Parallel and Perpendicular Lines

Parallel lines have the same gradient

Two straight lines are parallel when they have the same gradient and different intercepts. The lines $y=3x-4$ and $y=3x+1$ both have gradient 3, so they are parallel, and the only difference between them is where they cross the $y$-axis. Lines that share a gradient but not an intercept never meet.

The test works in both directions. To decide whether two lines are parallel, put both into the form $y=mx+c$ and compare the gradients; to write down the equation of a parallel line, keep the gradient exactly as it is and change only the intercept.

Vertical lines are parallel when they have the same form $x=k$, so $x=2$ and $x=5$ are parallel. Their gradients are not numbers you can compare, so they are handled by looking at the equations instead.

The equation of a parallel line through a point

This is a two-step calculation. Step one: read the gradient of the given line, because a parallel line has that same gradient. Step two: put the coordinates of the point into $y=mx+c$ with that gradient and solve for $c$.

Find the line parallel to $y=4x-1$ that passes through $(1,-3)$. The gradient is 4, so the new line has the form $y=4x+c$. Substituting the point: $-3=4\times 1+c$, so $-3=4+c$ and $c=-7$. The equation is $y=4x-7$.

Check it on the point you were given: $4\times 1-7=-3$. The intercept is not copied from the original line - $y=4x-1$ would pass through $(0,-1)$ and miss the point $(1,-3)$ completely.

• Keep the gradient; it is the whole point of the question
• Substitute the point to find the new $c$
• Check the answer passes through the given point
• A line parallel to $y=mx+c$ has the form $y=mx+k$ for some different $k$

When the given line is not in the form $y=mx+c$

Some questions give the line as $ax+by=c$ or with a number in front of $y$. The coefficient of $x$ is not the gradient until $y$ is on its own, so rearrange first, then read the gradient.

For $3x+2y=12$: subtract $3x$ to get $2y=12-3x$, then divide every term by 2 to get $y=6-\frac{3}{2}x$. The gradient is $-\frac{3}{2}$, so a parallel line has the same gradient. Through the point $(0,5)$: $5=-\frac{3}{2}\times 0+c$, so $c=5$ and the line is $y=-\frac{3}{2}x+5$.

The same care is needed with $2y=3x+1$. Dividing by 2 gives $y=\frac{3}{2}x+\frac{1}{2}$, so the gradient is $\frac{3}{2}$ and not 3. Every later step of the question depends on this one number being right.

• Move the $x$-term first, then divide every term by the number in front of $y$
• $3x+2y=12$ has gradient $-\frac{3}{2}$, not 3 and not $\frac{3}{2}$
• $2y=3x+1$ has gradient $\frac{3}{2}$, not 3

Extended only: perpendicular lines and the negative reciprocal

Perpendicular lines meet at a right angle. Their gradients multiply to $-1$, so if one gradient is $m$, the other is $-\frac{1}{m}$. In practice: turn the gradient upside down and change its sign. The negative reciprocal of 2 is $-\frac{1}{2}$, of $-2$ is $\frac{1}{2}$, of $\frac{3}{4}$ is $-\frac{4}{3}$, and of $-\frac{1}{3}$ is 3.

The product is the check. $2\times (-\frac{1}{2})=-1$, and $\frac{3}{4}\times (-\frac{4}{3})=-1$. If the product of the two gradients is not $-1$, the lines are not perpendicular, however they look on the page.

Two cases need care. A horizontal line such as $y=5$ has gradient 0, and the line perpendicular to it is vertical, so it has an equation of the form $x=k$ and no gradient you can write down. And a vertical line has no negative reciprocal either, so a line perpendicular to $x=4$ is horizontal, with gradient 0.

Extended only: the perpendicular line through a point

This is the parallel-line method with one extra step. First work out the perpendicular gradient as the negative reciprocal. Then substitute the given point into $y=mx+c$ and solve for $c$, exactly as before.

Find the line perpendicular to $y=2x+3$ that passes through $(4,1)$. The gradient of the given line is 2, so the perpendicular gradient is $-\frac{1}{2}$. Substituting the point: $1=-\frac{1}{2}\times 4+c$, so $1=-2+c$ and $c=3$. The equation is $y=-\frac{1}{2}x+3$, which is the same line as $y=-0.5x+3$.

Check both properties: the line passes through $(4,1)$ because $-\frac{1}{2}\times 4+3=1$, and the two gradients multiply to $-1$ because $2\times (-\frac{1}{2})=-1$.

If the given line needs rearranging, do that before anything else. For $2y=3x+1$ the gradient is $\frac{3}{2}$, so the perpendicular gradient is $-\frac{2}{3}$, not $-\frac{1}{3}$.

• Find the negative reciprocal of the given gradient
• Substitute the point to find $c$
• Check the product of the gradients is $-1$ and the point satisfies the equation

Extended only: the perpendicular bisector of a segment

The perpendicular bisector of a segment $AB$ is the line that cuts $AB$ in half at right angles. Every point on it is the same distance from $A$ as it is from $B$. Finding its equation takes three steps: the midpoint of $AB$, then the negative reciprocal of the gradient of $AB$, then the intercept.

Find the perpendicular bisector of the segment joining $A(-3,8)$ and $B(9,-2)$. The midpoint is $(\frac{-3+9}{2},\frac{8+(-2)}{2})=(3,3)$. The gradient of $AB$ is $\frac{-2-8}{9-(-3)}=\frac{-10}{12}=-\frac{5}{6}$, so the perpendicular gradient is $\frac{6}{5}$. Substituting $(3,3)$: $3=\frac{6}{5}\times 3+c$, so $3=\frac{18}{5}+c$ and $c=-\frac{3}{5}$. The perpendicular bisector is $y=\frac{6}{5}x-\frac{3}{5}$.

Check the arithmetic as you go, because an error in any one of the three steps spoils the answer. The midpoint must sit halfway along $AB$, the two gradients must multiply to $-1$, and the finished line must pass through the midpoint.

• Step 1: midpoint of the segment, using the average of the coordinates
• Step 2: negative reciprocal of the segment's gradient
• Step 3: substitute the midpoint to find $c$
• Check: $-\frac{5}{6}\times \frac{6}{5}=-1$, and the line passes through $(3,3)$