Maths · 4 Coordinate geometry

4.6 Perpendicular lines

Before you start

Calculating the gradient of a line from two points on it, $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ (4.2).
Finding the midpoint of a line segment (4.3).
Rearranging a linear equation into the form $y = m x + c$ and reading off the gradient $m$ and the $y$ -intercept $c$ (4.4).
Parallel lines have equal gradients (4.5).
Multiplying fractions, turning a fraction upside down, and substituting negative numbers into an expression.

Syllabus objectives covered (1)

Find the gradient and equation of a straight line perpendicular to a given line.

Part 1 of 7

Why perpendicular gradients multiply to $- 1$

Two lines are perpendicular if they meet at a right angle, $90 °$ . In 4.5 you saw that parallel lines have equal gradients. Perpendicular lines also have gradients that are linked, and this lesson is about that link.
Start with the signs. If one line goes uphill from left to right, a line at right angles to it must go downhill. So perpendicular gradients have opposite signs, one positive and one negative.
Now the steepness. A gentle uphill line, turned through a right angle, becomes a steep downhill line. So a small gradient goes with a large perpendicular gradient.
To see the exact link, draw the gradient as a step. A line with gradient $\frac{2}{3}$ goes 3 right for every 2 up.
Turn that step through a quarter turn ( $90 °$ ) anticlockwise. The 3 right turns into 3 up. The 2 up turns into 2 left.
graph diagram — not drawn yet
A coordinate grid from -3 to 4 across and -2 to 4 up with the origin O; a solid line through O and (3, 2) with a dashed step from O 3 right to (3, 0) then 2 up to (3, 2), labelled '3 right, 2 up'; a second solid line through O and (-2, 3) with a dashed step from O 2 left to (-2, 0) then 3 up to (-2, 3), labelled '2 left, 3 up'; a right-angle mark at O between the two solid lines.
Turning the step through $90 °$ swaps the run and the rise and makes one of them negative, so the gradient $\frac{2}{3}$ becomes $- \frac{3}{2}$ .
The new step is 2 left and 3 up. Going left is a negative run, so the gradient of the turned line is
$\frac{3}{- 2} = - \frac{3}{2}$
The whole step turned through $90 °$ , so the new line is perpendicular to the first. Compare the gradients: the fraction has turned upside down and the sign has changed. Multiply them:
$\frac{2}{3} \times (- \frac{3}{2}) = - 1$
The same happens for any step. A step of $b$ right and $a$ up has gradient $\frac{a}{b}$ . Turned through $90 °$ it becomes $a$ left and $b$ up, with gradient $- \frac{b}{a}$ .
$m_{1} \times m_{2} = \frac{a}{b} \times (- \frac{b}{a}) = - 1$
So the rule is: two lines, neither of them horizontal or vertical, are perpendicular exactly when the product of their gradients is $- 1$ . Horizontal and vertical lines are the awkward case, and they have their own section below.

Part 2 of 7

Flip and change the sign: the negative reciprocal

Rearrange $m_{1} \times m_{2} = - 1$ by dividing both sides by $m_{1}$ . This gives the gradient you need directly.
$m_{2} = - \frac{1}{m_{1}}$
In words: turn the gradient upside down (its reciprocal) and change its sign. The result is called the negative reciprocal.
A whole number is a fraction over 1. Gradient $4$ is $\frac{4}{1}$ . Flip it to $\frac{1}{4}$ , then change the sign: the perpendicular gradient is $- \frac{1}{4}$ .
A negative gradient gives a positive answer. Gradient $- \frac{2}{5}$ flips to $- \frac{5}{2}$ , and changing the sign gives $\frac{5}{2}$ .
A decimal is easiest as a fraction first. $0.25 = \frac{1}{4}$ , which flips to $4$ , so the perpendicular gradient is $- 4$ .
Do both moves every time. Flip without changing the sign and gradient $4$ gives $\frac{1}{4}$ , but $4 \times \frac{1}{4} = 1$ , not $- 1$ . Change the sign without flipping and you get $- 4$ , but $4 \times (- 4) = - 16$ .
So check every perpendicular gradient by multiplying the pair. If the product is not $- 1$ , one of the two moves is missing.

