Class 8 Maths Solution | Lesson 17 Coordinates | CDC

Exercise 17.1

Distance between two points

2. Find the value of x from the following right-angled triangles.

(a)

Given:

p = x

b = 12 cm

h = 13 cm

According to the Pythagoras Theorem:

h² = p² + b²

or, 13² = x² + 12²

or, 169 = x² + 144

or, x² = 25

∴ x = 5 cm

(b)

Given:

p = 14 cm

b = 48 cm

h = x cm

According to the Pythagoras Theorem:

h² = p² + b²

or, x² = 14² + 48²

or, x² = 196 + 2304

or, x² = 2500

∴ x = 50 cm

(c)

Given:

p = 21 cm

b = 72 cm

h = x

According to the Pythagoras Theorem:

h² = p² + b²

or, x² = 21² + 72²

or, x² = 441 + 5184

or, x² = 5625

∴ x = 75 cm

(d)

Given:

p = 6 m

b = 8 m

h = x

According to the Pythagoras Theorem:

h² = p² + b²

or, x² = 6² + 8²

or, x² = 36 + 64

or, x² = 100

∴ x = 10 m

(e)

Given:

p = x

b = 15 ft

h = 25 ft

According to the Pythagoras Theorem:

h² = p² + b²

or, 25² = x² + 15²

or, 625 = x² + 225

or, x² = 400

∴ x = 20 ft

(f)

Given:

p = x

b = √5 cm

h = √8 cm

According to the Pythagoras Theorem:

h² = p² + b²

or, (√8)² = x² + (√5)²

or, 8 = x² + 5

or, x² = 3

∴ x = √3 cm

(g)

Given:

p = 15 cm

b = 8 cm

h = x

According to the Pythagoras Theorem:

h² = p² + b²

or, x² = 15² + 8²

or, x² = 225 + 64

or, x² = 289

∴ x = 17 cm

(h)

Given:

p = 24 cm

b = 7 cm

h = x

According to the Pythagoras Theorem:

h² = p² + b²

or, x² = 24² + 7²

or, x² = 576 + 49

or, x² = 625

∴ x = 25 cm

1. Find the distance between two points given below:

(a) (4, -7) and (-1, 5)

Solution:

x₁ = 4, x₂ = -1, y₁ = -7, y₂ = 5

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √{(-1) - 4}² + {5 - (-7)}²

= √(-5)² + (12)²

= √(25 + 144)

= √169

= 13 units

(b) (-3, 4) and (4, 3)

Solution:

x₁ = -3, x₂ = 4, y₁ = 4, y₂ = 3

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √{4 - (-3)}² + (3 - 4)²

= √(7)² + (-1)²

= √(49 + 1)

= √50

= 5√2 units

(c) (1, -2) and (5, -6)

Solution:

x₁ = 1, x₂ = 5, y₁ = -2, y₂ = -6

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √(5 - 1)² + {(-6) - (-2)}²

= √(4)² + (-4)²

= √(16 + 16)

= √32

= 4√2 units

(d) (1, 7) and (1, 1)

Solution:

x₁ = 1, x₂ = 1, y₁ = 7, y₂ = 1

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √(1 - 1)² + (1 - 7)²

= √(0 + (-6)²

= √36

= 6 units

(e) (2, 7) and (4, 9)

Solution:

x₁ = 2, x₂ = 4, y₁ = 7, y₂ = 9

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √(4 - 2)² + (9 - 7)}²

= √(2)² + (2)²

= √(4 + 4)

= √8

= 2√2 units

(f) (-8, 7) and (-3, 4)

Solution:

x₁ = -8, x₂ = -3, y₁ = 7, y₂ = 4

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √{(-3) - (-8)}² + {4 - (7)}²

= √(5)² + (-3)²

= √(25 + 9)

= √34

= 5.83 units

5. If the point A intersects X-axis at -8 and point B intersects Y-axis at 6, find the distance of AB.

Solution:

x₁ = -8, x₂ = 0, y₁ = 0, y₂ = 6

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √{0 - (-8)}² + {6 - 0}²

= √(8)² + (6)²

= √(64 + 36)

= √100

= 10 units

6. The position of cat and mouse are given in the graph paper. Find the coordinates of their position and find the distance between them.

Let the coordinates of the cat be (-5, 0) and the mouse be (5, 5).

Solution:

x₁ = -5, x₂ = 5, y₁ = 0, y₂ = 5

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √{5 - (-5)}² + {5 - (0)}²

= √(10)² + (5)²

= √(100 + 25)

= √125

= 5√5 units

7. Prove that the points A (-4, 0), B (-4, -4), C (2, -4), and D (2, 0) are the vertices of a rectangle.

Solution:

The distances between the points:

AB = √{(-4) - (-4)}² + {(-4) - 0}² = 4 units

BC = √{2 - (-4)}² + {(-4) - (-4)}² = 6 units

CD = √(2 - 2)² + {0 - (-4)}² = 4 units

DA = √{(-4) - 2}² + (0 - 0)² = 6 units

Since opposite sides are equal, ABCD forms a rectangle.

