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:

.....(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.

.....(1)

Length AB = = = 4

Length BC = = = 4

Length AC = = = 4

Hence The given points form an equilateral triangle.