Perpendicular Lines and the equation of a tangent

What you'll learn

Read gradients and intercepts from $y = m x + c$ .
Find equations of straight lines from points and gradients.
Use negative reciprocals for perpendicular lines.
Find tangent equations using the radius of a circle.

1. Straight lines in $y = m x + c$ form

In the equation $y = m x + c$ , the variables $x$ and $y$ represent coordinates of any point on the line.

Definition

Gradient and y-intercept

The gradient is the steepness of a line; in $y = m x + c$ it is $m$ . The y-intercept is where the line crosses the y-axis; in $y = m x + c$ it is $c$ , so the line passes through $(0, c)$ .

A straight line in y=mx+c form showing the gradient as rise over run and the y-intercept at (0,c).

A positive gradient slopes upwards from left to right. A negative gradient slopes downwards.

Parallel lines

Parallel lines are straight lines that never meet. Non-vertical parallel lines have the same gradient.

Key Idea

Parallel lines

To write a line parallel to $y = m x + c$ , keep the same gradient $m$ but choose a different y-intercept.

Example

Writing a parallel line and a line through a given y-intercept

Suppose a line has equation $y=34x+2y=\frac{3}{4}x+2$ .

The original line, a parallel line with the same gradient, and a different line through the same y-intercept.

The gradient is $34\frac{3}{4}$ and the y-intercept is 2.
A different line parallel to it must keep gradient $34\frac{3}{4}$ , so one possible answer is $y=34x−5y=\frac{3}{4}x-5$ .
A different line through (0, 2) must keep y-intercept 2, but can have a different gradient, so one possible answer is $y = - 2 x + 2$ .

2. Finding a line from two points

If a line passes through two points, the gradient is:

m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}

This means:

change in y goes on top;
change in x goes on the bottom.

Then use $y = m x + c$ to find $c$ .

Example

Finding the equation of a line through two points

Find the equation of the line through (0, 6) and (4, 18).

The line through the two given points with the change in x and change in y highlighted for the gradient calculation.

Use the gradient formula:
$m=18−64−0=124=3m=\frac{18-6}{4-0}=\frac{12}{4}=3$
The point (0, 6) shows the y-intercept directly, so $c = 6$ .
Substitute $m = 3$ and $c = 6$ into $y = m x + c$ :
$y = 3 x + 6$

3. Perpendicular gradients

Perpendicular lines meet at a right angle, 90°.

Key Idea

Perpendicular gradients

For two non-vertical straight lines, perpendicular gradients multiply to -1. If one gradient is $m$ , the perpendicular gradient is $−1m-\frac{1}{m}$ .

Examples:

gradient 4 has perpendicular gradient $−14-\frac{1}{4}$ ;
gradient $−23-\frac{2}{3}$ has perpendicular gradient $32\frac{3}{2}$ .

To show two lines are perpendicular, rearrange both into $y = m x + c$ and check their gradients multiply to -1.

Common Mistake

Forgetting the negative sign

The reciprocal alone is not enough: a line with gradient $25\frac{2}{5}$ has perpendicular gradient $−52-\frac{5}{2}$ , not $52\frac{5}{2}$ .

Example

A perpendicular line through a point

A line passes through (6, 7) and is perpendicular to $y = 2 x - 4$ . Find its equation.

The given line and the required perpendicular line through (6,7) meet at a right angle.

The given line has gradient 2.
The perpendicular gradient is $−12-\frac{1}{2}$ .
Start with $y=−12x+cy=-\frac{1}{2}x+c$ .
Substitute the point (6, 7) to find $c$ :
$7=−12(6)+c=−3+c⇒c=107=-\frac{1}{2}(6)+c=-3+c \Rightarrow c=10$
Write the equation:
$y=−12x+10y=-\frac{1}{2}x+10$

4. Perpendicular bisectors

A line segment is the finite piece of a line between two endpoints. The midpoint is exactly halfway between the endpoints.

For endpoints $x_1,y_1)$ and $x_2,y_2)$ , the midpoint is:

(x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

Definition

Perpendicular bisector

The perpendicular bisector of a line segment passes through the midpoint of the segment and meets the segment at 90°.

A perpendicular bisector passes through the midpoint of a line segment and forms a right angle with it.

Example

Finding a perpendicular bisector

Points A and B are (2, 3) and (8, 5). Find the perpendicular bisector of AB.

Segment AB, its midpoint (5,4), and the perpendicular bisector to be found.

Find the midpoint:
$(2+82,3+52)=(5,4)\left(\frac{2+8}{2},\frac{3+5}{2}\right)=(5,4)$
Find the gradient of AB:
$m=5−38−2=26=13m=\frac{5-3}{8-2}=\frac{2}{6}=\frac{1}{3}$
The perpendicular gradient is -3.
Use the midpoint (5, 4) in $y = - 3 x + c$ :
$\Rightarrow c=19$
Write the perpendicular bisector:
$y = - 3 x + 19$

5. Tangents to circles

A circle is the set of points a fixed distance from a centre.

