) = (5, 4)

Coordinates of the mid-point of BD are (7+3/2 , 6+2/2) = (5, 4)

Thus, it’s clearly seen that the mid-point of AC and BD are the same.

So, ABCD is a parallelogram.

Now,

AC = √[(6 – 4)^{2} + (3 – 5)^{2}] = √[(2)^{2} + (-2)^{2}] = √[4 + 4] = √8 units

And,

BD = √[(7 – 3)^{2} + (6 – 2)^{2}] = √[(4)^{2} + (4)^{2}] = √[16 + 16] = √32 units

Since AC ≠ BD

*Therefore, ABCD is not a rectangle.*

**16. Prove that (4, 3), (6, 4), (5, 6) and (3, 5) are the angular points of a square.**

**Solution: **

Let A(4,3) , B(6,4) , C(5,6) and D(3,5) be the given points.

The distance formula is

It’s seen that the lengths of all the sides are the same.

Now, the length of diagonals are

Also, the lengths of both diagonals are the same.

*Therefore, we can conclude that the given points are the angular points of a square.*

**17. Prove that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices of a rectangle.**

**Solution: **

Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) be the given points.

Now,

Coordinates of the mid-point of AC are (-4 + 4/ 2, -1 + 0/2) = (0, -1/2)

Coordinates of the mid-point of BD are (-2 + 2/2, -4 + 3/2) = (0, -1/2)

Thus, it’s seen that AC and BD have the same point.

And we have diagonals

The lengths of the diagonals are also the same.

*Therefore, the given points are the vertices of a rectangle.*

**18. Find the length of the medians of a triangle whose vertices are A(-1, 3), B(1, -1) and C(5, 1).**

**Solution: **

Let AD, BF and CE be the medians of ΔABC.

Coordinates of D are (5 + 1/ 2, 1 – 1/ 2) = (3, 0)

Coordinates of E are (-1 + 1/ 2, 3 – 1/ 2) = (0, 1)

Coordinates of F are (5 – 1/ 2, 1 + 3/ 2) = (2, 2)

Now,

Finding the length of the respective medians,

**19. Find the ratio in which the line segment joining the points A (3, -3) and B (-2, 7) is divided by the x-axis. Also, find the coordinates of the point of division.**

**Solution:**

Let the point on the x-axis be (x, 0). [y-coordinate is zero].

And, let this point divides the line segment AB in the ratio of k : 1.

Now, using the section formula for the y-coordinate, we have

0 = (7k – 3)/(k + 1)

7k – 3 = 0

k = 3/7

*Therefore, the line segment AB is divided by the x-axis in the ratio 3: 7*

**20. Find the ratio in which point P(x, 2) divides the line segment joining the points A (12, 5) and B (4, -3). Also, find the value of x.**

**Solution: **

Let P divide the line joining A and B and let it divide the segment in the ratio k: 1

Now, using the section formula for the y-coordinate, we have

2 = (-3k + 5)/ (k + 1)

2(k + 1) = -3k + 5

2k + 2 = -3k + 5

5k = 3

k = 3/5

Thus, P divides the line segment AB in the ratio of 3: 5

Using the value of k, we get the x-coordinate as

x = 12 + 60/ 8 = 72/8 = 9

*Therefore, the coordinates of point P are (9, 2).*

**21. Find the ratio in which the point P(-1, y) lying on the line segment joining A(-3, 10) and B(6, -8) divides it. Also, find the value of y.**

**Solution: **

Let P divide A(-3, 10) and B(6, -8) in the ratio of k: 1.

Given coordinates of P as (-1, y).

Now, using the section formula for the x-coordinate, we have

-1 = 6k – 3/ k + 1

-(k + 1) = 6k – 3

7k = 2

k = 2/7

Thus, point P divides AB in the ratio of 2: 7.

Using the value of k to find the y-coordinate, we have

y = (-8k + 10)/ (k + 1)

y = (-8(2/7) + 10)/ (2/7 + 1)

y = -16 + 70/ 2 + 7 = 54/9

y = 6

*Therefore, the y-coordinate of P is 6.*

**22. Find the coordinates of point A, where AB is the diameter of the circle whose centre is (2, -3) and B is (1, 4).**

**Solution: **

Let the coordinates of point A be (x, y).