Part 3 of 7

Get the given line into $y = m x + c$ first

The gradient is the number multiplying $x$ only when the equation reads $y = \dots$ with $y$ on its own. In $y = 3 x - 7$ the gradient is $3$ .
If $y$ has a number in front of it, divide every term by that number first. In $5 y = 2 x + 3$ , the $5$ is multiplying $y$ , so divide every term by $5$ :
$y = \frac{2}{5} x + \frac{3}{5}$
Now the gradient is $\frac{2}{5}$ , and the perpendicular gradient is $- \frac{5}{2}$ . Reading the $2$ straight off $5 y = 2 x + 3$ would give a wrong gradient before you even started.
For the form $a x + b y = c$ , move the $x$ term across first, then divide. From $3 x + 4 y = 12$ , subtract $3 x$ from both sides, then divide by $4$ :
$4 y = - 3 x + 12 \Rightarrow y = - \frac{3}{4} x + 3$
The gradient is $- \frac{3}{4}$ , so the perpendicular gradient is $\frac{4}{3}$ . Watch the sign as the $x$ term crosses the equals sign: $+ 3 x$ on the left becomes $- 3 x$ on the right.
A quick check for any line $a x + b y = c$ : its gradient is $- \frac{a}{b}$ . For $3 x + 4 y = 12$ that is $- \frac{3}{4}$ , which agrees with the rearrangement.

Part 4 of 7

The equation of the perpendicular line through a point

A straight line is fixed by two facts: its gradient and one point on it. 'Perpendicular to the given line' gives you the gradient. The question gives you the point.
Example: find the equation of the line through $(8, - 1)$ that is perpendicular to $y = 4 x + 3$ . Step 1 is the gradient. The given gradient is $4$ , so the new gradient is $- \frac{1}{4}$ .
Step 2: write the new line as $y = - \frac{1}{4} x + c$ . You do not know $c$ yet, but $(8, - 1)$ is on the line, so its coordinates must fit the equation. Substitute $x = 8$ and $y = - 1$ :
$- 1 = - \frac{1}{4} (8) + c \Rightarrow - 1 = - 2 + c \Rightarrow c = 1$
Step 3: write the whole equation, $y = - \frac{1}{4} x + 1$ . The answer is the equation, not the value of $c$ .
Check by putting the point back in: $- \frac{1}{4} (8) + 1 = - 2 + 1 = - 1$ , which is the $y$ -coordinate of the point.
The syllabus notes say equations of lines are expected in fully simplified form. If the question asks for the form $a x + b y = c$ with integers, clear the fraction by multiplying every term by $4$ , then move the $x$ term across:
$4 y = - x + 4 \Rightarrow x + 4 y = 4$
You may have met the form $y - y_{1} = m (x - x_{1})$ . It gives exactly the same line. Use whichever method you are sure of.

Part 5 of 7

The perpendicular bisector of a line segment

To bisect means to cut into two equal parts. The perpendicular bisector of a line segment $A B$ is the line that cuts $A B$ in half and crosses it at a right angle.
Those two words give you the two facts you need. 'Bisector': the line passes through the midpoint of $A B$ . 'Perpendicular': its gradient is the negative reciprocal of the gradient of $A B$ .
Recall the midpoint from 4.3: add the $x$ -coordinates and halve, then do the same for the $y$ -coordinates.
$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
The method has four steps, in this order: (1) the midpoint of $A B$ ; (2) the gradient of $A B$ ; (3) its negative reciprocal; (4) substitute the midpoint into $y = m x + c$ to find $c$ .
graph diagram — not drawn yet
A coordinate grid from -4 to 6 on both axes showing the segment from A(-3, 4) to B(5, -2) with its midpoint M(1, 1) marked, equal tick marks on AM and MB, and the line y = (4/3)x - 1/3 drawn through (-2, -3), M and (4, 5), with a right-angle mark where it crosses AB at M.
The bisector passes through the midpoint M, not through A or B, and it crosses AB at a right angle.
The trap is the wrong point or the wrong gradient. The bisector goes through the midpoint, not through $A$ or $B$ . It has the perpendicular gradient, not the gradient of $A B$ .
Every point on the perpendicular bisector is the same distance from $A$ as from $B$ . This gives you a check at the end: pick a point on your line and compare its distances to $A$ and to $B$ .

Part 6 of 7

Horizontal and vertical lines

A horizontal line has an equation $y = k$ and gradient $0$ . A vertical line has an equation $x = k$ . Its run is $0$ , and you cannot divide by $0$ , so a vertical line has no gradient.
A horizontal line and a vertical line are always perpendicular. The rule $m_{1} \times m_{2} = - 1$ cannot show this, because no number multiplied by $0$ gives $- 1$ . Deal with these lines by sight instead.
Perpendicular to a horizontal line means vertical. The line through $(2, 5)$ perpendicular to $y = 3$ is $x = 2$ , because every point on a vertical line has the same $x$ -coordinate as the point you were given.
Perpendicular to a vertical line means horizontal. The line through $(4, 6)$ perpendicular to $x = - 1$ is $y = 6$ .
The axes count too. The $x$ -axis is the line $y = 0$ and the $y$ -axis is the line $x = 0$ . So a line perpendicular to the $x$ -axis is vertical, and a line perpendicular to the $y$ -axis is horizontal.

Part 7 of 7

Showing that two lines are perpendicular

A Show (that) question gives you the result, and your job is to provide the evidence that leads to it. So every gradient must be worked out on the page, not just stated.
For two lines given as equations, rearrange each one to $y = m x + c$ and write down both gradients. For two sides of a shape given by vertices, work out each gradient from two points.
Then multiply the gradients. For gradients $\frac{3}{2}$ and $- \frac{2}{3}$ :
$\frac{3}{2} \times (- \frac{2}{3}) = - 1$
Finish with a sentence: 'The product of the gradients is $- 1$ , so the lines are perpendicular.' The sentence links your arithmetic to the result you were asked to show.
To show that a triangle has a right angle, test the two sides that meet at the angle named. For angle $P Q R$ the angle is at $Q$ , the middle letter, so the sides are $P Q$ and $Q R$ .

Worked examples

Example 1. Find the gradient of a line perpendicular to the line $4 x + 3 y = 9$ .

Show the working

The gradient is not the $4$ . First get $y$ on its own. Subtract $4 x$ from both sides, so the $x$ term moves to the right with its sign changed.
$3 y = - 4 x + 9$
Divide every term by $3$ , because $3$ is what is multiplying $y$ .
$y = - \frac{4}{3} x + 3$
Now read the gradient: the number multiplying $x$ is $- \frac{4}{3}$ .
Flip it and change the sign. Upside down, $- \frac{4}{3}$ becomes $- \frac{3}{4}$ . Changing the sign gives $\frac{3}{4}$ .
Check that the product is $- 1$ .
$- \frac{4}{3} \times \frac{3}{4} = - 1$

Answer: $\frac{3}{4}$

Example 2. Find the equation of the line that passes through $(4, - 3)$ and is perpendicular to the line $2 x - 5 y = 10$ . Give your answer in the form $a x + b y = c$ , where $a$ , $b$ and $c$ are integers.

Show the working

Rearrange the given line to find its gradient. Subtract $2 x$ from both sides.
$- 5 y = - 2 x + 10$
Divide every term by $- 5$ . Dividing by a negative number changes the sign of every term.
$y = \frac{2}{5} x - 2$
The given gradient is $\frac{2}{5}$ , so the perpendicular gradient is $- \frac{5}{2}$ . Check: $\frac{2}{5} \times (- \frac{5}{2}) = - 1$ .
Write the new line as $y = - \frac{5}{2} x + c$ and substitute the point, $x = 4$ and $y = - 3$ .
$- 3 = - \frac{5}{2} (4) + c \Rightarrow - 3 = - 10 + c \Rightarrow c = 7$
So the line is $y = - \frac{5}{2} x + 7$ . The question asks for integers, so multiply every term by $2$ to clear the fraction.
$2 y = - 5 x + 14$
Add $5 x$ to both sides to reach the form asked for.
$5 x + 2 y = 14$
Check with the point: $5 (4) + 2 (- 3) = 20 - 6 = 14$ , which matches.

Answer: $5 x + 2 y = 14$

Example 3. Find the equation of the perpendicular bisector of the line joining $A (- 3, 4)$ and $B (5, - 2)$ .

Show the working

Step 1, the midpoint $M$ . Add the $x$ -coordinates and halve, then do the same for $y$ .
$M = (\frac{- 3 + 5}{2}, \frac{4 + (- 2)}{2}) = (1, 1)$
Step 2, the gradient of $A B$ . Use rise over run, taking both coordinates in the same order ( $B$ minus $A$ ).
$m_{A B} = \frac{- 2 - 4}{5 - (- 3)} = \frac{- 6}{8} = - \frac{3}{4}$
Step 3, the perpendicular gradient. Flip and change the sign: $- \frac{3}{4}$ becomes $\frac{4}{3}$ .
Step 4, substitute the midpoint, because the bisector passes through $M$ , not through $A$ or $B$ .
$1 = \frac{4}{3} (1) + c \Rightarrow c = 1 - \frac{4}{3} = - \frac{1}{3}$
So the bisector is $y = \frac{4}{3} x - \frac{1}{3}$ . Multiplying every term by $3$ gives $3 y = 4 x - 1$ , which rearranges to $4 x - 3 y = 1$ : the same line without fractions.
Check with the equal-distance property. The point $(- 2, - 3)$ is on the line, since $4 (- 2) - 3 (- 3) = 1$ . Its squared distance to $A$ is $1^{2} + 7^{2} = 50$ and to $B$ is $7^{2} + 1^{2} = 50$ , so it is the same distance from both ends.

