Practice Set 5.1
- Find the distance between each of the following pairs of points. (1) A(2, 3), B(4, 1)…
- Determine whether the points are collinear. (1) A(1, -3), B(2, -5), C(-4, 7) (2) L(-2,…
- Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).…
- Verify that points P(-2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled…
- Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a…
- Show that points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are vertices of a rhombus…
- Find x if distance between points L(x, 7) and M(1, 15) is 10.
- Show that the points A(1, 2), B(1, 6), C(1 + 2 √3, 4) are vertices of an equilateral…
Practice Set 5.1
Question 1.
Find the distance between each of the following pairs of points.
(1) A(2, 3), B(4, 1)
(2) P(-5, 7), Q(-1, 3)
(3) R(0, -3), S(0, 5/2)
(4) L(5, -8), M(-7, -3)
(5) T(-3, 6), R(9, -10)
(6) 
, X(11, 4)
Answer:
The distance between points A(x1, y1) and B(x2, y2) is given by,
1. Given Points: A(2, 3) and B(4, 1)
we can see that,
x1 = 2
x2 = 4
y1 = 3
y2 = 1
Putting the values in the distance formula we get,
d = 
⇒ d = 
⇒ d = √8
2. Given Points: P(-5, 7) and Q(-1, 3)
we can see that,
x1 = -5
x2 = -1
y1 = 7
y2 = 3
Putting these values in distance formula we get,
d = 
d = √32
3. Given Points: R(0, -3), S(0, 5/2)
we can see that,
x1 = 0
x2 = 0
y1 = -3
y2 = 5/2
On putting these values in distance formula we get,
d = 
d = 
d = 
4. Given Points: L(5, -8), M(-7, -3)
we can see that,
x1 = 5
x2 = -7
y1 = -8
y2 = -3
On putting these values in distance formula we get,
d = 
d = 
d = √169 = 13
5. Given Points: T(-3, 6), R(9, -10)
we can see that,
x1 = -3
x2 = 9
y1 = 6
y2 = -10
On putting these values in distance formula we get,
d = 
d = 
d = 20
6. Given Points: W(
), X(11, 4)
we can see that,
x1 = -7/2
x2 = 11
y1 = 4
y2 = 4
On putting these values in distance formula we get,
d = 
d = 
d = 
Question 2.
Determine whether the points are collinear.
(1) A(1, -3), B(2, -5), C(-4, 7)
(2) L(-2, 3), M(1, -3), N(5, 4)
(3) R(0, 3), D(2, 1), S(3, -1)
(4) P(-2, 3), Q(1, 2), R(4, 1)
Answer:
If Three points (a,b), (c,d), (e,f) are collinear then the area formed by the triangle by the three points is zero.
Area of a triangle = 
...(1)
1.
(a,b) = (1,-3)
(c,d) = (2,-5)
(e,f) = (-4,7)
Area = 
Area = 
= 0
Hence the points are collinear.
2.
(a,b) = (-2,3)
(c,d) = (1,-3)
(e,f) = (5,4)
Area = 
Area = 
Hence the points are not collinear.
3.
(a,b) = (0,3)
(c,d) = (2,1)
(e,f) = (3,-1)
Area = 
Area = 
Hence the points are non collinear.
4.
(a,b) = (-2,3)
(c,d) = (1,2)
(e,f) = (4,1)
Area = 
Area = 
Hence the points are collinear.
Question 3.
Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).
Answer:
A point in the x = axis is of the form (a,0)
Distance d between two points(a,b) and (c,d)is given by
Distance between (-3,4) and (a,0) =
D = 
D 
Distance between (1,-4) and (a,0)
D = 
D = 
As the two points are equidistant from the point (a.0)
= 
Squaring both sides, we get
(1-a)2 + 16 = (3 + a)2 + 16
1 + a2 -2a = 9 + a2 + 6a
8a = -8
a = -1
Hence the point is (-1,0)
Question 4.
Verify that points P(-2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.
Answer:
In a right angles triangle ABC, right angled at B, according to the pythagoras theorem
AB2 + BC2 = AC2
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by
.....(1)
For the given points Distance between P and Q is
PQ = 
= 
QR = 
= 
PR = 
= 
= 
PQ2 = 16
QR2 = 25
PR2 = 41
As PQ2 + QR2 = PR2
Hence the given points form a right angled triangle.
Question 5.
Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram.
Answer:
In a parallelogram, opposite sides are equal and parallel.
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by .....(1)
For the given points, length PQ = 
PQ = 
Length QR = 
QR = 
= 
Length RS = 
RS = 
= 
Length SP = 
SP = = 
As PQ = RS and QR = SP
Checking for slopes
Slope of a line between two points (a,b) and (c,d) is
Slope PQ = 
= 1
Slope QR = 
= 
Slope RS = 
Slope SP = 
As PQ = RS and their slope = 1
And
QR = SP and their slope = -1.
Hence the given points form a parallelogram.
Question 6.
Show that points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are vertices of a rhombus ABCD.
Answer:
In a Rhombus the sides are equal and the diagonals bisect each other at 90°
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by .....(1)
Length AB = 
= 
Length BC = 
=
Length CD = 
=
Length AD = 
=
Slope of a line between two points (a,b) and (c,d) is
Slope of Diagonal AC = 
= 1
Slope of diagonal BD = 
= -1
Note: If the Product of slopes of two lines = -1 then they are perpendicular to each other.
As the product of slopes pf two diagonals = -1. Hence they're perpendicular to each other.
Hence The given points form a rhombus.
Question 7.
Find x if distance between points L(x, 7) and M(1, 15) is 10.
Answer:
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by .....(1)
Distance between LM = 
= 10
Squaring both sides, we get
(x-1)2 + 64 = 100
(x-1)2 = 36
x-1 = ±6
Hence x = 7 or -5
Question 8.
Show that the points A(1, 2), B(1, 6), C(1 + 2 √3, 4) are vertices of an equilateral triangle.
Answer:
For an equilateral triangle, all its sides are equal.
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by .....(1)
Length AB = 
= 
= 4
Length BC = 
= 
= 4
Length AC = 
= = 4
Hence The given points form an equilateral triangle.
