Perpendicular lines

Perpendicular gradients Extended

Two lines are perpendicular when they meet at a right angle. If neither line is horizontal or vertical, their gradients $m_1$ and $m_2$ are linked by a simple rule:

$$m_1\times m_2=-1,\text{so}{m}_2=-\frac{1}{m_1}$$

In words: the perpendicular gradient is the negative reciprocal. Turn the gradient upside down and change its sign.

• Gradient $2$ (which is $\frac{2}{1}$): perpendicular gradient $-\frac{1}{2}$.
• Gradient $-\frac{3}{4}$: perpendicular gradient $\frac{4}{3}$.
• Gradient $1$: perpendicular gradient $-1$.
• Gradient $0.4$: perpendicular gradient $-\frac{1}{0.4}=-2.5$.

Always check by multiplying: $2\times -\frac{1}{2}=-1$. If the product is not $-1$, the lines are not perpendicular.

Why it works. Take a gradient triangle with a run of 3 and a rise of 2, so the gradient is $\frac{2}{3}$. Turn the whole triangle through a right angle. The side that went 3 across now goes 3 up, and the side that went 2 up now goes 2 back to the left. The turned line rises 3 for a run of $-2$, so its gradient is $\frac{3}{-2}=-\frac{3}{2}$. The rise and run swap places (the reciprocal) and one of them changes direction (the minus sign).

The rule does not apply to horizontal and vertical lines, because a vertical line has no numerical gradient. A horizontal line, such as $y=4$, is simply perpendicular to every vertical line, such as $x=-1$.

Turning a gradient triangle Extended

The first line goes 3 across and 2 up: gradient $\frac{2}{3}$. Turned through ${90}^{°}$, the same triangle goes 3 up and 2 to the left: gradient $-\frac{3}{2}$. And $\frac{2}{3}\times -\frac{3}{2}=-1$.

Finding a perpendicular gradient from an equation Extended

First find the gradient of the given line, which means making $y$ the subject if it is not already. Then take the negative reciprocal.

For example, to find the gradient of a line perpendicular to $2y=3x+1$:

1. Divide every term by 2: $y=\frac{3}{2}x+\frac{1}{2}$, so the gradient is $\frac{3}{2}$.
2. Negative reciprocal: $-\frac{2}{3}$.

The usual error is to take the reciprocal of the coefficient of $x$ before making $y$ the subject: in $2y=3x+1$ the gradient is not 3, so the perpendicular gradient is not $-\frac{1}{3}$.

Equation of a perpendicular line through a point Extended

The method is the same as for a parallel line, except for the gradient.

1. Find the gradient of the given line.
2. Take the negative reciprocal to get the gradient $m$ of the new line.
3. Substitute the given point into $y=mx+c$ and solve for $c$.
4. Write the equation in the form the question asks for, and check that the point fits.

When the perpendicular gradient is a fraction, the arithmetic for $c$ is easier if you multiply out carefully: with $m=-\frac{1}{3}$ and the point $(6,1)$, $-\frac{1}{3}\times 6=-2$.

Perpendicular bisectors Extended

The perpendicular bisector of a line segment $AB$ is the line that cuts $AB$ in half and crosses it at a right angle. Every point on it is the same distance from $A$ as from $B$, which is why it appears in questions about loci and about the centre of a circle.

To find its equation, put together three skills from this topic:

1. Midpoint of $AB$ (subtopic 3.4): the bisector passes through it.
2. Gradient of $AB$ (subtopic 3.3).
3. Perpendicular gradient: the negative reciprocal of the gradient of $AB$.

Then substitute the midpoint into $y=mx+c$. The key point to remember is that the bisector passes through the midpoint, not through $A$ or $B$.

Right angles in shapes Extended

The gradient rule gives a quick test for a right angle when you know coordinates.

• A right-angled triangle: find the gradients of two sides that meet at a vertex. If they multiply to $-1$, the angle there is ${90}^{°}$.
• A rectangle or a square: adjacent sides are perpendicular.
• A rhombus or a kite: the diagonals are perpendicular, and in a rhombus each diagonal is the perpendicular bisector of the other.
• The shortest distance from a point to a line is along the perpendicular. Find the equation of the perpendicular through the point, solve simultaneously to find where it meets the line, then use the length formula.

In a "show that" question, write each gradient, show the product equals $-1$, and say in words what that means.