Answer: $y = \frac{4}{3} x - \frac{1}{3}$ , which is the same line as $4 x - 3 y = 1$

Example 4. Triangle $P Q R$ has vertices $P (- 1, 1)$ , $Q (3, 7)$ and $R (9, 3)$ . Show that angle $P Q R$ is a right angle.

Show the working

Angle $P Q R$ is at $Q$ , the middle letter, between the sides $P Q$ and $Q R$ . Those are the two gradients to test.
Gradient of $P Q$ :
$m_{P Q} = \frac{7 - 1}{3 - (- 1)} = \frac{6}{4} = \frac{3}{2}$
Gradient of $Q R$ :
$m_{Q R} = \frac{3 - 7}{9 - 3} = \frac{- 4}{6} = - \frac{2}{3}$
Multiply the two gradients.
$\frac{3}{2} \times (- \frac{2}{3}) = - 1$
Finish with the sentence that links the product to the angle: the product of the gradients is $- 1$ , so $P Q$ is perpendicular to $Q R$ and angle $P Q R = 90 °$ .

Answer: $m_{P Q} = \frac{3}{2}$ and $m_{Q R} = - \frac{2}{3}$ ; their product is $- 1$ , so $P Q$ is perpendicular to $Q R$ and angle $P Q R = 90 °$ .

Where people go wrong

Mistake: Flipping the gradient but keeping its sign, e.g. giving $\frac{1}{3}$ as the gradient perpendicular to $3$ .

Instead: Flip and change the sign: the perpendicular gradient is $- \frac{1}{3}$ .

$3 \times \frac{1}{3} = 1$ , not $- 1$ . Two lines that both slope upwards cannot meet at a right angle.

Mistake: Changing the sign but not flipping, e.g. giving $- 3$ as the gradient perpendicular to $3$ .

Instead: Flip as well: $- \frac{1}{3}$ .

$3 \times (- 3) = - 9$ , not $- 1$ . The steepness has to change too: a steep line turned through $90 °$ becomes a gentle one.

Mistake: Taking the gradient of $5 y = 2 x + 3$ to be $2$ .

Instead: Divide every term by $5$ first: $y = \frac{2}{5} x + \frac{3}{5}$ , so the gradient is $\frac{2}{5}$ and the perpendicular gradient is $- \frac{5}{2}$ .

The number multiplying $x$ is the gradient only when $y$ stands on its own.

Mistake: Putting the perpendicular bisector of $A B$ through $A$ or $B$ , or giving it the gradient of $A B$ .

Instead: It passes through the midpoint of $A B$ and has the negative reciprocal of the gradient of $A B$ .

'Bisector' means it cuts $A B$ in half, so it goes through the halfway point. 'Perpendicular' means it crosses $A B$ at a right angle, so its gradient is the perpendicular one.

Mistake: Stopping at $c = 1$ , or at the gradient $- \frac{1}{4}$ , when the question asks for the equation of the line.

Instead: Write the whole equation: $y = - \frac{1}{4} x + 1$ .

An equation of a line connects $x$ and $y$ . A value of $c$ or of the gradient on its own is only one part of it.