8. When presented on the maps, if the coordinates of A and B are (4,7) and (7,3) respectively, what is the distance between these two points on the map? If 1 unit is equal to 55 km, find the actual distance between A and B.

Solution:

x₁ = 4, x₂ = 7, y₁ = 7, y₂ = 3

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

Distance = √(7 - 4)² + (3 - 7)²

= √(3)² + (-4)²

= √(9 + 16)

= √25

= 5 units

Actual distance = 5 × 55 = 275 km

9. Prove that the points P (1, 6), Q (4, 1), and R (-4, 3) are the vertices of an isosceles triangle.

Solution:

Here, P (1, 6) and Q (4, 1)

x₁ = 1, x₂ = 4, y₁ = 6, y₂ = 1

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

PQ = √(4 - 1)² + (1 - 6)²

= √(3)² + (-5)²

= √(9 + 25)

= √34 units

Q (4, 1) and R (-4, 3)

x₁ = 4, x₂ = -4, y₁ = 1, y₂ = 3

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √{(-4) - 4}² + (3 - 1)²

= √(-8)² + (2)²

= √(64 + 4)

= √68 units

P (1, 6) and R (-4, 3)

x₁ = 1, x₂ = -4, y₁ = 6, y₂ = 3

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √{(-4) - 1}² + (3 - 6)²

= √(-5)² + (-3)²

= √(25 + 9)

= √34 units

Since PQ = PR = √34, two sides are equal.

Thus, PQR is an isosceles triangle.

10. If the points A (2, -1), B (3, 4), C (-2, 3), and D (-3, -2) are the vertices of a rhombus ABCD, then find the distance of its diagonals AC and BD.

Solution:

A (2, -1) and C (-2, 3)

x₁ = 2, x₂ = -2, y₁ = -1, y₂ = 3

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

AC = √{(-2) - 2}² + {3 - (-1)}²

= √(-4)² + (4)²

= √(16 + 16)

= √32

= 4√2 units

B (3, 4), D (-3, -2)

x₁ = 3, x₂ = -3, y₁ = 4, y₂ = -2

BD = √{(-3) - 3}² + {(-2) - 4}²

= √(-6)² + (-6)²

= √(36 + 36)

= √72

= 6√2 units

Thus, the lengths of the diagonals AC and BD are 4√2 units and 6√2 units, respectively.

11. If the points P (9, 12) and Q (1, 6) are on the circumference of the circle, what will be the radius of that circle? Does the point (-7, 0) lie on the circumference?

Solution:

Finding the radius of the circle using the distance formula between P and Q:

The distance between P and Q (Diameter of the circle):

P (9, 12), Q (1, 6)

x₁ = 9, x₂ = 1, y₁ = 12, y₂ = 6

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √(1 - 9)² + (6 - 12)²

= √(-8)² + (-6)²

= √(64 + 36)

= √100

= 10 units

Since PQ is the diameter, the radius of the circle is: 10 / 2 = 5 units

Now, checking if the point (-7, 0) lies on the circumference.

Finding the distance from (-7, 0) to the center of the circle (midpoint of PQ).

Midpoint of PQ:

= (9 + 1)/2 , (12 + 6)/2 )

= (10/2 , 18/2)

= (5, 9)

Distance from (-7, 0) to (5, 9):

= √{5 - (-7)}² + (9 - 0)²

= √(12)² + (9)²

= √(144 + 81)

= √225

= 15 units

Since 15 ≠ 5, the point (-7, 0) does not lie on the circumference of the circle.

12. Find the distance from the origin O to point A and point B, where A = (-7, 7) and B = (7, -7).

Solution:

Here,

O (0, 0) and A (-7, 7)

x₁ = 0, x₂ = -7, y₁ = 0, y₂ = 7

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

OA = √{(-7) - 0}² + (7 - 0}²

= √(-7)² + 7²

= √(49 + 49)

= √98

= 7√2 units

Now,

O (0, 0) and B (7, -7)

x₁ = 0, x₂ = 7, y₁ = 0, y₂ = -7

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

OB = √{7 - 0}² + {(-7) - 0}²

= √(7)² + (-7)²

= √(49 + 49)

= √98

= 7√2 units

Thus, the distances from the origin O to points A and B are both 7√2 units.

13. If the distance between P (0, 6) and Q (a, 0) is 6 units, what is the value of a?

Solution:

P (0, 6) and Q (a, 0)

x₁ = 0, x₂ = a, y₁ = 6, y₂ = 0 and distance= 6 units

We know that,

Distance = √(x₂ - x₁)² + (y₂ - y₁)²

= √(a - 0)² + (0 - 6)²

= √a² + (-6)²

= √a² + 36

= √a² + 36

∴ Distance = √a² + 36

Now,

6 = √a² + 36

Squaring both sides:

36 = a² + 36

or, a² = 36 - 36

or, a² = 0

a = 0

Thus, the value of a is 0.