# Equation of a straight line

The general of a straight line is $$y = mx + c$$, where $$m$$ is the gradient and $$(0,c)$$ the coordinates of the y-intercept.

Look at the National 4 straight line section before continuing.

We can find the equation of a straight line when given the gradient and a point on the line by using the :

$y - b = m(x - a)$

where $$m$$ is the gradient and $$(a,b)$$ is on the line.

### Example

Find the equation of the line with gradient 3, passing through (4, 1).

Using $$y - b = m(x - a)$$ with m = 3, a = 4 and b = 1.

$y - 1 = 3(x - 4)$

$y - 1 = 3x - 12$

$y = 3x - 11$

Now try the example question below.

Question

Find the equation of the line which passes through the points A(-2, 0) and B(1, 6) and state the gradient and .

Since we are given 2 points that already lie on the line, using the formula as before, we need to calculate the gradient first.

${m_{AB}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{6 - 0}}{{1 - ( - 2)}} = \frac{6}{3} = \frac{2}{1} = 2$

Now we can use the formula when m = 2, a = 1 and b = 6 (You can use any of the 2 points given in the question).

$y - b = m(x - a)$

$y - 6 = 2(x - 1)$

$y - 6 = 2x - 2$

$y = 2x + 4$

Therefore the gradient equals 2 and y-intercept equals (0, 4).