Exercise 7.1
Multiple Choice Questions (Q. No. 1-9)
Question. 1 The distance of the point P (2,3) from the x-axis is
(a) 2    
(b) 3    
(c) 1    
(d) 5
Answer. (b)   
Explanation:
Suppose P (a, b) is any point on the cartesian plane in first quadrant as shown below:

Then,
• a is the perpendicular distance from the y-axis
• b is the perpendicular distance from the x-axis
So, distance of the point P(2, 3) from the x-axis = 3
Distance of the point P(2, 3) from the y-axis = 2.
Question. 2 The distance between the points A(0, 6) and B (0, ‒2) is
(a) 6
(b) 8
(c) 4
(d) 2
Answer. (b)   
Explanation:

Question. 3 The distance of the point P(‒6, 8) from the origin is

Question. 4 The distance between the points (0, 5) and (-5, 0) is
(a) 5
(b)
(c)            
(d) 10
Answer. (b)
Explanation:

Question. 5 If AOBC is a rectangle whose three vertices are A (0, 3), O(0, 0) and B(5, 0), then the length of its diagonal is

Answer. (c)
Explanation:

Question. 6 The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

Answer. (b)   
Explanation:

Distance between A (0, 4) and O (0, 0) + Distance between O (0, 0) and B (3, 0) + Distance between A (0, 4) and B (3, 0) = 3 + 4 + 5 = 12.
Question. 7 The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
(a) 14
(b) 28
(c) 8
(d) 6
Answer. (c)    
Explanation:

Question. 8 The points (−4, 0), (4, 0) and (0, 3) are the vertices of a
(a) right angled triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Answer. (b)   
Explanation:

Question. 9 The point which divides the line segment joining the points (7, -6) and (3, 4) in ratio 1 : 2 internally lies in the
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant
Answer. (d)   
Explanation:

Exercise 7.1
Multiple Choice Questions (Q. No. 10-18)
Question. 10 The point which lies on the perpendicular bisector of the line segment joining the points A (-2, -5) and B (2, 5)is
(a) (0, 0)         
(b) (0, 2)
(c) (2, 0)         
(d) (-2, 0)
Answer. (a)   
Explanation:
The perpendicular bisector of any line segment always passes through the mid-point of that line segment.

Hence, (0, 0) is the required point which lies on the perpendicular bisector of the line segment joining the points A (-2, -5) and B (2, 5).
Question. 11 The fourth vertex D of a parallelogram ABCD whose three vertices are A (-2, 3), B (6, 7) and C (8, 3) is
(a) (0, 1)         
(b) (0, -1)
(c) (-1, 0)
(d) (1, 0)
Answer. (b)   
Explanation:
Suppose the co-ordinate of fourth vertex or point D of parallelogram be(x₄, y₄)and P, Q be the middle points of AC and BD, respectively.

Question. 12 If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then

Answer. (d)   
Explanation:
A rough diagram of the point P (2, 1) on the line segment joining the points A (4, 2) and B (8, 4) is shown in the figure given below:

Question. 13 If P (a/3, 4) is the mid-point of the line segment joining the points Q (-6, 5) and R (-2, 3), then the value of a is
(a) – 4
(b) -12
(c) 12
(d) -6
Answer. (b)   
Explanation:
Given that, P (a/3, 4) is the mid-point of the line segment joining the points Q(-6, 5) and R (-2, 3).

Question. 14 The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the Y-axis at
(a) (0, 13)
(b) (0, -13)
(c) (0, 12)
(d) (13, 0)
Answer. (a)
Explanation:

It is given that the perpendicular bisector of the line segment is perpendicular on the line segment.
If two lines having slope m₁ and m₂ are perpendicular to each other, then,

Putting x = 0 in above equation we get:
            3 × 0 + y = 13
⟹        y = 13.
Hence, the required point is (0, 13).
Question. 15 The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is

Answer. (a)   
Explanation:
Form the given figure, coordinates of three vertices of a triangle are: O (0, 0), A (0, 2y), B (2x, 0).
Suppose the required point be P whose coordinates are (h, k)
Now, P is equidistant from the three vertices of DAOB, therefore, PO = PA = PB or (PO)² = (PA)² = (PB)²
By using distance formula, we get:

Question. 16 If a circle drawn with origin as the centre passes through (13/2, 0), then the point which does not lie in the interior of the circle is



Question. 17 A line intersects the Y-axis and X-axis at the points P and Q, respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(a) (0, -5) and (2, 0)
(b) (0, 10) and (-4, 0)
(c) (0, 4) and (-10, 0)
(d) (0, -10) and (4, 0)
Answer. (d)   
Explanation:
Suppose the coordinates of P are(0, y) and Q are (x, 0).
The given situation can be represented by the following diagram:

