**Coordinate Geometry Exercise 14.1 – Questions:**

**Find the slope of the curve whose inclination is**

**(i) 90˚**

**(ii) 45˚**

**(iii) 30˚**

**(iv) 0˚**

**Find the angles of inclination of straight lines whose slopes are**

**(i)√3**

**(ii) 1**

**(iii) ^{1}/_{√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)**

**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)**

**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) **

**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)**

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

**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

**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˚

**Find the slope of the line joining the points**

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

Solution:

Let (x_{1} ,y_{1}) = (-4, 1) and (x_{2} ,y_{2}) (-5, 2)

Slope = ^{y2-y1}/_{x2-x1} = ^{2-1}/_{-5-(-4)} = ^{1}/_{-5+4} = -1

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

Solution:

Let (x_{1} ,y_{1}) = (4, -8) and (x_{2} ,y_{2}) = (5, -2)

Slope = ^{y2-y1}/_{x2-x1} = ^{-2-(-8)}/_{5-4} = ^{-2+8}/_{1} = 6

** (iii) (0, 0) and (√3, 3)**

Solution:

Let (x_{1} ,y_{1}) = (0, 0) and (x_{2} ,y_{2}) = (√3, 3)

Slope = ^{y2-y1}/_{x2-x1} = ^{3-0}/_{√3-0} = ^{3}/_{√3 }= √3

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

Solution:

Let (x_{1} ,y_{1}) = (-5, 0) and (x_{2} ,y_{2}) = (0, -7)

Slope = ^{y2-y1}/_{x2-x1} = ^{-7-0}/_{0-(-5)} = ^{-7}/_{5}

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

Solution:

Let (x_{1} ,y_{1}) = (2a, 3b) and (x_{2} ,y_{2}) = (a, -b)

Slope = ^{y2-y1}/_{x2-x1} = ^{-b-3b}/_{a-2a} = ^{-4b}/_{-a} = ^{4b}/_{a}

**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 (x_{1} ,y_{1}) = (5, 2) and (x_{2} ,y_{2})= (0, 5)

Slope = m_{1 }= ^{y2-y1}/_{x2-x1} = ^{5-2}/_{0-5} = ^{3}/_{-5}

For line 2:

Let (x_{1} ,y_{1}) = (0, 0) and (x_{2} ,y_{2})= (-5, 3)

Slope = m_{2 }= ^{y2-y1}/_{x2-x1} = ^{3-0}/_{-5-0} = ^{3}/_{-5}

We have m_{1} = m_{2}

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 (x_{1} ,y_{1}) = (3, 3) and (x_{2} ,y_{2})= (4, 6)

Slope = m_{1 }= ^{y2-y1}/_{x2-x1} = ^{6-3}/_{4-3} = ^{3}/_{1} = 3

For line 2:

Let (x_{1} ,y_{1}) = (4, 1) and (x_{2} ,y_{2})= (6, 7)

Slope = m_{2 }= ^{y2-y1}/_{x2-x1} = ^{7-1}/_{6-4} = ^{6}/_{2} = 3

We have m_{1} = m_{2}

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 (x_{1} ,y_{1}) = (4, 7) and (x_{2} ,y_{2})= (3, 5)

Slope = m_{1 }= ^{y2-y1}/_{x2-x1} = ^{5-7}/_{3-4} = ^{-2}/_{-1} = 2

For line 2:

Let (x_{1} ,y_{1}) = (-1, 7) and (x_{2} ,y_{2})= (1, 6)

Slope = m_{2 }= ^{y2-y1}/_{x2-x1} = ^{6-7}/_{1-(-1)} = ^{-1}/_{2}

We have m_{1} ≠ m_{2}

m_{1}.m_{2} = 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 (x_{1} ,y_{1}) = (-1, -2) and (x_{2} ,y_{2})= (1, 6)

Slope = m_{1 }= ^{y2-y1}/_{x2-x1} = ^{6-(-2)}/_{1-(-1)} = ^{8}/_{2} = 4

For line 2:

Let (x_{1} ,y_{1}) = (-1, 1) and (x_{2} ,y_{2})= (-2, -3)

Slope = m_{2 }= ^{y2-y1}/_{x2-x1} = ^{-3-1}/_{-2-(-1)} = ^{-4}/_{-1} = 4

We have m_{1} = m_{2}

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

**Find the slope of the line perpendicular to the line joining the points**

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

Solution:

Let (x_{1} ,y_{1}) = (1, 7) and (x_{2} ,y_{2})= (-4, 3)

Slope = m_{1 }= ^{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 m_{1}.m_{2} = -1]

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

Solution:

Let (x_{1} ,y_{1}) = (2, 3) and (x_{2} ,y_{2})= (1, 4)

Slope = m_{1 }= ^{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 m_{1}.m_{2} = -1]

**Find the slope of the line parallel to the line joining the points**

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

Solution:

Let (x_{1} ,y_{1}) = (-4, 3) and (x_{2} ,y_{2})= (2, 5)

Slope = m_{1 }= ^{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 (x_{1} ,y_{1}) = (1, -5) and (x_{2} ,y_{2})= (7, 1)

Slope = m_{1 }= ^{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.

**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 (x_{1} ,y_{1}) = (2, 7) and (x_{2} ,y_{2})= (3, 6)

Slope = m_{1 }= ^{y2-y1}/_{x2-x1} = ^{6-7}/_{3-2} = ^{-1}/_{1} = -1

Line 2:

Let (x_{1} ,y_{1}) = (9, a) and (x_{2} ,y_{2})= (11, 3)

Slope = m_{2 }= ^{y2-y1}/_{x2-x1} = ^{3-a}/_{11-9} = ^{3-a}/_{2}

Since m_{1} = m_{2} 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

**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 (x_{1} ,y_{1}) = (1, 0) and (x_{2} ,y_{2})= (4, 3)

Slope = m_{1 }= ^{y2-y1}/_{x2-x1} = ^{3-0}/_{4-1} = ^{3}/_{3} = 1

Line 2:

Let (x_{1} ,y_{1}) = (-2, -1) and (x_{2} ,y_{2})= (m, 0)

Slope = m_{2 }= ^{y2-y1}/_{x2-x1} = ^{0-(-1)}/_{m-(-2)} = ^{1}/_{m+2}

Since m_{1.}m_{2} = -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, m_{1}.m_{2} = -1

1.(^{1}/_{m+2}) = -1

^{1}/_{m+2 }= -1

1 = -(m+2)

1 = -m – 2

-m = 1 + 2

m = -3