**NCERT Exercise Solutions** include input from the best Maths teachers in the country who are part of BYJUâ€™S team. This solution has answers to the questions given in Exercise 7.4 of NCERT textbooks. To score good marks in the Class 10 board examination, students must practise different kinds of questions. Exercise questions are provided in NCERT Class 10 Maths Solutions to make students solve a variety of questions.

Maths is a complex subject; if students leave practising, they will lose hold of formulas and methodology to solve the problems. Hence, students are advised to solve these exercises using NCERT Solutions often to get a good grasp of the concepts.

## Topics Covered in Exercise 7.4

- Section formula
- Solved examples

### How Is It Helpful?

- Solutions have long answer questions provided in Exercise 7.4.
- Solving exercises using NCERT Class 10 Solutions will provide students with unlimited practice.
- Gain knowledge on methodology to solve problems in the area of triangles.

## NCERT Solutions for Class 10 Maths Chapter 7- Coordinate Geometry Exercise 7.4

### Access Other Exercise Solutions of Class 10 Maths Chapter 7 â€“ Coordinate Geometry

The other exercises of NCERT Class 10 Maths Chapter 7 Solutions can be accessed below.

Exercise 7.1â€“ 10 Questions (8 practical-based questions, 2 reasoning questions)

Exercise 7.2â€“ 10 Questions (8 long answer questions, 2 short answer questions)

Exercise 7.3â€“ 5 Questions (3 long answer questions, 2 practical-based questions)

### Access Answers to NCERT Class 10 Maths Chapter 7 â€“ Coordinate Geometry Exercise 7.4

## Exercise 7.4 Page No: 171

**1. Determine the ratio in which the line 2x + y â€“ 4 = 0 divides the line segment joining the points A(2, â€“2) and B(3, 7). **

**Solution:**

Consider line 2x + y â€“ 4 = 0 divides line AB, joined by the two points A(2, -2) and B(3, 7) in k : 1 ratio.

Coordinates of point of division can be given as follows:

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

Substituting the values of x and y given equation, i.e., 2x + y â€“ 4 = 0, we have

2{(2 + 3k)/(k + 1)} + {(-2 + 7k)/(k + 1)} â€“ 4 = 0

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

4 + 6k â€“ 2 + 7k = 4(k+1)

-2 + 9k = 0

Or k = 2/9

Hence, the ratio is 2:9.

**2. Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear. **

**Solution:**

If given points are collinear, then the area of the triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) be the vertices of a triangle.

Area of a triangle =Â 1/2 Ã— [x_{1}(y_{2} â€“ y_{3}) + x_{2}(y_{3} â€“ y_{1}) + x_{3}(y_{1} â€“ y_{2})] = 0

2x â€“ y + 7y â€“ 14 = 0

2x + 6y â€“ 14 = 0

x + 3y â€“ 7 = 0.

Which is the required result.

**3. Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).**

**Solution:**

Let A = (6, -6), B = (3, -7), C = (3, 3) be the points on a circle.

If O is the centre, then OA = OB = OC (radii are equal).

If O = (x, y) then

OA = âˆš[(x â€“ 6)^{2Â }+ (y + 6)^{2}]

OB =Â âˆš[(x â€“ 3)^{2Â }+ (y + 7)^{2}]

OC =Â âˆš[(x â€“ 3)^{2Â }+ (y â€“ 3)^{2}]

Choose, OA = OB; we have

After simplifying, we get -6x = 2y â€“ 14 â€¦.(1)

Similarly, OB = OC

(x â€“ 3)^{2 }+ (y + 7)^{2} = (x â€“ 3)^{2} + (y â€“ 3)^{2}

(y + 7)^{2} = (y â€“ 3)^{2}

y^{2 }+ 14y + 49 = y^{2 }â€“ 6y + 9

20y =-40

or y = -2

Substituting the value of y in equation (1), we get

-6x = 2y â€“ 14

-6x = -4 â€“ 14 = -18

x = 3

Hence, the centre of the circle is located at point (3,-2).

**4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.**

**Solution: **

Let ABCD is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD.

To find the coordinates of points B and D,

**Step 1: Find the distance between A and C and the coordinates of point O.**

We know that the diagonals of a square are equal and bisect each other.

AC =Â âˆš[(3 + 1)^{2Â }+ (2 â€“ 2)^{2}] = 4

Coordinates of O can be calculated as follows:

x = (3 â€“ 1)/2 = 1 and y = (2 + 2)/2 = 2

So, O(1,2)

**Step 2: Find the side of the square using Pythagorasâ€™ theorem.**

Let a be the side of the square, and AC = 4

From the right triangle ACD,

a = 2âˆš2

Hence, each side of square = 2âˆš2

**Step 3: Find the coordinates of point D.**

Equate the length measure of AD and CD.