If AB is the diameter, then the centre in the mid-point of the diameter,

So,

(2, -3) = (x + 1/ 2, y + 4/ 2)

2 = x + 1/2 and -3 = y + 4/ 2

4 = x + 1 and -6 = y + 4

x = 3 and y = -10

*Therefore, the coordinates of A are (3, -10).*

**23. If the points (-2, 1), (1, 0), (x, 3) and (1, y) form a parallelogram, find the values of x and y. **

**Solution:**

Let A(-2, 1), B(1, 0), C(x , 3) and D(1, y) be the given points of the parallelogram.

We know that the diagonals of a parallelogram bisect each other.

So, the coordinates of the mid-point of AC = Coordinates of the mid-point of BD

*Therefore, the value of x is 4, and the value of y is 2.*

**24. The points A(2, 0), B(9, 1), C(11, 6) and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not. **

**Solution: **

Given points are A(2, 0), B(9, 1), C(11, 6) and D(4, 4).

Coordinates of mid-point of AC are (11+2/ 2, 6+0/ 2) = (13/2, 3)

Coordinates of mid-point of BD are (9+4/ 2, 1+4/ 2) = (13/2, 5/2)

As the coordinates of the mid-point of AC ≠ coordinates of the mid-point of BD, ABCD is not even a parallelogram.

*Therefore, ABCD cannot be a rhombus too.*

**25. In what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3,-8)?**

**Solution: **

Let the point (-4, 6) divide the line segment AB in the ratio k: 1.

So, using the section formula, we have

The same can be checked for the y-coordinate also.

*Therefore, the ratio in which the point (-4, 6) divides the line segment AB is 2: 7*

**26. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also, find the coordinates of the point of division. **

**Solution: **

Let P(5, -6) and Q(-1, -4) be the given points.

Let the y-axis divide the line segment PQ in the ratio k: 1.

Then, by using the section formula for the x-coordinate (as it’s zero), we have

Thus, the ratio in which the y-axis divides the given 2 points is 5: 1.

Now, to find the coordinates of the point of division,

Putting k = 5, we get

*Hence, the coordinates of the point of division are (0, -13/3).*

**27. Show that A(-3, 2), B(-5, 5), C(2, -3) and D(4, 4) are the vertices of a rhombus. **

**Solution: **

Given points are A(-3, 2), B(-5, 5), C(2, -3) and D(4, 4).

Now,

Coordinates of the mid-point of AC are (-3+2/ 2, 2-3/ 2) = (-1/2, -1/2).

And,

Coordinates of mid-point of BD are (-5+4/ 2, -5+4/ 2) = (-1/2, -1/2)

Thus, the mid-point for both diagonals is the same. So, ABCD is a parallelogram.

Next, the sides

It’s seen that ABCD is a parallelogram with adjacent sides equal.

*Therefore, ABCD is a rhombus.*

**28. Find the lengths of the medians of a ΔABC having vertices at A(0, -1), B(2, 1) and C(0, 3).**

**Solution: **

Let AD, BE and CF be the medians of ΔABC.

Then,

Coordinates of D are (2+0/ 2, 1+3/ 2) = (1, 2)

Coordinates of E are (0/2, 3-1/ 2) = (0, 1)

Coordinates of F are (2+0/ 2, 1-1/ 2) = (1, 0)

Now, the length of the medians

**29. Find the lengths of the median of a ΔABC having vertices at A(5, 1), B(1, 5) and C(-3, -1).**

**Solution:**

Given vertices of ΔABC as A(5, 1), B(1, 5) and C(-3, -1).

Let AD, BE and CF be the medians.

Coordinates of D are (1-3/ 2, 5-1/ 2) = (-1, 2)

Coordinates of E are (5-3/ 2, 1-1/2) = (1, 0)

Coordinates of F are (5+1/ 2, 1+5/ 2) = (3, 3)

Now, the length of the medians

**30. Find the coordinates of the point which divide the line segment joining the points (-4, 0) and (0, 6) in four equal parts. **

**Solution:**

Let A(-4, 0) and B(0, 6) be the given points

And let P, Q and R be the points which divide AB is four equal points.

