NCERT Solutions for Class 10 Maths Exercise 7.4 Chapter 7 Coordinates Geometry

NCERT exercise Solutions includes input from the best Maths teachers of the country who are part of BYJU’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 first and second term examination students must practice different kinds of questions. Exercise questions are provided here to make you solve a variety of questions.

Maths is a volatile subject, if you leave practicing you will lose hold on formulas and methodology to solve the problems. Hence students are advised to solve this NCERT exercises often to get comprehended with questions

Topics covered in Exercise 7.4

  1. Section formula
  2. Solved examples

How it is helpful

  • Has solution to long answer questions provided in Exercise 7.4.
  • Solving NCERT exercises will provide you unlimited practice.
  • Gain knowledge on methodology to solve problems on area of triangle.

Download PDF of NCERT Solutions for Class 10 Maths Chapter 7- Coordinate Geometry Exercise 7.4

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Access Other Exercise Solutions of Class 10 Maths Chapter 7- Coordinate Geometry

Exercise 7.1– 10 Questions (8 practical based questions, 2 reasoning questions)

Exercise 7.2– 10 Questions (8 long answer questions, 2 short answer questions)

Exercise 7.3– 5 Questions (3 long answer questions, 2 practical based questions)

Access Answers to NCERT Class 10 Maths Chapter 7 –Coordinate Geometry Exercise 7.4

Exercise 7.4 Page No: 171

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

Solution:

Consider line 2x + y – 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 – 4 = 0, we have

2{(2 + 3k)/(k + 1)} + {(-2 + 7k)/(k + 1)} – 4 = 0

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

4 + 6k – 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 area of triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

Area of a triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

[x(2 – 0) + 1 (0 – y) + 7( y – 2)] = 0

2x – y + 7y – 14 = 0

2x + 6y – 14 = 0

x + 3y – 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) are the points on a circle.

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

If O = (x, y) then

OA = √[(x – 6)+ (y + 6)2]

OB = √[(x – 3)+ (y + 7)2]

OC = √[(x – 3)+ (y – 3)2]

Choose: OA = OB, we have

After simplifying above, we get -6x = 2y – 14 ….(1)

Similarly: OB = OC

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

(y + 7)2 = (y – 3)2

y2 + 14y + 49 = y2 – 6y + 9

20y =-40

or y = -2

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

-6x = 2y – 14

-6x = -4 – 14 = -18

x = 3

Hence, the centre of the circle 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: Coordinate of points B and D.

NCERT Solutions for Class 10 Chapter 7-29

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

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

AC = √[(3 + 1)+ (2 – 2)2] = 4

Coordinates of O can be calculated as follows:

x = (3 – 1)/2 = 1 and y = (2 + 2)/2 = 2

So, O(1,2)

Step 2: Find the side of the square using Pythagoras theorem

Let a be the side of square and AC = 4

From right triangle, ACD,

a = 2√2

Hence, each side of square = 2√2

Step 3: Find coordinates of point D

Equate length measure of AD and CD

Say, if coordinate of D are (x1, y1)

AD = √[(x1 + 1)+ (y1 – 2)2]

Squaring both sides,

AD2 = (x1 + 1)+ (y1 – 2)2

Similarly, CD2 = (x1 – 3)+ (y1 – 2)2

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

(x1 + 1)+ (y1 – 2)2 = (x1 – 3)+ (y1 – 2)2

x12 + 1 + 2x1 = x12 + 9 – 6x1

8x1 = 8

x1 = 1

Value of y1 can be calculated as follows by using the value of x.

From step 2: each side of square = 2√2

CD2 = (x1 – 3)+ (y1 – 2)2

8 = (1 – 3)+ (y1 – 2)2

8 = 4 + (y1 – 2)2

y1 – 2 = 2

y1 = 4

Hence, D = (1, 4)

Step 4: Find 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 = (x2, y2)

For BD;

1 = (x2 + 1)/2

x2 = 1

And 2 = (y2 + 4)/2

=> y2 = 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 the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.

(i) Taking A as 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?

NCERT Solutions for Class 10 Chapter 7-30

Solution:

(i) Taking A as origin, 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 origin,

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 formula: Area of a triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

= ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

= ½ (- 12 – 3 + 24 )

= 9/2 sq unit

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

Area of a triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

= ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

= ½ ( 36 + 13 – 40)

= 9/2 sq unit

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

Area is same in both case because triangle remains the same no matter which point is considered as origin.

6. The vertices of a ∆ 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 ∆ ADE and compare it with area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)

Solution:

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

NCERT Solutions for Class 10 Chapter 7- 31

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 ratio 1 : 3.

Coordinates of D can be calculated as follows:

x = (m1x2 + m2x1)/(m1 + m2) and y = (m1y2 + m2y1)/(m1 + m2)

Here m1 = 1 and m2 = 3

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

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

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

Similarly, 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 Area of triangle:

Using formula: Area of a triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of triangle ∆ ABC can be calculated as follows:

= ½ [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]

= ½ (12 – 4 + 7) = 15/2 sq unit

Area of ∆ ADE can be calculated as follows:

= ½ [4(23/4 – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4)]

= ½ (3 – 13/4 + 19/16)

= ½ ( 15/16 ) = 15/32 sq unit

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

7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ 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 point 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 the three medians is called the centroid

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

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.

Solution:

NCERT Solutions for Class 10 Chapter 7-32

(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)

Point Q and P would be coincident because medians of a triangle intersect each other at a common point called 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 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 same which shows that medians intersect each other at a common point, i.e. centroid of the triangle.

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, the coordinates of centroid can be given as follows:

x = (x1 + x2 + x3)/3 and y = (y1 + y2 + y3)/3

8. ABCD is a rectangle formed by the points A (-1, – 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:

NCERT Solutions for Class 10 Chapter 7-33

P id the mid-point of side AB,

Coordinate of P = ( (-1 – 1)/2, (-1 + 4)/2 ) = (-1, 3/2)

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

Coordinate of Q = (2, 4)

Coordinate of R = (5, 3/2)

Coordinate of S = (2, -1)

Now,

Length of PQ = √[(-1 – 2)2 + (3/2 – 4)2] = √(61/4) = √61/2

Length of SP = √[(2 + 1)2 + (-1 – 3/2)2] = √(61/4) = √61/2

Length of QR = √[(2 – 5)2 + (4 – 3/2)2] = √(61/4) = √61/2

Length of RS = √[(5 – 2)2 + (3/2 + 1)2] = √(61/4) = √61/2

Length of PR (diagonal) = √[(-1 – 5)2 + (3/2 – 3/2)2] = 6

Length of QS (diagonal) = √[(2 – 2)2 + (4 + 1)2] = 5

The above values show that, PQ = SP = QR = RS = √61/2, i.e. all sides are equal.

But PR ≠ QS i.e. diagonals are not of equal measure.

Hence, the given figure is a rhombus.


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