Say, if the coordinates of D are (x_{1}, y_{1}).

AD =Â âˆš[(x_{1} + 1)^{2Â }+ (y_{1} â€“ 2)^{2}]

Squaring both sides,

AD^{2} = (x_{1} + 1)^{2Â }+ (y_{1} â€“ 2)^{2}

Similarly, CD^{2} =Â (x_{1}Â â€“ 3)^{2Â }+ (y_{1} â€“ 2)^{2}

Since all sides of a square are equal, which means AD = CD

(x_{1} + 1)^{2Â }+ (y_{1} â€“ 2)^{2} =Â (x_{1}Â â€“ 3)^{2Â }+ (y_{1} â€“ 2)^{2}

x_{1}^{2} + 1 + 2x_{1} = x_{1}^{2} + 9 â€“ 6x_{1}

8x_{1} = 8

x_{1} = 1

The value of y_{1} can be calculated as follows by using the value of x.

From step 2, each side of the square = 2âˆš2

CD^{2} =Â (x_{1}Â â€“ 3)^{2Â }+ (y_{1} â€“ 2)^{2}

8 =Â (1Â â€“ 3)^{2Â }+ (y_{1} â€“ 2)^{2}

8 = 4 + (y_{1} â€“ 2)^{2}

y_{1} â€“ 2 = 2

y_{1} = 4

Hence, D = (1, 4)

**Step 4: Find the coordinates of point B.**

From line segment BOD,

Coordinates of B can be calculated using coordinates of O as follows:

Earlier, we had calculated O = (1, 2)

Say B = (x_{2}, y_{2})

For BD;

1 = (x_{2} + 1)/2

x_{2 } = 1

And 2 = (y_{2} + 4)/2

=> y_{2} = 0

Therefore, the coordinates of required points are B = (1,0) and D = (1,4)

**5. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot, as shown in fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot. **

**(i) Taking A as the origin, find the coordinates of the vertices of the triangle. **

**(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?**

**Also, calculate the areas of the triangles in these cases. What do you observe? **

**Solution: **

(i) Taking A as the origin, the coordinates of the vertices P, Q and R are

From figure: P = (4, 6), Q = (3, 2), R (6, 5)

Here AD is the x-axis, and AB is the y-axis.

(ii) Taking C as the origin,

The coordinates of vertices P, Q and R are ( 12, 2), (13, 6) and (10, 3), respectively.

Here, CB is the x-axis, and CD is the y-axis.

Find the area of triangles:

Area of triangle PQR in case of origin A.

Using the formula Area of a triangle = 1/2 Ã— [x_{1}(y_{2} â€“ y_{3}) + x_{2}(y_{3} â€“ y_{1}) + x_{3}(y_{1} â€“ y_{2})]

= Â½ [4(2 â€“ 5) + 3 (5 â€“ 6) + 6 (6 â€“ 2)]

= Â½ (- 12 â€“ 3 + 24 )

= 9/2 sq unit

(ii) Area of triangle PQR in case of origin C.

Area of a triangle =Â 1/2 Ã— [x_{1}(y_{2} â€“ y_{3}) + x_{2}(y_{3} â€“ y_{1}) + x_{3}(y_{1} â€“ y_{2})]

= Â½ [ 12(6 â€“ 3) + 13 ( 3 â€“ 2) + 10( 2 â€“ 6)]

= Â½ ( 36 + 13 â€“ 40)

= 9/2 sq unit

This implies that the Area of triangle PQR at origin A = Area of triangle PQR at origin C

The area is the same in both cases because the triangle remains the same no matter which point is considered the origin.

**6. The vertices of âˆ† ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E, respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of the âˆ† ADE and compare it with the area of âˆ† ABC. (Recall Theorem 6.2 and Theorem 6.6)**

**Solution:**

Given: The vertices of âˆ† ABC are A (4, 6), B (1, 5) and C (7, 2).

AD/AB = AE/AC = 1/4

AD/(AD + BD) = AE/(AE + EC) = 1/4

Point D and Point E divide AB and AC, respectively, in the ratio 1 : 3.