Now, we know that AP: PB = 1: 3

Using the section formula, the coordinates of P are

And it’s seen that Q is the mid-point of AB

So, the coordinates of Q are

Finally, the ratio of AR: BR is 3: 1

Then by using the section formula, the coordinates of R are

### RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.4 Page No: 14.37

**1. Find the centroid of the triangle whose vertices are:**

**(i) (1, 4), ( -1, -1) and (3, -2) (ii) (-2, 3), (2, -1) and (4, 0)**

**Solution:**

We know that the coordinates of the centroid of a triangle whose vertices are

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) are

(i) So, the coordinates of the centroid of a triangle whose vertices are

(1, 4), (-1, -1) and (3, -2) are

(1, 1/3)

Thus, the centroid of the triangle is (1, 1/3).

(ii) So, the coordinates of the centroid of a triangle whose vertices are

(-2, 3), (2, -1) and (4, 0).

(4/3, 2/3)

*Thus, the centroid of the triangle is (4/3, 2/3)*

**2. Two vertices of a triangle are (1, 2), (3, 5), and its centroid is at the origin. Find the coordinates of the third vertex.**

**Solution:**

Let the coordinates of the third vertex be (x, y).

Then, we know that the coordinates of the centroid of the triangle are

Given that the centroid for the triangle is at the origin (0, 0).

⇒ x + 4 = 0 ⇒ y + 7 = 0

⇒ x = -4 ⇒ y = -7

*Therefore, the coordinates of the third vertex are (-4, -7).*

**3. Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin. **

**Solution: **

Let the coordinates of the third vertex be (x, y).

Then, we know that the coordinates of the centroid of the triangle are

Given that the centroid for the triangle is at the origin (0, 0).

⇒ x – 3 = 0 ⇒ y – 1 = 0

⇒ x = 3 ⇒ y = 1

*Therefore, the coordinates of the third vertex are (3, 1).*

**4. A(3, 2) and B(-2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates (5/3, -1/3). Find the coordinates of the third vertex C of the triangle. **

**Solution: **

Let the coordinates of the third vertex C be (x, y).

Given, A(3, 2) and B(-2, 1) are two vertices of a triangle ABC.

Then, we know that the coordinates of the centroid of the triangle are

Given that the centroid for the triangle is **(**5/3, -1/3).

⇒ x + 1 = 5 ⇒ y + 3 = -1

⇒ x = 4 ⇒ y = -4

*Therefore, the coordinates of the third vertex C are (4, -4)*

**5. If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of its centroid. **

**Solution: **

Let A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) be the vertices of triangle ABC.

Let D (-2, 3), E (4, -3) and F (4, 5) be the mid-points of sides BC, CA and AB, respectively.

As D is the mid-point of BC

……. (1)

Similarly, E and F are the mid-points of AC and AB.

…….. (2)

And,

…… (3)

From (1), (2) and (3), we have

…….. (4)

Forms (1) and (4), we get

*Thus, the coordinates of A are (10, -1)*

From (2) and (4), we get

*Thus, the coordinates of B are (-2, 11).*

From (3) and (4), we get

*Thus, the coordinates of C are (-2, -5).*

Hence, the vertices of triangle ABC are A (10, -1), B (-2, 11) and C (-2, -5).

Therefore, the coordinates of the centroid of triangle ABC are

### RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.5 Page No: 14.53

**1. Find the area of a triangle whose vertices are**

**(i) (6, 3), (-3, 5) and (4, – 2)**

**(ii) [(at _{1}^{2}, at_{1}),( at_{2}^{2}, 2at_{2})( at_{3}^{2}, 2at_{3})]**

**(iii) (a, c + a), (a, c) and (-a, c – a)**

**Solution:**

(i) Let A(6, 3), B(-3, 5) and C(4,-2) be the given points

We know that the area of a triangle is given by

1/2[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1 }+ y_{2})]

