Coordinate Geometry Exercise 14.1 – Class 10

Coordinate Geometry Exercise 14.1 – Questions:

1. Find the slope of the curve whose inclination is

(i) 90˚

(ii) 45˚

(iii) 30˚

(iv) 0˚

2. Find the angles of inclination of straight lines whose slopes are

(i)√3

(ii) 1

(iii) ¹/_√3

3. Find the slope of the line joining the points

(i) (-4, 1) and (-5, 2)

(ii) 4, -8) and (5, -2)

(iii) (0, 0) and (√3, 3)

(iv) (-5, 0) and (0, -7)

(v) (2a, 3b) and (a, -b)

4. Find whether the lines drawn through the two pairs of points are parallel or perpendicular

(i)(5, 2), (0, 5) and (0, 0), (-5, 3)

(ii) (3, 3), (4, 6) and (4, 1), (6, 7)

(iii) (4, 7), (3, 5) and (-1, 7), (1, 6)

(iv) (-1, -2), (1, 6) and (-1, 1), (-2, -3)

5. Find the slope of the line perpendicular to the line joining the points

(i)(1, 7) and (-4, 3)

(ii) (2, -3) and (1, 4)

6. Find the slope of the line parallel to the line joining the points

(i) (-4, 3) and (2, 5)

(ii) (1, -5) and (7, 1)

7. A line passing through the points (2, 7) and (3, 6) is parallel to a line joining (9, a) and (11, 3). Find a.
8. A line joining through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0). Find the value of m.

 

---

Coordinate Geometry Exercise 14.1 – Solutions:

1. Find the slope of the curve whose inclination is

(i) 90˚

Solution:

θ = 90˚

Slope = m = tan θ

m = tan 90˚

m = undefined

 

 (ii) 45˚

Solution:

θ = 45˚

Slope = m = tan θ

m = tan 45˚

m = 1

 

 (iii) 30˚

Solution:

θ = 30˚

Slope = m = tan θ

m = tan 30˚

m = ¹/_√3

 

 (iv) 0˚

Solution:

θ = 0˚

Slope = m = tan θ

m = tan 0˚

m = 0

 

2. Find the angles of inclination of straight lines whose slopes are

(i)√3

Solution:

Slope = tan θ

√3 = tanθ

We know that, tan 60˚ = √3

θ = 60˚

 

 (ii) 1

Solution:

Slope = tan θ

1 = tan θ

We know that, tan 45˚ = 1

θ = 45˚

 

 (iii) ¹/_√3

Solution:

Slope = tan θ

¹/_√3 = tanθ

We know that, tan 30˚ = ¹/_√3

θ = 30˚

 

3. Find the slope of the line joining the points

(i) (-4, 1) and (-5, 2)

Solution:

Let (x₁ ,y₁) = (-4, 1) and (x₂ ,y₂) (-5, 2)

Slope = ^y2-y1/_x2-x1 = ^2-1/_-5-(-4) = ¹/_-5+4 = -1

 

 (ii) 4, -8) and (5, -2)

Solution:

Let (x₁ ,y₁) = (4, -8) and (x₂ ,y₂) = (5, -2)

Slope = ^y2-y1/_x2-x1 = ^-2-(-8)/_5-4 = ^-2+8/₁ = 6

 

 (iii) (0, 0) and (√3, 3)

Solution:

Let (x₁ ,y₁) = (0, 0) and (x₂ ,y₂) = (√3, 3)

Slope = ^y2-y1/_x2-x1 = ^3-0/_√3-0 = ³/_√3 = √3

 

 (iv) (-5, 0) and (0, -7)

Solution:

Let (x₁ ,y₁) = (-5, 0) and (x₂ ,y₂) = (0, -7)

Slope = ^y2-y1/_x2-x1 = ^-7-0/_0-(-5) = ^-7/₅

 

 (v) (2a, 3b) and (a, -b)

Solution:

Let (x₁ ,y₁) = (2a, 3b) and (x₂ ,y₂) = (a, -b)

Slope = ^y2-y1/_x2-x1 = ^-b-3b/_a-2a = ^-4b/_-a = ^4b/_a

 

4. Find whether the lines drawn through the two pairs of points are parallel or perpendicular

(i)(5, 2), (0, 5) and (0, 0), (-5, 3)

Solution:

For line 1:

Let (x₁ ,y₁) = (5, 2) and (x₂ ,y₂)= (0, 5)

Slope = m₁ = ^y2-y1/_x2-x1 = ^5-2/_0-5 = ³/_-5

For line 2:

Let (x₁ ,y₁) = (0, 0) and (x₂ ,y₂)= (-5, 3)

Slope = m₂ = ^y2-y1/_x2-x1 = ^3-0/_-5-0 = ³/_-5

We have m₁ = m₂

Therefore, (5, 2), (0, 5) and (0, 0), (-5, 3) lines are parallel.

 

 (ii) (3, 3), (4, 6) and (4, 1), (6, 7)

Solution:

For line 1:

Let (x₁ ,y₁) = (3, 3) and (x₂ ,y₂)= (4, 6)

Slope = m₁ = ^y2-y1/_x2-x1 = ^6-3/_4-3 = ³/₁ = 3

For line 2:

Let (x₁ ,y₁) = (4, 1) and (x₂ ,y₂)= (6, 7)

Slope = m₂ = ^y2-y1/_x2-x1 = ^7-1/_6-4 = ⁶/₂ = 3

We have m₁ = m₂

Therefore, (3, 3), (4, 6) and (4, 1), (6, 7) parallel.