Coordinates of D can be calculated as follows:

x = (m_{1}x_{2} + m_{2}x_{1})/(m_{1} + m_{2}) and y =Â (m_{1}y_{2} + m_{2}y_{1})/(m_{1} + m_{2})

Here m_{1} = 1 and m_{2} = 3

Consider line segment AB which is divided by point D at the ratio 1:3.

x = [3(4) + 1(1)]/4 = 13/4

y = [3(6) + 1(5)]/4 = 23/4

Similarly, the coordinates of E can be calculated as follows:

x = [1(7) + 3(4)]/4 = 19/4

y = [1(2) + 3(6)]/4 = 20/4 = 5

**Find the area of the triangle.**

Using the formula, Area of a triangle = 1/2 Ã— [x_{1}(y_{2} â€“ y_{3}) + x_{2}(y_{3} â€“ y_{1}) + x_{3}(y_{1} â€“ y_{2})]

Area of triangle âˆ† ABC can be calculated as follows:

= Â½ [4(5 â€“ 2) + 1( 2 â€“ 6) + 7( 6 â€“ 5)]

= Â½ (12 â€“ 4 + 7) = 15/2 sq unit

Area of âˆ† ADE can be calculated as follows:

= Â½ [4(23/4 â€“ 5) + 13/4 (5 â€“ 6) + 19/4 (6 â€“ 23/4)]

= Â½ (3 â€“ 13/4 + 19/16)

= Â½ ( 15/16 ) = 15/32 sq unit

Hence, the ratio of the area of triangle ADE to the area of the triangle ABC = 1 : 16.

**7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of âˆ† ABC. **

**(i) The median from A meets BC at D. Find the coordinates of point D. **

**(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1. **

**(iii) Find the coordinates of points Q and R on medians BE and CF, respectively, such that BQ : QE = 2:1 and CR : RF = 2 : 1.**

**(iv) What do you observe? **

**[Note: The point which is common to all three medians is called the centroid,**

**and this point divides each median in the ratio 2 : 1.]**

**(v) If A (x _{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.**

**Solution:**

(i) Coordinates of D can be calculated as follows:

Coordinates of D = ( (6+1)/2, (5+4)/2 ) = (7/2, 9/2)

So, D is (7/2, 9/2)

(ii) Coordinates of P can be calculated as follows:

Coordinates of P = ( [2(7/2) + 1(4)]/(2 + 1), [2(9/2) + 1(2)]/(2 + 1) ) = (11/3, 11/3)

So, P is (11/3, 11/3)

(iii) Coordinates of E can be calculated as follows:

Coordinates of E = ( (4+1)/2, (2+4)/2 ) = (5/2, 6/2) = (5/2 , 3)

So, E is (5/2 , 3)

Points Q and P would be coincident because the medians of a triangle intersect each other at a common point called the centroid. Coordinate of Q can be given as follows:

**Coordinates of Q** =( [2(5/2) + 1(6)]/(2 + 1), [2(3) + 1(5)]/(2 + 1) ) = (11/3, 11/3)

F is the mid-point of the side AB

Coordinates of F = ( (4+6)/2, (2+5)/2 ) = (5, 7/2)

Point R divides the side CF in the ratio 2:1

**Coordinates of R** = ( [2(5) + 1(1)]/(2 + 1), [2(7/2) + 1(4)]/(2 + 1) ) = (11/3, 11/3)

(iv) Coordinates of P, Q and R are the same, which shows that medians intersect each other at a common point, i.e., the centroid of the triangle.

(v) If A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of triangle ABC, the coordinates of the centroid can be given as follows:

x = (x_{1} + x_{2} + x_{3})/3 and y = (y_{1} + y_{2} + y_{3})/3

**8. ABCD is a rectangle formed by the points A (-1, â€“ 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA, respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.**

**Solution:**

P id the mid-point of side AB,

Coordinate of P = ( (-1 â€“ 1)/2, (-1 + 4)/2 ) = (-1, 3/2)

Similarly, Q, R and S are, (As Q is the mid-point of BC, R is the midpoint of CD and S is the midpoint of AD.)

Coordinate of Q = (2, 4)

Coordinate of R = (5, 3/2)

Coordinate of S = (2, -1)

Now,

Length of PQ = âˆš[(-1 â€“ 2)^{2} + (3/2 â€“ 4)^{2}]Â = âˆš(61/4) =Â âˆš61/2

Length of SP =Â âˆš[(2 + 1)^{2} + (-1 â€“ 3/2)^{2}]Â = âˆš(61/4) =Â âˆš61/2

Length of QR =Â âˆš[(2 â€“ 5)^{2} + (4 â€“ 3/2)^{2}]Â = âˆš(61/4) =Â âˆš61/2

Length of RS =Â âˆš[(5 â€“ 2)^{2} + (3/2 + 1)^{2}]Â = âˆš(61/4) =Â âˆš61/2

Length of PR (diagonal) =Â âˆš[(-1 â€“ 5)^{2} + (3/2 â€“ 3/2)^{2}]Â = 6

Length of QS (diagonal) = âˆš[(2 â€“ 2)^{2} + (4 + 1)^{2}] = 5

The above values show that PQ = SP = QR = RS = âˆš61/2, i.e., all sides are equal.

But PR â‰ QS, i.e., diagonals, are not of equal measure.

Hence, the given figure is a rhombus.

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