GRADIENT OF A STRAIGHT LINE WORKSHEET

Problem 1 :

Find the angle of inclination of the straight line whose gradient is 1/√3.

Problem 2 :

Find the gradient of the straight line passing through the points (3, -2) and (-1, 4). 

Problem 3 :

Using the concept of gradient, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

Problem 4 :

Find the gradient of the line 3x - 2y + 7 = 0. 

Problem 5 :

If the straight line 5x + ky - 1 = 0 has the gradient 5, find the value of 'k'.  

Solutions

Problem 1 :

Find the angle of inclination of the straight line whose gradient is 1/√3.

Solution :

Let θ be the angle of inclination of the line. 

Then, gradient of the line,  m  = tan θ.

Given : Gradient  =  1/√3.

So, we have 

tan θ  =  1/√3

θ  =  30°

So, the angle of inclination is 30°.

Problem 2 :

Find the gradient of the straight line passing through the points (3, -2) and (-1, 4). 

Solution :

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

Then, the formula to find the gradient, 

m  =  (y₂ - y₁) / (x₂ - x₁)

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

m  =  (4 + 2) / (-1 - 3)

m  =  - 6 / 4

m  =  - 3 / 2

So, the gradient is -3/2.

Problem 3 :

Using the concept of gradient, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

Solution :

Slope of the line joining (x₁, y₁) and (x₂, y₂) is, 

m  =  (y₂ - y₁) / (x₂ - x₁)

Using the above formula,

Gradient of the line AB joining the points A (5, - 2) and B (4- 1) is 

 =  (-1 + 2) / (4 - 5)

=  - 1

Gradient of the line BC joining the points B (4- 1) and C (1, 2) is 

 =  (2 + 1) / (1 - 4)

=  - 1

Thus,

gradient of AB  =  gradient of BC

Also, B is the common point.

So, the points A , B and C are collinear.

Problem 4 :

Find the gradient of the line 3x - 2y + 7 = 0. 

Solution :

When the general form of equation of a straight line is given, the formula to find gradient is

m  =  -  coefficient of x / coefficient of y 

In the given line 3x - 2y + 7 = 0,

coefficient of x  = 3 and coefficient of y  =  - 2

Gradient,  m  =  (-3) / (-2)  =  3/2 

So, the gradient of the given line is 3/2. 

Problem 5 :

If the straight line 5x + ky - 1 = 0 has the gradient 5, find the value of 'k'.  

Solution :

When the general form of equation of a straight line is given, the formula to find gradient is

m  =  -  coefficient of x / coefficient of y 

In the given line 3x - 2y + 7 = 0,

coefficient of x  = 3 and coefficient of y  =  k

Slope,  m  =  -5 / k

Given : Gradient  =  5

So, we have  5  =  -5/k

5k  =  -5

k  =  -1

So, the value of 'k' is  -1. 

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