Here,

x_{1 }= 6, y_{1 }= 3, x_{2} = -3, y_{2 }= 5, x_{3 }= 4, y_{3} = -2

So,

Area of ∆ABC = 1/2 [6(5+2)+(-3)(- 2 -3)+ 4(3 – 5)]

=1/2 [6 × 7- 3 × ( – 5) + 4( – 2)]

= 1/2[42 +15 – 8]

*= 49/2 sq. units*

(ii) Let A = (x_{1}, y_{1}) = (at_{1}^{2}, 2at_{1}), B = (x_{2},y_{2}) = (at_{2}^{2}, 2at_{2}), C= (x_{3}, y_{3}) = (at_{3}^{2}, 2at_{2}) be the given points.

Then,

The area of ∆ABC is given by

(iii) Let A = (x_{1},y_{1}) = (a, c + a), B = (x_{2}, y_{2}) = (a, c) and C = (x_{3}, y_{3}) = (- a, c **– **a) be the given points

Then,

The area of ∆ABC is given by

= 1/2[a ( – {c – a}) + a(c – a – (c + a)) +( – a)(c + a – a)]

= 1/2 [a(c – c + a) + a(c – a – c – a) – a(c + a – c)]

= 1/2[a × a + ax( – 2a) – a × a]

= 1/2[a^{2 }– 2a^{2 }– a^{2}]

= 1/2×(-2a)^{2 }

*= – a ^{2}*

**2. Find the area of the quadrilaterals, the coordinates of whose vertices are**

**(i) (-3, 2), (5, 4), (7, -6) and (-5, – 4)**

**(ii) (1, 2), (6, 2), (5, 3) and (3, 4)**

**(iii) (-4, -2), (-3, -5), (3, -2), (2, 3)**

**Solution:**

(i)

Let A(-3, 2), B(5, 4), C(7,- 6) and D ( -5, – 4) be the given points.

The area of ∆ABC is given by

= 1/2[-3(4 + 6) + 5(- 6 – 2) + 7(2 – 4)]

= 1/2[-3×1 + 5×(-8) + 7(-2)]

= 1/2[- 30 – 40 -14]

= – 42

As the area cannot be negative,

The area of ∆ADC = 42 square units

Now, the area of ∆ADC is given by

= 1/2[-3( – 6 + 4) + 7(- 4 – 2) + (- 5)(2 + 6)]

= 1/2[- 3( – 2) + 7(- 6) – 5 × 8]

= 1/2[6 – 42 – 40]

= 1/2 × – 76

= – 38

But, as the area cannot be negative,

The area of ∆ADC = 38 square units

Thus, the area of quadrilateral ABCD = Ar. of ABC+ Ar. of ADC

= (42 + 38)

*= 80 sq. units*

(ii)

Let A(1, 2), B (6, 2), C (5, 3) and (3, 4) be the given points.

Firstly, the area of ∆ABC is given by

= 1/2[1(2 – 3) + 6(3 – 2) + 5(2 – 2)]

= 1/2[ -1 + 6 × (1) + 0]

= 1/2[ – 1 + 6]

= 5/2

Now, the area of ∆ADC is given by

= 1/2[1(3 – 4) + 5(4 – 2) + 3(2 – 3)]

= 1/2[-1 × 5 × 2 + 3(-1)]

= 1/2[-1 + 10 – 3]

= 1/2[6]

= 3

Thus, the area of quadrilateral ABCD = Area of ABC + Area of ADC

(iii)

Let A (- 4, 2), B( – 3, – 5), C (3,- 2) and D(2, 3) be the given points.