Question. 18 The area of a triangle with vertices (a, b + c), (b, c + a)and (c, a + b)is
(a) (a + b + c)²
(b) 0
(c) (a + b + c)
(d) abc
Answer. (b)
Explanation:

Exercise 7.2
Very Short Answer Type Questions (Q. No. 1-6)
Write whether True or False and justify your answer
Question. 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:

Here, we see that sides of ΔABC and ΔFDE are proportional
Therefore, by SSS similarity rule both the triangles are similar.        
Question. 2 The point P (-4, 2) lies on the line segment joining the points A (-4, 6) and B (-4, -6).
Solution. True
Explanation:
If we plot all the points P (-4, 2), A (-4, 6) and B (-4, -6) on the graph paper then, we will find, point P (-4, 2) lies on the line segment joining the points A (-4, 6) and B (-4, -6),

Question. 3 The points (0, 5), (0, -9) and (3, 6) are collinear.
Solution. False
Given, x₁ = 0, x₂ = 0 , x₃ = 3 and  y₁ = 5, y₂ = −9, y₃ = 6
Three points will be collinear if the area of triangle formed by these three points is zero.
Now, area of triangle formed by three points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given as,            

As the area of triangle formed by the points (0, 5), (0, ‒ 9) and (3, 6) is non-zero, hence the points are non-collinear.
Question. 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
For P(0, 2) to be the point of intersection of y–axis and perpendicular bisector of the line joining the points A(–1, 1) and B(3, 3), P must be equidistant from A and B.
i.e., PA = PB

Question. 5 The points A (3, 1), B (12, -2) and C (0, 2) cannot be vertices of a triangle.
Solution. True
For any three points to be the vertices of a triangle, the area of that triangke must be non-zero.
Now, area of triangle formed by three points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given as,            

Since, area of ΔABC = 0, therefore, the points, A (3, 1), B (12, ‒2) and C (0, 2) are collinear and can’t be the vertices of a triangle.

Question. 6 The points A (4, 3), B (6, 4), C (5, – 6) and D (-3, 5) are vertices of a parallelogram.
Solution. False
Given points will be the vertices of a parallelogram, if the opposite sides of that parallelogram come out to be equal.
By using distance formula, distance between A (4, 3) and B (6, 4), 

In parallelogram, opposite sides are equal. But, in this case all sides AB, BC, CD and DA are different.
Therefore, given coordinates are not the vertices of a parallelogram.
Exercise 7.2
Very Short Answer Type Questions (Q. No. 7-12)
Write whether True or False and justify your answer
Question. 7 A circle has its centre at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle.
Solution. True
Point Q (6, 8) will lie outside the circle if its distance from the centre of circle is greater than the radius of that circle.

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

Question. 9 The point P(5, -3) is one of the two points of trisection of line segment joining the points A(7, -2) and B(1, -5).
Solution. True
Suppose, P (5, -3) divides the line segment joining the points A (7, -2) and B (1, -5) in the ratio m : 1internally.

So the point P divides the line segment AB in ratio 1 : 2. Therefore, point P in the point of trisection of AB.
Question. 10 The points A(–6,10), B(-4, 6) and C(3, -8) are collinear such that AB =  AC.
Solution. True
For points A(–6,10), B(-4, 6) and C(3, -8) to be collinear, the area of triangle formed by these points must be zero.
Now, area of triangle formed by three points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given as,            
Here, x₁ = -6, x₂ = -4, x₃ = 3 and y₁ = 10, y₂ = 6, y₃ = -8

Question. 11 The point P(-2, 4) lies on a circle of radius 6 and centre (3, 5).
Solution. False
The point P(-2, 4) will lie on given circle if its distance from centre (3, 5) equal to the radius of this circle.

Question. 12 The points A (-1, -2), B (4, 3), C (2, 5) and D (-3, 0) in that order form a rectangle.
Solution. True
Given points will be the vertices of a parallelogram, if the opposite sides and diagonals of that rectangle come out to be equal.


As, diagonals AC and SD are equal, therefore, the points A (-1, -2), B (4, 3), C (2, 5) and D (-3, 0) form a rectangle.
Exercise 7.3
Short Answer Type Questions (Q. No. 1-7)
Question. 1 Name the type of triangle formed by the points A (-5, 6), B (-4, -2) and C (7, 5).
Solution.


Question. 3 What type of quadrilateral do the points A (2, -2), B (7, 3), C (11, -1) and D (6, -6)
taken in that order form?
Solution.        
To find the type of quadrilateral, we will first find the length of all four sides and diagonals of
the quadrilateral formed by the points A (2, -2), B (7, 3), C (11, -1) and D (6, -6).