Check yourself

Write down the gradient of a line perpendicular to a line with gradient (a) $5$ (b) $- \frac{3}{4}$ (c) $0.4$ .
Answer
(a) $- \frac{1}{5}$ (b) $\frac{4}{3}$ (c) $0.4 = \frac{2}{5}$ , so the perpendicular gradient is $- \frac{5}{2}$ , which is $- 2.5$ .
Find the gradient of a line perpendicular to $3 y = 2 x - 7$ .
Answer
Divide by $3$ : $y = \frac{2}{3} x - \frac{7}{3}$ . The gradient is $\frac{2}{3}$ , so the perpendicular gradient is $- \frac{3}{2}$ .
Are the lines $y = 3 x - 2$ and $x + 3 y = 6$ perpendicular? Show how you know.
Answer
Yes. $x + 3 y = 6$ rearranges to $y = - \frac{1}{3} x + 2$ . The gradients are $3$ and $- \frac{1}{3}$ , and $3 \times (- \frac{1}{3}) = - 1$ , so the lines are perpendicular.
Write down the equation of the line through $(5, 1)$ that is perpendicular to the line $y = - 2$ .
Answer
$x = 5$ . The line $y = - 2$ is horizontal, so the perpendicular line is vertical, and every point on it has $x$ -coordinate $5$ .
Find the equation of the line through $(2, 7)$ that is perpendicular to $y = \frac{1}{2} x - 3$ .
Answer
The perpendicular gradient is $- 2$ . Then $7 = - 2 (2) + c$ , so $c = 11$ , and the line is $y = - 2 x + 11$ .
Find the equation of the perpendicular bisector of the line joining $(1, 2)$ and $(7, 6)$ .
Answer
Midpoint $(4, 4)$ . Gradient of the segment $\frac{4}{6} = \frac{2}{3}$ , so the perpendicular gradient is $- \frac{3}{2}$ . Then $4 = - \frac{3}{2} (4) + c$ gives $c = 10$ : $y = - \frac{3}{2} x + 10$ , which is the same line as $3 x + 2 y = 20$ .
Explain why a line with gradient $2$ and a line with gradient $\frac{1}{2}$ are not perpendicular.
Answer
$2 \times \frac{1}{2} = 1$ , not $- 1$ . Both lines slope upwards, and two lines that both slope upwards cannot meet at a right angle.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The line $L_{1}$ has equation $3 x - 4 y = 8$ . The point $P$ is $(1, 5)$ .

(a) Find the gradient of $L_{1}$ . 2 marks No calculator

Mark scheme

M1 correct first step in rearranging towards $y = m x + c$ , e.g. $4 y = 3 x - 8$ or $y = \frac{3}{4} x - 2$ oe, or uses gradient $= - \frac{a}{b}$ with $\frac{- 3}{- 4}$ ; implied by a correct answer
A1 $\frac{3}{4}$ oe

(b) Write down the gradient of a line perpendicular to $L_{1}$ . 1 mark No calculator

Mark scheme

B1 $- \frac{4}{3}$ oe, FT $- 1 \div$ their (a)

(c) The line $L_{2}$ passes through $P$ and is perpendicular to $L_{1}$ . Find the equation of $L_{2}$ . Give your answer in the form $a x + b y = c$ , where $a$ , $b$ and $c$ are integers. 3 marks No calculator

Mark scheme

M1 substitutes $(1, 5)$ into $y = - \frac{4}{3} x + c$ oe, FT their (b)
A1 $c = \frac{19}{3}$ oe, or $y = - \frac{4}{3} x + \frac{19}{3}$
A1 $4 x + 3 y = 19$ oe with integer coefficients

(d) Write down the equation of the line through $P$ that is perpendicular to the $x$ -axis. 1 mark No calculator

Mark scheme

B1 $x = 1$ cao

Question 2. $A$ is the point $(- 5, 3)$ and $B$ is the point $(3, - 1)$ .

(a) Find the coordinates of the midpoint of $A B$ . 1 mark No calculator

Mark scheme

B1 $(- 1, 1)$ cao

(b) Find the equation of the perpendicular bisector of $A B$ . Give your answer in the form $y = m x + c$ . 4 marks No calculator

Mark scheme

M1 gradient of $A B = \frac{- 1 - 3}{3 - (- 5)}$ or $- \frac{4}{8}$ oe
M1 perpendicular gradient $2$ , FT $- 1 \div$ their gradient of $A B$
M1 substitutes their midpoint into $y = (their perpendicular gradient) x + c$
A1 $y = 2 x + 3$ cao

(c) The perpendicular bisector crosses the $y$ -axis at $C$ . Show that $C$ is the same distance from $A$ as from $B$ . 2 marks No calculator

Mark scheme

M1 uses $C = (0, 3)$ with Pythagoras for $C A$ or $C B$ , e.g. $C B^{2} = 3^{2} + 4^{2}$
A1 $C A = 5$ and $C B = 5$ , so $C A = C B$

Question 3. Triangle $P Q R$ has vertices $P (- 2, 0)$ , $Q (2, 2)$ and $R (4, - 2)$ .

(a) Show that angle $P Q R = 90 °$ . 3 marks No calculator

Mark scheme

B1 gradient of $P Q = \frac{2}{4}$ or $\frac{1}{2}$ . Alternative method (lengths), marked B1 B1 B1 in the same way: $P Q^{2} = 20$ ; then $Q R^{2} = 20$ and $P R^{2} = 40$ ; then $20 + 20 = 40$ , so angle $P Q R = 90 °$ by the converse of Pythagoras
B1 gradient of $Q R = \frac{- 4}{2}$ or $- 2$
B1 dep on both previous marks: $\frac{1}{2} \times (- 2) = - 1$ , so $P Q$ is perpendicular to $Q R$ and angle $P Q R = 90 °$

(b) Find the equation of the line through $P$ that is perpendicular to $P R$ . 3 marks No calculator

Mark scheme

M1 gradient of $P R = \frac{- 2}{6}$ or $- \frac{1}{3}$
M1 gradient $3$ (FT $- 1 \div$ their gradient of $P R$ ) used with $(- 2, 0)$ to find $c$
A1 $y = 3 x + 6$ oe

How this is examined

Finding the perpendicular bisector of the line joining two given points: A Find worth about 4 marks, with the two end points given as coordinates; a part asking for the midpoint may come before it Four steps written in order: the midpoint, e.g. $(1, 1)$ ; the gradient of the segment from rise over run, e.g. $\frac{- 6}{8}$ ; the perpendicular gradient from $- 1 \div$ that gradient, e.g. $\frac{4}{3}$ ; and the midpoint substituted into $y = m x + c$ . The first three are method steps, and the last mark is for the whole equation, e.g. $y = \frac{4}{3} x - \frac{1}{3}$ .
Writing the equation of the line through a given point, perpendicular to a given line: A 3-mark Find; the given line may be written as $a x + b y = c$ , and the answer may have to be given in a stated form The perpendicular gradient, then the point substituted to find $c$ , e.g. $- 1 = - \frac{1}{4} (8) + c$ , then the whole equation, fully simplified: $y = - \frac{1}{4} x + 1$ . A value of $c$ on its own is not an equation. If the question asks for $a x + b y = c$ with integers, the final line must be in that form: $x + 4 y = 4$ .
Proving that two lines, or two sides of a shape, are perpendicular: A Show (that) of 2 or 3 marks, about two given equations or about an angle of a triangle whose vertices are given Both gradients worked out on the page, their product shown to be $- 1$ , and the sentence 'the product of the gradients is $- 1$ , so the lines are perpendicular'. One gradient alone, or 'it looks like a right angle', does not show it.
Reading the gradient of a line that is not written as $y = m x + c$ , then finding the perpendicular gradient: A 2-mark Find the gradient of a line perpendicular to a line such as $5 y = 2 x + 3$ or $3 x + 4 y = 12$ The rearrangement with $y$ on its own, e.g. $y = \frac{2}{5} x + \frac{3}{5}$ , which shows where the gradient comes from; then the negative reciprocal, $- \frac{5}{2}$ .
A line perpendicular to a horizontal line, a vertical line or an axis: A 1-mark Write down the equation of the line through a given point The other kind of line through the point, using the coordinate that stays fixed: perpendicular to $y = 3$ through $(2, 5)$ is $x = 2$ ; perpendicular to the $x$ -axis through $(1, 5)$ is $x = 1$ . The answer is an equation, $x = 2$ , not a number.

Words to know

Perpendicular: Meeting at a right angle ( $90 °$ ).
Gradient: The steepness of a line: the rise divided by the run, $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
Reciprocal: A number turned upside down. The reciprocal of $\frac{a}{b}$ is $\frac{b}{a}$ , and the reciprocal of $4$ is $\frac{1}{4}$ .
Negative reciprocal: The reciprocal with its sign changed. A line perpendicular to a line with gradient $m$ has gradient $- \frac{1}{m}$ .
Bisect: Cut into two equal parts.
Line segment: The part of a line between two end points.
Perpendicular bisector: The line through the midpoint of a line segment that is perpendicular to it. Every point on it is the same distance from both ends of the segment.

All Maths lessons Next: 5.1 Geometrical terms

4.6 Perpendicular lines

Why perpendicular gradients multiply to −1

Flip and change the sign: the negative reciprocal

Get the given line into y=mx+c first

The equation of the perpendicular line through a point

The perpendicular bisector of a line segment

Horizontal and vertical lines

Showing that two lines are perpendicular

Worked examples

Where people go wrong

Check yourself

Exam questions

How this is examined

Words to know

Why perpendicular gradients multiply to $- 1$

Get the given line into $y = m x + c$ first