Definition

Radius and tangent

A radius is a line segment from the centre of a circle to its edge. A tangent is a straight line that touches a circle at exactly one point.

The radius to the point of contact is perpendicular to the tangent.

The key fact is: the radius to the point of contact is perpendicular to the tangent.

So to find a tangent:

Find the gradient from the centre to the point on the circle.
Take the perpendicular gradient.
Use the point on the circle in $y = m x + c$ .

For a circle centred at the origin, $x^2+y^2=r^2$ . So $x^2+y^2=20$ has centre (0, 0) and radius $20=25\sqrt{20}=2\sqrt{5}$ .

Example

Tangent to a circle centred at the origin

The circle $x^2+y^2=20$ has point P(2, 4) on it. Find the tangent at P.

The radius from the origin to P(2,4) is perpendicular to the tangent at P.

The centre is (0, 0).
Find the gradient of the radius from the centre to P:
$m=4−02−0=2m=\frac{4-0}{2-0}=2$
The tangent is perpendicular to the radius, so its gradient is $−12-\frac{1}{2}$ .
Use point P(2, 4) in $y=−12x+cy=-\frac{1}{2}x+c$ :
$4=−12(2)+c=−1+c⇒c=54=-\frac{1}{2}(2)+c=-1+c \Rightarrow c=5$
Write the tangent equation:
$y=−12x+5y=-\frac{1}{2}x+5$

Tip

Tangent sanity check

Your tangent gradient should be the negative reciprocal of the radius gradient. If it is the same gradient, you have found a line parallel to the radius, not a tangent.

Exam technique

In the exam

Rearrange any straight-line equation into $y = m x + c$ before comparing gradients.
For perpendicular lines, flip the fraction and change the sign.
For a tangent, draw or imagine the radius from the centre to the point of contact first.
Use substitution carefully to find $c$ , then give your final answer as a full equation.

Self review

Check yourself

If a line has gradient $47\frac{4}{7}$ , what is the perpendicular gradient?
What two things does a perpendicular bisector always do?
Why is the radius useful when finding the equation of a tangent?

Perpendicular Lines and the equation of a tangent

What you'll learn

Read gradients and intercepts from $y = m x + c$ .
Find equations of straight lines from points and gradients.
Use negative reciprocals for perpendicular lines.
Find tangent equations using the radius of a circle.

1. Straight lines in $y = m x + c$ form

In the equation $y = m x + c$ , the variables $x$ and $y$ represent coordinates of any point on the line.

Definition

Gradient and y-intercept

The gradient is the steepness of a line; in $y = m x + c$ it is $m$ . The y-intercept is where the line crosses the y-axis; in $y = m x + c$ it is $c$ , so the line passes through $(0, c)$ .

A straight line in y=mx+c form showing the gradient as rise over run and the y-intercept at (0,c).

A positive gradient slopes upwards from left to right. A negative gradient slopes downwards.

Parallel lines

Parallel lines are straight lines that never meet. Non-vertical parallel lines have the same gradient.

Key Idea

Parallel lines

To write a line parallel to $y = m x + c$ , keep the same gradient $m$ but choose a different y-intercept.

Example

Writing a parallel line and a line through a given y-intercept

Suppose a line has equation $y=34x+2y=\frac{3}{4}x+2$ .

The original line, a parallel line with the same gradient, and a different line through the same y-intercept.

The gradient is $34\frac{3}{4}$ and the y-intercept is 2.
A different line parallel to it must keep gradient $34\frac{3}{4}$ , so one possible answer is $y=34x−5y=\frac{3}{4}x-5$ .
A different line through (0, 2) must keep y-intercept 2, but can have a different gradient, so one possible answer is $y = - 2 x + 2$ .

2. Finding a line from two points

If a line passes through two points, the gradient is:

m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}

This means:

change in y goes on top;
change in x goes on the bottom.

Then use $y = m x + c$ to find $c$ .

Example

Finding the equation of a line through two points

Find the equation of the line through (0, 6) and (4, 18).

The line through the two given points with the change in x and change in y highlighted for the gradient calculation.

Use the gradient formula:
$m=18−64−0=124=3m=\frac{18-6}{4-0}=\frac{12}{4}=3$
The point (0, 6) shows the y-intercept directly, so $c = 6$ .
Substitute $m = 3$ and $c = 6$ into $y = m x + c$ :
$y = 3 x + 6$

3. Perpendicular gradients

Perpendicular lines meet at a right angle, 90°.

Key Idea

Perpendicular gradients

For two non-vertical straight lines, perpendicular gradients multiply to -1. If one gradient is $m$ , the perpendicular gradient is $−1m-\frac{1}{m}$ .

