## Equation of a straight line

### Gradient form

The general equation for a straight line in its gradient form is:

$y = mx + c$
where = the gradient of the line
c = a constant

If we let x = 0, then it is found that y = c. Therefore is equal to the y-intercept.

#### Application

Sketch the graph of the equation y = 2x + 3.

Firstly, c = 3 and therefore the y-intercept is at the point (0, 3).
To find the x-intercept, let y = 0. The x-intercept is therefore at (-3/2, 0).
Plot both of these points and join them to create the line.

### Intercept form

Another way to express a linear relation is in the form:

$ax + by = c$
where a, b, and c are all constants

This is referred to as the intercept form as it is convenient in determining the x and y intercepts.

#### Application

Sketch the graph 8x + 11y = 88.

When x = 0, y = 8. Therefore the y-intercept is at (0, 8).
When y = 0x = 11. Therefore the x-intercept is at (11, 0).

## Finding the equation of a straight line

The gradient, m, of the line can be rearranged using a general point (x, y) on the same line

$y - y_1 = m(x - x_1)$

This form is most convenient when determining the equation of a straight line.

### Given two points

When two points of a line are found, they can be substituted into the gradient formula in order to find the slope, m. After that, the re-arranged gradient formula can be used for convenience.

Find the equation of the straight line passing through the points (6, 4) and (2, -4).

The gradient must first be found:
$m = \frac{-4 - 4}{2 - 6}$
$m = 2$

This value, as well as any of the given points can be substituted into the formula:
$y - 4 = 2(x - (-4))$
$y = 2x + 12$

### Given a point, and the gradient

Using the same re-arranged formula, the equation can be quickly deduced.

Find the equation of the line that passes through the point (2,5) and has a gradient of -3.

Using $y - y_1 = m(x - x_1)$
$y - 5 = -3(x - 2)$
$y = -3x + 11$

### Given a graph

In a graph, the value of c reads as the y-intercept. m is the gradient and can be calculated using two points on the line. The equation can therefore be written in the form y = mx + c

See also: