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Parallel and perpendicular lines

Covers AS91256

Parallel lines

Parallel lines have the same gradient.

$m_1 = m_2$

They point in the same direction, so they never meet, no matter how far they are extended.
The $y$ -intercepts must differ — if both the gradient and the intercept match, the two equations describe the same line, not a parallel pair.
$y = 3x + 1$ and $y = 3x - 8$ are parallel. $y = 3x + 1$ and $2y = 6x + 2$ are the same line.

Perpendicular lines

Perpendicular lines meet at a right angle, and their gradients multiply to $-1$ :

$m_1 \times m_2 = -1$

Rearranged, this says:

$m_2 = -\frac{1}{m_1}$

The perpendicular gradient is the negative reciprocal — flip the fraction and change the sign. Both steps are needed.

Worked examples of the flip:
- $m = 2$ → perpendicular gradient $-\dfrac{1}{2}$
- $m = -\dfrac{3}{4}$ → perpendicular gradient $\dfrac{4}{3}$
- $m = -5$ → perpendicular gradient $\dfrac{1}{5}$
A quick check: the two gradients always have opposite signs, unless one is zero.

The special case

A horizontal line and a vertical line are perpendicular, but the rule $m_1m_2 = -1$ cannot be used — a vertical line has no gradient.
$y = 3$ and $x = 7$ meet at a right angle, and you say so directly rather than trying to multiply gradients.
This is the one case where the formula does not apply, and questions do test it.

Finding a parallel line through a given point

Take the gradient from the given line.
Use the point-gradient form with the new point.
Worked through — the line through $(4, 1)$ parallel to $y = 2x - 5$ :
- Gradient is 2 (unchanged)
- $y - 1 = 2(x - 4)$ , so $y = 2x - 7$

Finding a perpendicular line through a given point

Take the gradient of the given line.
Find the negative reciprocal.
Use the point-gradient form.
Worked through — the line through $(6, 2)$ perpendicular to $y = \dfrac{3}{2}x + 1$ :
- Original gradient $\dfrac{3}{2}$ , so the perpendicular gradient is $-\dfrac{2}{3}$
- $y - 2 = -\dfrac{2}{3}(x - 6)$ , so $y = -\dfrac{2}{3}x + 6$
Rearrange the given line into $y = mx + c$ first if it is in general form. You cannot read the gradient off $3x + 5y = 20$ directly.

The perpendicular bisector

The perpendicular bisector of a segment is the line that cuts it in half at right angles.
It is the set of all points equidistant from the two endpoints.
The method:
- Find the midpoint of the segment — the bisector passes through it.
- Find the gradient of the segment.
- Take the negative reciprocal for the bisector's gradient.
- Use the point-gradient form with the midpoint and the new gradient.
Both steps are essential. A line through the midpoint that is not perpendicular is not a perpendicular bisector, and neither is a perpendicular line that misses the midpoint.

Using perpendicularity to prove a right angle

To show a triangle is right-angled, or a quadrilateral is a rectangle:
- Find the gradients of the two sides meeting at the corner in question.
- Multiply them.
- If the product is $-1$ , the sides are perpendicular and the angle is $90°$ .
State the conclusion, not just the arithmetic: " $m_1m_2 = -1$ , so $AB$ is perpendicular to $BC$ and the angle at $B$ is a right angle."

Tips and Tricks

Negative reciprocal means both steps. From $m = \frac{2}{5}$ , the perpendicular gradient is $-\frac{5}{2}$ — flipped and negated. Doing only one gives $\frac{5}{2}$ or $-\frac{2}{5}$ , both wrong.

Rearrange to $y = mx + c$ before reading a gradient. You cannot see the gradient of $4x + 3y = 12$ until it becomes $y = -\frac{4}{3}x + 4$ .

A horizontal and a vertical line are perpendicular, but the product rule does not apply because one gradient is undefined. Say this explicitly rather than forcing the formula.

A perpendicular bisector needs both the midpoint and the perpendicular gradient. Missing either one makes it a different line entirely.

State the geometric conclusion. At Merit the marker wants "therefore the lines are perpendicular, so the angle is $90°$ " — not just $-1$ written at the end of a calculation.

Check the signs. Perpendicular gradients always have opposite signs. If both of yours are positive, one of them is wrong.

Worked ExampleFinding a perpendicular bisector

Find the equation of the perpendicular bisector of the segment joining $A(-1, 6)$ and $B(5, -2)$ .

Step 1 — Understand what is needed

A perpendicular bisector must satisfy two conditions:

it passes through the midpoint of $AB$
it is perpendicular to $AB$

Both must be found, and both must be used.

Step 2 — Find the midpoint of $AB$

Average each coordinate:

$x = \frac{-1 + 5}{2} = \frac{4}{2} = 2$

$y = \frac{6 + (-2)}{2} = \frac{4}{2} = 2$

$M = (2, 2)$

Step 3 — Find the gradient of $AB$

$m_{AB} = \frac{-2 - 6}{5 - (-1)} = \frac{-8}{6} = -\frac{4}{3}$

Simplify before going further — $\frac{-8}{6}$ reduces to $-\frac{4}{3}$ .

Step 4 — Find the perpendicular gradient

Flip the fraction and change the sign:

$m_{\perp} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4}$

Check: $-\frac{4}{3} \times \frac{3}{4} = -1$ ✓ and the two gradients have opposite signs ✓

Step 5 — Build the equation

Use the point-gradient form with the midpoint $(2, 2)$ and the perpendicular gradient $\frac{3}{4}$ :

$y - 2 = \frac{3}{4}(x - 2)$

Step 6 — Rearrange

$y - 2 = \frac{3}{4}x - \frac{3}{2}$

$y = \frac{3}{4}x - \frac{3}{2} + 2$

$y = \frac{3}{4}x + \frac{1}{2}$

Step 7 — Check both defining properties

Does it pass through the midpoint?

$\frac{3}{4}(2) + \frac{1}{2} = 1.5 + 0.5 = 2 \;\checkmark$

Is it perpendicular to $AB$ ?

$-\frac{4}{3} \times \frac{3}{4} = -1 \;\checkmark$

Step 8 — Confirm with the equidistance property

Every point on a perpendicular bisector is equidistant from the two endpoints. Testing the point $(6, 5)$ , which lies on the bisector since $\frac{3}{4}(6) + \frac{1}{2} = 5$ :

$\text{distance to } A(-1,6) = \sqrt{7^2 + (-1)^2} = \sqrt{50}$

$\text{distance to } B(5,-2) = \sqrt{1^2 + 7^2} = \sqrt{50} \;\checkmark$

Equal distances, confirming the line is genuinely the perpendicular bisector.

Parallel and perpendicular lines

Covers AS91256

Parallel lines

Parallel lines have the same gradient.

$m_1 = m_2$

They point in the same direction, so they never meet, no matter how far they are extended.
The $y$ -intercepts must differ — if both the gradient and the intercept match, the two equations describe the same line, not a parallel pair.
$y = 3x + 1$ and $y = 3x - 8$ are parallel. $y = 3x + 1$ and $2y = 6x + 2$ are the same line.

Perpendicular lines

Perpendicular lines meet at a right angle, and their gradients multiply to $-1$ :

$m_1 \times m_2 = -1$

Rearranged, this says:

$m_2 = -\frac{1}{m_1}$

The perpendicular gradient is the negative reciprocal — flip the fraction and change the sign. Both steps are needed.

Worked examples of the flip:
- $m = 2$ → perpendicular gradient $-\dfrac{1}{2}$
- $m = -\dfrac{3}{4}$ → perpendicular gradient $\dfrac{4}{3}$
- $m = -5$ → perpendicular gradient $\dfrac{1}{5}$
A quick check: the two gradients always have opposite signs, unless one is zero.

The special case

A horizontal line and a vertical line are perpendicular, but the rule $m_1m_2 = -1$ cannot be used — a vertical line has no gradient.
$y = 3$ and $x = 7$ meet at a right angle, and you say so directly rather than trying to multiply gradients.
This is the one case where the formula does not apply, and questions do test it.

Finding a parallel line through a given point

Take the gradient from the given line.
Use the point-gradient form with the new point.
Worked through — the line through $(4, 1)$ parallel to $y = 2x - 5$ :
- Gradient is 2 (unchanged)
- $y - 1 = 2(x - 4)$ , so $y = 2x - 7$

Finding a perpendicular line through a given point

Take the gradient of the given line.
Find the negative reciprocal.
Use the point-gradient form.
Worked through — the line through $(6, 2)$ perpendicular to $y = \dfrac{3}{2}x + 1$ :
- Original gradient $\dfrac{3}{2}$ , so the perpendicular gradient is $-\dfrac{2}{3}$
- $y - 2 = -\dfrac{2}{3}(x - 6)$ , so $y = -\dfrac{2}{3}x + 6$
Rearrange the given line into $y = mx + c$ first if it is in general form. You cannot read the gradient off $3x + 5y = 20$ directly.

The perpendicular bisector

The perpendicular bisector of a segment is the line that cuts it in half at right angles.
It is the set of all points equidistant from the two endpoints.
The method:
- Find the midpoint of the segment — the bisector passes through it.
- Find the gradient of the segment.
- Take the negative reciprocal for the bisector's gradient.
- Use the point-gradient form with the midpoint and the new gradient.
Both steps are essential. A line through the midpoint that is not perpendicular is not a perpendicular bisector, and neither is a perpendicular line that misses the midpoint.

