9.1.1 Equation of a Straight Line - Mathematics

9.1.1 Equation of a Straight Line: y = mx + c

1. Given the value of the gradient, m, and the y-intercept, c, an equation of a straight line
y = mx + c can be formed.

2. If the equation of a straight line is written in the form y = mx + c, the gradient, m, and the y-intercept, c, can be determined directly from the equation.

Example 1:

Given that the equation of a straight line is y = 3 – 4x. Find the gradient and y-intercept of the line?

Solution:

y= 3 – 4x

y= – 4x + 3 ← (y = mx + c)

Therefore, gradient, m = – 4

y-intercept, c = 3

3. If the equation of a straight line is written in the form ax + by + c = 0, change it to the form y = mx + c before finding the gradient and the y-intercept.

Example 2:

Given that the equation of a straight line is 4x + 6y– 3 = 0. What is the gradient and y-intercept of the line?

Solution:

4x + 6y – 3 = 0

6y = –4x + 3

$\begin{array}{l} y = - \frac{2}{3} x + \frac{1}{2} \leftarrow y = m x + c \\ ∴ Gradient m = - \frac{2}{3} \\ y - intercept, c = \frac{1}{2} \end{array}$

9.1.2 Equation of a Straight Line (Sample Questions)

Question 1:

Given that the equation of a straight line is 4x + 6y – 3 = 0. What is the gradient of the line?

Solution:

4x + 6y – 3 = 0

6y = – 4x + 3

$\begin{array}{l} y = - \frac{4 x}{6} + \frac{3}{6} \\ y = - \frac{2}{3} x + \frac{1}{2} \\ y = m x + c \\ g r a d i e n t, m = - \frac{2}{3} \end{array}$

Question 2:

Given that the equation of a straight line is y = – 7x + 3. Find the y-intercept of the line?

Solution:

y = mx + c, c is y-intercept of the straight line.

Therefore for the straight line y = – 7x + 3,

y-intercept is 3

Question 3:

Find the equation of the straight line MN if its gradient is equal to 3.

Solution:

Given m = 3

Substitute m = 3 and (-2, 5) into y = mx + c.

5 = 3 (-2) + c

5 = -6 + c

c = 11

Therefore the equation of the straight line MN is y = 3x + 11

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