NCERT Exemplar Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2

Chapter 7 Coordinate Geometry NCERT Exemplar Solutions Exercise 7.2 Class 10 Maths

Chapter Name	NCERT Maths Exemplar Solutions for Chapter 7 Coordinate Geometry Exercise 7.2
Book Name	NCERT Exemplar for Class 10 Maths
Other Exercises	Exercise 7.1 Exercise 7.3 Exercise 7.4
Related Study	NCERT Solutions for Class 10 Maths

Exercise 7.2 Solutions

Short Answer Questions with Reasoning

State whether the following statements are true or false. Justify your answer.

1. ΔABC with vertices A (-2,0), B(2, 0) and C(0, 2) is similar to ΔDEF with vertices D(-4, 0), E(4, 0) and F(0, 4).

Solution

True. 
Explanation : 

By distance formula , 
d = √[(x₂ – x₁)² + (y₂ – y₁)² ]
we can find, 

Also,  

FD = 4√2 
So, 
AB/DE = BC/EF = CA/FD = 1/2
Hence, triangle ABC and DEF are similar.

2. Point P(-4, 2) lies on the line segment joining the points A(-4, 6) and B(-4, -6).

Solution

True, 
Explanation :

We will plot the points P(-4, 2), A(-4, 6) and B(-4, -6)on a graph paper and connect the points. 

So, from the graph it is clear that, point P(-4, 2) lies on the line segment joining the points A(-4, 6) and B(-4, -6)

3. The points (0, 5), (0, –9) and (3, 6) are collinear.

Solution

False 

Explanation,

The points are collinear if area of a triangle formed by its points is equals to the zero.
Given, 
x₁ = 0, 
x₂ = 0, 
x₃ =  3 and 
y₁ = 5, 
y₂ = -9, 
y₃ = 6 

Area of trianlge,
= ½ [ x₁(y₂ – y₃ ) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) ]
= 1/2 [0(-9 - 6) - 0(6 - 5) + 4(5 + 9)]
= (1/2) (42)
= 42/2 ≠ 0 
So, the points are not collinear. 
From the above equation, it is clear that the points are not collinear.

4. Point P (0, 2) is the point of intersection of y–axis and perpendicular bisector of line segment joining the points A (–1, 1) and B (3, 3).

Solution

False 
Explanation : 
The points lying on perpendicular bisector of the line segment joining the two points is equidistant from the two points. 
So, 
PA should be equals to the PB. 
Using distance formula, 
d = √[(x₂ – x₁)² + (y₂ – y₁)² ]

So, PA is not equal to PB.

5. Points A (3, 1), B (12, –2) and C (0, 2) cannot be the vertices of a triangle.

Solution

True, 
Explanation : 
Coordinates of A = (x₁ , y₁ )
= (3, 1)
Coordinates of B = (x₂ , y₂ )
= (12, -2)
Coordinates of C = (x₃ , y₃ )
= (0, 2) 
Area of ΔABC = ½ [ x₁(y₂ – y₃ ) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) ] 
= 1/2 [3 (-2-2) + 12(2 - 1) + 0{1 - (-2)}]
= 1/2 [3(-4) + 12(1) + 0(3)]
= 1/2 [-12 + 12 + 0]
= 1/2 × 0 = 0
Area of ΔABC = 0 
As, the points A(3, 1), B(12, -2) and C(0, 2) are collinear. 
So, the points A(3, 1), B(12, -2) and C(0, 2) can't be the vertices of a triangle.

6. Points A (4, 3), B (6, 4), C (5, –6) and D (–3, 5) are the vertices of a parallelogram.

Solution

False 
Explanation : 

The given points are A(4, 3), B(6, 4), C(5, -6) and D(-3, 5). 
We will find the distances AB, BC, CD and AD respectively by distance formula. 
We will see that, 
Every distance is different from one another. 
As the distances are different, we can conclude that the points are not the vertices of a parallelogram. 

7. A circle has its center at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle.

Solution

True 

At first, we draw a circle and a point 

Also, if the distance of any point from the centre is less than/equal to/ more than the radius, then the point is inside/on/outside the circle, respectively
OP = √[(5-0)² + (0 - 0)² ]
⇒ OP = √25
⇒ OP = 5 units 
which is radius of the circle 

and, 
OQ = √[(6-0)² + (8 - 0)² ]
⇒ OQ = √(100)
⇒ OQ = 10 units 

We see that, OQ > OP 
So, it is true that point Q(6, 8), lies outside the circle. 

8. The point A (2, 7) lies on the perpendicular bisector of line segment joining the points P (6, 5) and Q (0, – 4).

Solution

False 
If A(2, 7) lies on perpendicular bisector of P(6, 5) and Q(0, -4) then AP = AQ 
AP = √[(6-2)² + (5 - 7)² ]
⇒ AP = √(16 + 4)
⇒ AP = √20 units 
and, 
AQ = √[(0-2)² + (-4 - 7)² ]
⇒ AQ = √125 units 

Therefore, A does not lies on the perpendicular bisector of PQ. 

9. Point P(5, -3) is one of the two points of trisection of the line segment joining the point A(7, -2) and B(1, -5).

Solution

Let the point P(5, -3) divides the line segment joining the points A(7, -2) and B(1, -5) in the ratio k : 1 internally. 

By section formula, the coordinate of point P will be, 

So, 
(k + 7)/(k + 1) = 5 
⇒ k + 7 = 5(k + 1)
⇒ -4k = -2
⇒ k = 1/2 
The point P divides the line segment AB in ratio 1 : 2.
Therefore, point P in the point of trisection of AB. 

10. Points A(-6, 10), B(-4, 6) and C(3, -8) are collinear such that AB = (2/9)AC.

Solution

True 

If the area of triangle formed by the points (x₁ , y₁ ),  (x₂ , y₂ ) and  (x₃ , y₃ ) is zero, then the points are collinear, 

ar(ABC) = ½ [ x₁(y₂ – y₃ ) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) ] = 0 
= 1/2 [-6{6 - (-8)} -4(-8 - 10) + 3(10 - 6)] = 0 
= -84 + 84 = 0 
So, it is true, 
And the points are collinear. 

Also, 

AB = 2√5 units 

Now, 

AB/AC = 2√5/9√5
⇒ AB = (2/9) AC 

Which is the required relation .

11. The point P (-2, 4) lies on a circle of radius 6 and centre C(3, 5).

Solution

False 

If the distance between the centre and any point is equal to the radius, then we say that point lie on the circle. 

So, distance between P(-2, 4) and centre (3, 5)
Distance  = √[(3 + 2)² + (5 - 4)²]
= √26

It is not equal to the radius of the circle. 

So, the point P(-2, 4) does not lies on the circle. 

12. The points A(-1, -2), B(4, 3) , C(2, 5) and D(-3, 0) in that order form a rectangle. 

Solution

True,

Distance between A(-1, -2) and B(4, 3) is : 

As, opposite sides and diagonals AC and BD are equal.

Hence, the points A(-1 , -2), B(4, 3), C(2, 5) and D(-3, 0) form a rectangle.