Question. 4 Find the value of a, if the distance between the points A (-3, -14) and B (a, -5) is 9 units.
Solution.        
Given, distance between A (-3, -14) and B (a, -5) is, AB = 9 units
By distance formula,

Question. 5 Find a point which is equidistant from the points A (- 5, 4) and B (- 1, 6). How
many such points are there?
Solution.                   
Let, M (h, k) be the point which is equidistant from the points A (-5, 4) and B (-1, 6).

Point (-3, 5) satisfies equation (i).
So, all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the points A and B.
Question. 6 Find the coordinates of the point Q on the X-axis which lies on the perpendicular bisector of the line segment joining the points A (- 5, -2) and B (4, -2). Name the type of triangle formed by the point Q, A and B.
Solution.                   
Situation given in question is shown in the figure given below,

As, the perpendicular bisector of the line segment AB bisects the segment AB, hence it will pass through the mid-point of AB i.e., R

Clearly BQ = QA ≠ AB, which means that the triangle formed by the points Q, A and B is an isosceles.
Question. 7 Find the value of m, if the points (5, 1), (- 2, – 3) and (8, 2m) are collinear.
Solution.                   

Exercise 7.3
Short Answer Type Questions (Q. No. 8-14)
Question. 8 If the point A (2, – 4) is equidistant from P (3, 8) and Q (-10, y), then find the value of y. Also, find distance PQ.
Solution:
Given, A (2, -4) is equidistant from P (3, 8) and Q(-10, y)
⟹          PA = QA     ….(i)
Since distance between two points (x₁, y₁) and (x₂, y₂) is given as,

Question. 9 Find the area of the triangle whose vertices are (-8, 4), (-6, 6) and (-3, 9)
Solution:

Question. 10 In what ratio does the X-axis divide the line segment joining the points (- 4, – 6) and (- 1, 7)? Find the coordinates of the points of division..
Solution:                  



Question. 12 If P (9a –2, –b)divides line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3:1, then find the values of a and b.
Solution:
Given, P (9a -2, –b) divides AB internally in the ratio 3:1.

Question. 13 If (a, b)is the mid-point of the line segment joining the points A (10, -6), B (k, 4) and a -2b = 18, then find the valueof k and thedistance AB.
Solution:
Given, (a, b) is the mid-point of line segment AB.
And, a -2b = 18                   ….(i)
As, mid-point of a line segment joining points



Exercise 7.3
Short Answer Type Questions (Q. No. 15-20)
Question. 15 The line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x -18y + k = 0. Find the value of k.
Solution:



Question. 17 If the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ΔABC right angled at B, then find the values of a and hence the area of ΔABC.
Solution:                  
As, the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ΔABC right angled at B.
By distance formula,




Question. 19 Find the values of k, if the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k − 1, 5k) are collinear.
Solution:
If three points are collinear, then the area of triangle formed by these points is zero.
Since, the points A (k + 1,2k), B (3k, 2k + 3) and C (5k – 1, 5k)are collinear, therefore area of ΔABC = 0

Question. 20 Find the ratio in which the line 2x+ 3y -5 = 0 divides the line segment joining the points (8, -9) and (2, 1). Also, find the coordinates of the point of division.
Solution:        
Let the line 2x + 3y -5 = 0 divides the line segment joining the points A (8, -9) and B (2, 1) in the ratio l: 1 at point P.

Exercise 7.4
Long Answer Type Questions:
Question. 1 If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, then find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
Solution.                   
Suppose the third vertex of an equilateral triangle be (x, y).
Let A (-4, 3), B (4, 3) and C (x, y).
In equilateral triangle all three sides are equal,


Question. 2 A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, then find the area of ΔADE.
Solution.                   
Here, A (6, 1), B(8, 2) and C (9, 4) are three vertices of a parallelogram ABCD.
Let the fourth vertex of parallelogram be D(x, y).
We know that, the diagonals of a parallelogram bisect each other.                                                              


Question. 3 The points A (x_1, y₁), B (x_2, y₂) and C (x₃, y₃) are the vertices of ΔABC.
(i) The median from A meets BC at D. Find the coordinates of the 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 are the coordinates of the centroid of the ΔABC?
Solution.
Here, coordinated of the points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ΔABC
are the vertices of ΔABC
(i) As, the median bisect the line segment into two equal parts i.e., here D is the mid-point of BC

(iii) Let the coordinates of a point Q be (p, q)



Question. 4 If the points A (1, -2),B (2, 3), C (a, 2) and D (-4, -3) form a parallelogram, then find the value of a and height of the parallelogram taking AB as base.
Solution.
In parallelogram, we know that diagonals are bisects each other i.e., mid-point of AC = mid-point of BD.