 

 (iii) (4, 7), (3, 5) and (-1, 7), (1, 6)

Solution:

For line 1:

Let (x₁ ,y₁) = (4, 7) and (x₂ ,y₂)= (3, 5)

Slope = m₁ = ^y2-y1/_x2-x1 = ^5-7/_3-4 = ^-2/_-1 = 2

For line 2:

Let (x₁ ,y₁) = (-1, 7) and (x₂ ,y₂)= (1, 6)

Slope = m₂ = ^y2-y1/_x2-x1 = ^6-7/_1-(-1) = ^-1/₂

We have m₁ ≠ m₂

m₁.m₂ = 2 x (¹/_-2) = – 1

Therefore, (4, 7), (3, 5) and (-1, 7), (1, 6) are perpendicular.

 

 (iv) (-1, -2), (1, 6) and (-1, 1), (-2, -3)

Solution:

For line 1:

Let (x₁ ,y₁) = (-1, -2) and (x₂ ,y₂)= (1, 6)

Slope = m₁ = ^y2-y1/_x2-x1 = ^6-(-2)/_1-(-1) = ⁸/₂ = 4

For line 2:

Let (x₁ ,y₁) = (-1, 1) and (x₂ ,y₂)= (-2, -3)

Slope = m₂ = ^y2-y1/_x2-x1 = ^-3-1/_-2-(-1) = ^-4/_-1 = 4

We have m₁ = m₂

Therefore, (3, 3), (4, 6) and (4, 1), (6, 7) are parallel.

 

5. Find the slope of the line perpendicular to the line joining the points

(i)(1, 7) and (-4, 3)

Solution:

Let (x₁ ,y₁) = (1, 7) and (x₂ ,y₂)= (-4, 3)

Slope = m₁ = ^y2-y1/_x2-x1 = ^3-7/_-4-1 = ^-4/_-5 = ⁴/₅

The slope of the line perpendicular to the line joining points (1, 7) and (-4, 3) is  ^-5/₄.[since m₁.m₂ = -1]

 

 (ii) (2, -3) and (1, 4)

Solution:

Let (x₁ ,y₁) = (2, 3) and (x₂ ,y₂)= (1, 4)

Slope = m₁ = ^y2-y1/_x2-x1 = ^4-3/_1-2 = ¹/_-1 = -1

The slope of the line perpendicular to the line joining points (2, -3) and (1, 4) is 1. [Since m₁.m₂ = -1]

 

6. Find the slope of the line parallel to the line joining the points

(i) (-4, 3) and (2, 5)

Solution:

Let (x₁ ,y₁) = (-4, 3) and (x₂ ,y₂)= (2, 5)

Slope = m₁ = ^y2-y1/_x2-x1 = ^5-3/_2-(-4) = ²/₆ = ¹/₃

The slope of the line parallel to the line joining points (-4, 3) and (2, 5) is ¹/₃.

 

 (ii) (1, -5) and (7, 1)

Solution:

Let (x₁ ,y₁) = (1, -5) and (x₂ ,y₂)= (7, 1)

Slope = m₁ = ^y2-y1/_x2-x1 = ^1-(-5)/_7-1 = ⁶/₆ = 1

The slope of the line parallel to the line joining points (1, -5) and (7, 1) is 1.

 

7. A line passing through the points (2, 7) and (3, 6) is parallel to a line joining (9, a) and (11, 3). Find a.

Solution:

Line 1:

Let (x₁ ,y₁) = (2, 7) and (x₂ ,y₂)= (3, 6)

Slope = m₁ = ^y2-y1/_x2-x1 = ^6-7/_3-2 = ^-1/₁ = -1

Line 2:

Let (x₁ ,y₁) = (9, a) and (x₂ ,y₂)= (11, 3)

Slope = m₂ = ^y2-y1/_x2-x1 = ^3-a/_11-9 = ^3-a/₂

Since m₁ = m₂ as line (2, 7) and (3, 6) is parallel to (9, a) and (11, 3).

We have, -1 = ^3-a/₂

-2 = 3 – a

-2 – 3 = – a

– 5 = -a

a = 5

 

8. A line joining through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0). Find the value of m.

Solution:

Line 1:

Let (x₁ ,y₁) = (1, 0) and (x₂ ,y₂)= (4, 3)

Slope = m₁ = ^y2-y1/_x2-x1 = ^3-0/_4-1 = ³/₃ = 1

Line 2:

Let (x₁ ,y₁) = (-2, -1) and (x₂ ,y₂)= (m, 0)

Slope = m₂ = ^y2-y1/_x2-x1 = ^0-(-1)/_m-(-2) = ¹/_m+2

Since m_1.m₂ = -1 as A line joining through the points (1, 0) and (4, 3) is perpendicular to the line joining (-2, -1) and (m, 0).

We have, m₁.m₂ = -1

1.(¹/_m+2) = -1

¹/_m+2 = -1

1 = -(m+2)

1 = -m – 2

-m = 1 + 2

m = -3