# 1.2 Lines

The equation of any line in ^{2}, except a line parallel to the *y*-axis, can be written in the form

where *m* is the *gradient* or *slope* of the line, and *c* is its *y-intercept*; that is, (0, *c*) is the point at which the line crosses the *y*-axis. (See the first sketch below.)

In the particular case that the line cuts the *y*-axis at the origin, its equation has the simple form

as *c* = 0 in this case. (See the second sketch below.)

Another special case occurs when *m* = 0. Then the line is parallel to the *x*-axis, and its equation is of the form

where *c* is the *y*-intercept. (See the third sketch below.)

Finally, the equation of a line parallel to the *y*-axis cannot be written in the form *y* = *mx* + *c*, but it can be written as

where (*a*, 0) is the point at which the line crosses the *x*-axis. (See the final sketch below.)

In both cases (1.1) and (1.2) above, the equation of a line in the plane can be written in the form

for some real numbers *a*, *b* and *c*, where *a* and *b* are not both zero.

Thus any line in ^{2} has an equation of the form (1.3); conversely, any equation of the form (1.3) represents a line in ^{2}.

## Equation of a line

The general equation of a line in ^{2} is

*ax* + *by* = *c*,

where *a*, *b* and *c* are real, and *a* and *b* are not both zero.

## Exercise 1

Determine the equation of the line with gradient −3 that passes through the point (2, −1).

### Answer

Using the formula for the equation of a line when given its gradient and one point on it, we find that the equation of this line is

We can rearrange this in the form

## Exercise 2

For each of the following pairs of points, determine the equation of the line through them.

**(a)**(1, 1) and (3, 5).**(b)**(0, 0) and (0, 8).**(c)**(0, 0) and (4, 2).**(d)**(4, −1) and (2, −1).

### Answer

**(a)** Since (1, 1) and (3, 5) lie on the line, its gradient is

Then, since the point (1, 1) lies on the line, its equation must be *y* − 1 = 2(*x* − 1), or *y* = 2*x* − 1.

**(b)** Both these points have *x*-coordinate 0, so they lie on the line with equation *x* = 0, the *y*-axis.

**(c)** Since the origin lies on the line, its equation must be of the form *y* = *mx*, where *m* is its gradient.

Since (4, 2) lies on the line, its coordinates must satisfy the equation of the line. Thus 2 = 4 *m*, so

Hence the equation of this line is

**(d)** Both these points have *y*-coordinate −1, so they lie on the line with equation *y* = −1.