Rectilinear shapes

Find the gradient, equations and intersections of medians, altitudes and perpendicular bisectors using our knowledge of the mid-point as well as parallel and perpendicular lines.

Part of

Maths

Algebraic and geometric skills

Basics of straight lines

There are several basic facts and equations connected with straight lines that you need to know by heart.

Distance formula

The distance between two points, $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by the formula:

$\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$

So the distance between (2,3) and (1,5) is:

$\sqrt {{{(1 - 2)}^2} + {{(5 - 3)}^2}}$

$= \sqrt {{{( - 1)}^2} + {{(2)}^2}}$

$= \sqrt 5$

Gradient

The gradient m between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by the formula:

$m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

This only applies where ${x_2} \ne {x_1}$ . If ${x_2} = {x_1}$ then the gradient is undefined.

The gradient between (2,3) and (1,5) is:

$m = \frac{{5 - 3}}{{1 - 2}} = - 2$

If a line with gradient makes an angle $\theta$ with the positive direction of the -axis then $m = \tan\theta$

Example 1

We can calculate the gradient of the line above by selecting two coordinate points that the straight line passes through.

(1,2) and (2,5)

$m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{5-2}{2-1}=\frac{3}{1}=3$

Now that we know the gradient, we can calculate the angle that the straight line makes with the positive direction of the axis.

$m=\tan\theta$

$3=\tan\theta$

$\theta =\tan^{-1}(3)$

$\theta =71.6^\circ$

Similarly, we can calculate the gradient of the straight line if we know the angle the line makes with the positive direction of the axis.

Example 2

This time you will notice, the line has a negative gradient, so you will need to use your knowledge of quartiles to calculate this gradient.

$m=\tan\theta$

$m = \tan 120^\circ$

$m=-\tan60^\circ$

$m = - \sqrt 3$

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