# Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

You are going to go through Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics. This post presents to the students a clear conception of how to move with the basics of Extra Questions and answers. The expert prepared The Extra Questions and And Answers. https://cbsencertanswers.com/is very much to make things way simpler and easier for the students. Especially those who are appearing for the board exams. We took every care to make sure that the effort serves the purpose. So, let us find out Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE MathematicsOn this page, you can find Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics.

## Solution.

(i) P (5, 0) lies on x – axis.

(ii) Q (0, -2) lies on y – axis (negation half).

(iii) R (-4, 0) lies on x – axis (negative half).

(iv) S (0, 5) lies on y – axis.

## Solution.

Let the given points be P (- 6, 7) and Q (- 1, – 5) . Here, x1 = – 6, y1 = 7 and x2 = -1, y2 = – 5 .

### YOU ARE READING: Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

So, the distance PQ is 13 which connects point P and Q.

## Solution.

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

Now we know, PR = QR .Then,

⇒ x+ 5 – 4x + y2 –2y = x+ 5 – 2x + y+ 4y

⇒ x+ 5 – 4x + y2 – 2y = x+ 5 – 2x + y+ 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.

## Solution.

It is given that, Length of the line segment is 10 units and 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.

### YOU ARE READING: Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

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

## Solution.

Given that Points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2).  It is required to prove that 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,

It’s observed that the opposite sides of the quadrilateral formed by the given four points are equal i.e. (AB = CD) & (DA = BC) and also, the diagonals BD & AC are found unequal. Hence Proved.

## Solution.

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

### YOU ARE READING: Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

As all the sides are equal the triangle is an equilateral triangle. Thus, the given vertices are of an equilateral triangle.

## 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.

## Solution.

Let A (2, 3) and B (-4, 1) be the given points. Let the point on y – axis equidistant from the above points be C (0, y). Now, we have

AC = √[(0 – 2)2 + (y – 3)2] = √[y2 – 6y + 9 + 4] = √[y2 – 6y + 13]

And,

BC = √[(0 – (-4))2 + (y – 1)2] = √y2 – 2y + 1 + 16] = √[y2 – 2y + 17]

As , AC = BC (given  in the condition)

So,  AC2 = BC2

y2 – 6y + 13 = y2 – 2y + 17

-4y = 4

y = -1

### YOU ARE READING: Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

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

## Solution.

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

So, PA = PB

PA2 = PB2

(3 -0)2 + (k -2)2 = (k – 0)2 + (5 – 2)2

9 + k2 + 4 – 4k – k2 – 9 = 0

4 – 4k = 0

-4k = -4

Therefore, the value of k = 1

## Solution.

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

So, PA2 = PB2

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

x2 + 49 – 14x + 36 = x2 + 9 + 6x + 16

-20x = -60

x = 3

### YOU ARE READING: Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

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

## Solution.

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

So, we have

Length of AB = √[(0 – (-2))2 + (1 – 5)2] = √[(2)2 + (-4)2] = √[4 + 16] = √20 = 2√5 units

Length of  BC = √[(2 – 0)2 + (-3 – 1)2] = √[(2)2 + (-4)2] = √[4 + 16] = √20 = 2√5 units

Length of AC = √[(2 – (-2))2 + (-3 – 5)2] = √[(4)2 + (-8)2] = √[16 + 64] = √80 = 4√5 units

Now, it is noticed 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.

## Solution.

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

So, the mid-point might be the required point

(x, y) = ( (-5 – 1/ 2), (4 + 6)/2 ) = (-6/2 , 10/2) = (-3, 5)

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

Now,

We also know that, AP = BP

So, AP2 = BP2

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

x2 + 25 + 10 + y2 – 8y + 16 = x2 + 2x + 1 + y2 – 12y + 36

10x + 41 – 8y = 2x + 37 – 12y

8x + 4y + 4 = 0

2x + y + 1 = 0

### YOU ARE READING: Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

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

## Solution.

Let P(x, y) be the required point. By section formula, we know that the coordinates are.

Here, x1 = – 1 y1 = 3 and x2  = 4 y2 = -7 , m: n = 3: 4

Then,

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

## Solution.

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

As, 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 :

## 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

Therefore , the value of k is 22 and value of length AB is 2√61 units.

## 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 value of k, we get the x – coordinate as:

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

### YOU ARE READING: Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

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

## Solution.

Let the coordinates of point A be (x, y). If AB is the diameter, then the center 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).

## Solution.

We know that the coordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2), (x3, y3)

So, the coordinates of the centroid of a triangle whose vertices are  (-2, 3), (2, -1) and (4, 0) is:

### YOU ARE READING: Coordinate Geometry Chapter 7 Extra Questions and Solutions For Class 10 CBSE Mathematics

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

## Solution.

Let the coordinates of the third vertex be (x, y). Then, we know that the coordinates of 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 is (3, 1).

## Solution.

Condition for  3 points to be collinear the area of the triangle formed with the 3 points has to be zero.  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.

.