Firstly, the area of ∆ABC is given by

= 1/2|(- 4)(- 5 + 2) – 3(-2 + 2) + 3(- 2 + 5)|

= 1/2|(-4)(-3) – 3(0) + 3(3)|

= 21/2

Now, the area of ∆ACD is given by

= 1/2|( – 4)(3 + 2) + 2( – 2 + 2) + 3( – 2 – 3)|

= 1/2|- 4(5) + 2(0) + 3(- 5)|= (- 35)/2

But, as the area can’t be negative,

The area of ∆ADC = 35/2

Thus, the area of quadrilateral (ABCD) = ar(∆ABC) + ar(∆ADC)

= 21/2 + 35/2

= 56/2

*= 28 sq. units*

**3. The four vertices of a quadrilateral are (1, 2), (-5, 6), (7, -4) and (k, -2), taken in order. If the area of the quadrilateral is zero, find the value of k. **

**Solution: **

Let A(1, 2), B(-5, 6), C(7, -4) and D(k, -2) be the given points

Firstly, the area of ∆ABC is given by

= 1/2|(1)(6 + 4) – 5(-4 + 2) + 7(2 – 6)|

= 1/2|10 + 30 – 28|

= ½ x 12

= 6

Now, the area of ∆ACD is given by

= 1/2|(1)(-4 + 2) + 7( – 2 – 2) + k(2 + 4)|

= 1/2|- 2 + 7x(-4) + k(6)|

= (- 30 + 6k)/2

= -15 + 3k

= 3k – 15

Thus, the area of quadrilateral (ABCD) = ar(∆ABC) + ar(∆ADC)

= 6 + 3k – 15

= 3k – 9

But, the given area of the quadrilateral is O.

So, 3k – 9 = 0

*k = 9/3 = 3*

**4. The vertices of ΔABC are (-2, 1), (5, 4) and (2, -3), respectively. Find the area of the triangle and the length of the altitude through A. **

**Solution:**

Let A(-2, 1), B(5, 4) and C(2, -3) be the vertices of ΔABC.

And let AD be the altitude through A.

The area of ΔABC is given by

= 1/2|(-2)(4 + 3) – 5(-3 – 1) + 2(1 – 4)|

= 1/2|-14 – 20 – 6|

= ½ x -40

= -20

But as the area cannot be negative,

The area of ΔABC = 20 sq. units

Now,

We know that the area of a triangle is

= ½ x Base x Altitude

20 = ½ x √58 x AD

AD = 40/ √58

*Therefore, the altitude AD = 40/ √58*

**5. Show that the following sets of points are collinear. **

**(a) (2, 5), (4, 6) and (8, 8) (ii) (1, -1), (2, 1) and (4, 5)**

**Solution: **

Condition: For the 3 points to be collinear, the area of the triangle formed with the 3 points has to be zero.

(a) Let A(2, 5), B(4, 6) and C(8, 8) be the given points.

Then, the area of ΔABC is given by

*Since the area (ΔABC) = 0, the given points (2, 5), (4, 6) and (8, 8) are collinear.*

(b) Let A(1, -1), B(2, 1) and C(4, 5) be the given points

Then, the area of ΔABC is given by

*Since the area (ΔABC) = 0, the given points (1, -1), (2, 1) and (4, 5) are collinear.*

**6. Find the area of a quadrilateral ABCD, the coordinates of whose vertices are A (-3, 2), B (5, 4), C (7, 6) and D (-5, -4).**

**Solution: **

Let’s join AC. So, we have 2 triangles formed.

Now, the ar (ABCD) = Ar (ΔABC) + Ar (ΔACD)

The area of ΔABC is given by

Next, the area of ΔACD is given by

*Thus, the area (ABCD) = 42 + 38 = 80 sq. units*

**7. In ⧍ABC, the coordinates of vertex A are (0, -1) and D(1, 0) and E(0, 1), respectively, the mid-points of the sides AB and AC. If F is the mid-point of side BC, find the area of ⧍DEF.**

**Solution: **

Let B(a, b) and C(p, q) be the other two vertices of the ⧍ABC.

Now, we know that D is the mid-point of AB.

So, coordinates of D = (0+a/ 2, -1+b/ 2)

(1, 0) = (a/2, b-1/2)

1 = a/2 and 0 = (b-1)/ 2

a = 2 and b = 1

Hence, the coordinates of B = (2, 1)

And, now

E is the mid-point of AC.

So, coordinates of E = (0+p/ 2, -1+q/ 2)

(0, 1) = (p/2 , (q -1)/ 2)

p/2 = 0 and 1 = (q – 1)/2

p = 0 and 2 = q -1

p = 0 and q = 3

Hence, the coordinates of C = (0, 3)

Again, F is the mid-point of BC.

