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### Practice Set 5.1 Co-ordinate Geometry Class 10th Mathematics Part 2 MHB Solution

Practice Set 5.1

1. Find the distance between each of the following pairs of points. (1) A(2, 3), B(4, 1)…
2. Determine whether the points are collinear. (1) A(1, -3), B(2, -5), C(-4, 7) (2) L(-2,…
3. Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).…
4. Verify that points P(-2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled…
5. Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a…
6. Show that points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are vertices of a rhombus…
7. Find x if distance between points L(x, 7) and M(1, 15) is 10.
8. 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)

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
y= 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)

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).

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 =

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.

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.

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.

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 =  =

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.

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.

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.