Using perpendicularity to prove a right angle

To show a triangle is right-angled, or a quadrilateral is a rectangle:
- Find the gradients of the two sides meeting at the corner in question.
- Multiply them.
- If the product is $-1$ , the sides are perpendicular and the angle is $90°$ .
State the conclusion, not just the arithmetic: " $m_1m_2 = -1$ , so $AB$ is perpendicular to $BC$ and the angle at $B$ is a right angle."

Tips and Tricks

Rearrange to $y = mx + c$ before reading a gradient. You cannot see the gradient of $4x + 3y = 12$ until it becomes $y = -\frac{4}{3}x + 4$ .

A horizontal and a vertical line are perpendicular, but the product rule does not apply because one gradient is undefined. Say this explicitly rather than forcing the formula.

A perpendicular bisector needs both the midpoint and the perpendicular gradient. Missing either one makes it a different line entirely.

State the geometric conclusion. At Merit the marker wants "therefore the lines are perpendicular, so the angle is $90°$ " — not just $-1$ written at the end of a calculation.

Check the signs. Perpendicular gradients always have opposite signs. If both of yours are positive, one of them is wrong.

Worked ExampleFinding a perpendicular bisector

Find the equation of the perpendicular bisector of the segment joining $A(-1, 6)$ and $B(5, -2)$ .

Step 1 — Understand what is needed

A perpendicular bisector must satisfy two conditions:

it passes through the midpoint of $AB$
it is perpendicular to $AB$

Both must be found, and both must be used.

Step 2 — Find the midpoint of $AB$

Average each coordinate:

$x = \frac{-1 + 5}{2} = \frac{4}{2} = 2$

$y = \frac{6 + (-2)}{2} = \frac{4}{2} = 2$

$M = (2, 2)$

Step 3 — Find the gradient of $AB$

$m_{AB} = \frac{-2 - 6}{5 - (-1)} = \frac{-8}{6} = -\frac{4}{3}$

Simplify before going further — $\frac{-8}{6}$ reduces to $-\frac{4}{3}$ .

Step 4 — Find the perpendicular gradient

Flip the fraction and change the sign:

$m_{\perp} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4}$

Check: $-\frac{4}{3} \times \frac{3}{4} = -1$ ✓ and the two gradients have opposite signs ✓

Step 5 — Build the equation

Use the point-gradient form with the midpoint $(2, 2)$ and the perpendicular gradient $\frac{3}{4}$ :

$y - 2 = \frac{3}{4}(x - 2)$

Step 6 — Rearrange

$y - 2 = \frac{3}{4}x - \frac{3}{2}$

$y = \frac{3}{4}x - \frac{3}{2} + 2$

$y = \frac{3}{4}x + \frac{1}{2}$

Step 7 — Check both defining properties

Does it pass through the midpoint?

$\frac{3}{4}(2) + \frac{1}{2} = 1.5 + 0.5 = 2 \;\checkmark$

Is it perpendicular to $AB$ ?

$-\frac{4}{3} \times \frac{3}{4} = -1 \;\checkmark$

Step 8 — Confirm with the equidistance property

Every point on a perpendicular bisector is equidistant from the two endpoints. Testing the point $(6, 5)$ , which lies on the bisector since $\frac{3}{4}(6) + \frac{1}{2} = 5$ :

$\text{distance to } A(-1,6) = \sqrt{7^2 + (-1)^2} = \sqrt{50}$

$\text{distance to } B(5,-2) = \sqrt{1^2 + 7^2} = \sqrt{50} \;\checkmark$

Equal distances, confirming the line is genuinely the perpendicular bisector.

Parallel and perpendicular lines

Parallel lines

Perpendicular lines

The special case

Finding a parallel line through a given point

Finding a perpendicular line through a given point

The perpendicular bisector

Using perpendicularity to prove a right angle

Step 1 — Understand what is needed

Step 2 — Find the midpoint of ABABAB

Step 3 — Find the gradient of ABABAB

Step 4 — Find the perpendicular gradient

Step 5 — Build the equation

Step 6 — Rearrange

Step 7 — Check both defining properties

Step 8 — Confirm with the equidistance property

Parallel and perpendicular lines

Parallel lines

Perpendicular lines

The special case

Finding a parallel line through a given point

Finding a perpendicular line through a given point

The perpendicular bisector

Using perpendicularity to prove a right angle

Step 1 — Understand what is needed

Step 2 — Find the midpoint of ABABAB

Step 3 — Find the gradient of ABABAB

Step 4 — Find the perpendicular gradient

Step 5 — Build the equation

Step 6 — Rearrange

Step 7 — Check both defining properties

Step 8 — Confirm with the equidistance property

Step 2 — Find the midpoint of $AB$

Step 3 — Find the gradient of $AB$

Step 2 — Find the midpoint of $AB$

Step 3 — Find the gradient of $AB$