Coordinates of F = (2+0/ 2, 1+3/ 2) = (1, 2)

Thus, the area of ⧍DEF is given by

**8. Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2).**

**Solution: **

Let the coordinates of P and R be (x_{1}, y_{1}) and (x_{2}, y_{2}), respectively.

And, let points E and F be the mid-points of PQ and QR, respectively.

x_{1 }+ 3 = 2, y_{1} + 2 = 4 and x_{2} + 3 = 4, y_{2 }+ 2 = -2

x_{1 }= -1, y_{1} = 2 and x_{2} = 1, y_{2} = -4

Hence, the coordinates of P and R are (-1, 2) and (1, 0), respectively.

Therefore, the area of ⧍PQR is given by

**9. If P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.**

**Solution:**

First, let’s join P and R.

Then,

The area of ⧍PSR is given by

And, now

The area of ⧍PQR is given by

Thus,

The area of the quad. PQRS = Area of ⧍PSR + Area of ⧍PQR

= 35/2 + 21/2

= 56/2

*= 28 sq. units*

**10. If A (-3, 5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.**

**Solution:**

Let’s join A and C.

So, we get ⧍ABC and ⧍ADC

Hence,

The area of the quad. ABCD = Area of ⧍ABC + Area of ⧍ADC

*Therefore, the area of the quadrilateral ABCD is 72 sq. units.*

**11. For what value of the points (a, 1), (1, -1) and (11, 4) are collinear?**

**Solution:**

Let A (a, 1), B (1, -1) and C (11, 4) be the given points.

Then the area of ⧍ABC is given by

We know that for the points to be collinear, the area of ⧍ABC has to be zero.

½(-5a + 25) = 0

5a = 25

*∴ a = 5*

**12. Prove that the points (a, b), (a _{1}, b_{1}) and (a-a_{1}, b-b_{1}) are collinear if ab_{1} = a_{1}b**

**Solution: **

Let A (a, b), B (a_{1}, b_{1}) and C (a-a_{1}, b-b_{1}) be the given points.

So, the area of ⧍ABC is given by

So, only if ab_{1} = a_{1}b, the area becomes zero.

⧍ABC = ½ (0) = 0

*Therefore, the given points are collinear if ab _{1} = a_{1}b*

**13. If the vertices of a triangle are (1,-3), (4,p) and (-9, 7) and its area is 15 sq. units, find the value (s) of p.**

**Solution: **

Let A(1,-3), B(4,p) and C(-9, 7) be the vertices of ⧍ABC.

Area of ⧍ABC = 15 sq. units

When the modulus is removed, two cases arise:

**14. If (x, y) be on the line joining the two points (1, -3) and (-4, 2). Prove that x + y + 2 = 0**

**Solution:**

Let A (x, y), B (1, -3) and C (-4, 2) be the given points.

The area of ⧍ABC is given by,

The three points lie on the same line (that means they are collinear).

Then, the area of ⧍ABC = 0

½ (-5x – 5y – 10) = 0

-5x – 5y – 10 = 0

-5(x + y + 2) = 0

x + y + 2 = 0

Hence, proved.

**15. Find the value of k if points (k, 3), (6, -2) and (-3, 4) are collinear. **

**Solution: **

Let A (k, 3), B (6, -2) and C (-3, 4) be the given points.

Then, the area of ⧍ABC is given by,

The points are collinear.

The area of ⧍ABC has to be zero.

½ x (-6k – 9) = 0

-6k – 9 = 0

k = -9/6

*∴ k = -3/2*

**16. Find the value of k, if points A(7, -2), B(5, 1) and C(3, 2k) are collinear. **

**Solution: **

Given,

Points A(7, -2), B(5, 1) and C(3, 2k)

Then, the area of ⧍ABC is given by,

The points are collinear.

The area of ⧍ABC has to be zero.

½ (-4k + 8) = 0

-4k + 8 = 0

-4k = -8

*∴ k = 2*

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