Examples:

gradient 4 has perpendicular gradient $−14-\frac{1}{4}$ ;
gradient $−23-\frac{2}{3}$ has perpendicular gradient $32\frac{3}{2}$ .

To show two lines are perpendicular, rearrange both into $y = m x + c$ and check their gradients multiply to -1.

Common Mistake

Forgetting the negative sign

The reciprocal alone is not enough: a line with gradient $25\frac{2}{5}$ has perpendicular gradient $−52-\frac{5}{2}$ , not $52\frac{5}{2}$ .

Example

A perpendicular line through a point

A line passes through (6, 7) and is perpendicular to $y = 2 x - 4$ . Find its equation.

The given line and the required perpendicular line through (6,7) meet at a right angle.

The given line has gradient 2.
The perpendicular gradient is $−12-\frac{1}{2}$ .
Start with $y=−12x+cy=-\frac{1}{2}x+c$ .
Substitute the point (6, 7) to find $c$ :
$7=−12(6)+c=−3+c⇒c=107=-\frac{1}{2}(6)+c=-3+c \Rightarrow c=10$
Write the equation:
$y=−12x+10y=-\frac{1}{2}x+10$

4. Perpendicular bisectors

A line segment is the finite piece of a line between two endpoints. The midpoint is exactly halfway between the endpoints.

For endpoints $x_1,y_1)$ and $x_2,y_2)$ , the midpoint is:

(x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

Definition

Perpendicular bisector

The perpendicular bisector of a line segment passes through the midpoint of the segment and meets the segment at 90°.

A perpendicular bisector passes through the midpoint of a line segment and forms a right angle with it.

Example

Finding a perpendicular bisector

Points A and B are (2, 3) and (8, 5). Find the perpendicular bisector of AB.

Segment AB, its midpoint (5,4), and the perpendicular bisector to be found.

Find the midpoint:
$(2+82,3+52)=(5,4)\left(\frac{2+8}{2},\frac{3+5}{2}\right)=(5,4)$
Find the gradient of AB:
$m=5−38−2=26=13m=\frac{5-3}{8-2}=\frac{2}{6}=\frac{1}{3}$
The perpendicular gradient is -3.
Use the midpoint (5, 4) in $y = - 3 x + c$ :
$\Rightarrow c=19$
Write the perpendicular bisector:
$y = - 3 x + 19$

5. Tangents to circles

A circle is the set of points a fixed distance from a centre.

Definition

Radius and tangent

A radius is a line segment from the centre of a circle to its edge. A tangent is a straight line that touches a circle at exactly one point.

The radius to the point of contact is perpendicular to the tangent.

The key fact is: the radius to the point of contact is perpendicular to the tangent.

So to find a tangent:

Find the gradient from the centre to the point on the circle.
Take the perpendicular gradient.
Use the point on the circle in $y = m x + c$ .

For a circle centred at the origin, $x^2+y^2=r^2$ . So $x^2+y^2=20$ has centre (0, 0) and radius $20=25\sqrt{20}=2\sqrt{5}$ .

Example

Tangent to a circle centred at the origin

The circle $x^2+y^2=20$ has point P(2, 4) on it. Find the tangent at P.

The radius from the origin to P(2,4) is perpendicular to the tangent at P.

The centre is (0, 0).
Find the gradient of the radius from the centre to P:
$m=4−02−0=2m=\frac{4-0}{2-0}=2$
The tangent is perpendicular to the radius, so its gradient is $−12-\frac{1}{2}$ .
Use point P(2, 4) in $y=−12x+cy=-\frac{1}{2}x+c$ :
$4=−12(2)+c=−1+c⇒c=54=-\frac{1}{2}(2)+c=-1+c \Rightarrow c=5$
Write the tangent equation:
$y=−12x+5y=-\frac{1}{2}x+5$

Tip

Tangent sanity check

Your tangent gradient should be the negative reciprocal of the radius gradient. If it is the same gradient, you have found a line parallel to the radius, not a tangent.

Exam technique

In the exam

Rearrange any straight-line equation into $y = m x + c$ before comparing gradients.
For perpendicular lines, flip the fraction and change the sign.
For a tangent, draw or imagine the radius from the centre to the point of contact first.
Use substitution carefully to find $c$ , then give your final answer as a full equation.

Self review

Check yourself

If a line has gradient $47\frac{4}{7}$ , what is the perpendicular gradient?
What two things does a perpendicular bisector always do?
Why is the radius useful when finding the equation of a tangent?

Perpendicular Lines and the equation of a tangent

What you'll learn

1. Straight lines in $y = m x + c$ form

Parallel lines

2. Finding a line from two points

3. Perpendicular gradients

4. Perpendicular bisectors

5. Tangents to circles

Recap questions

You've reached the end

Perpendicular Lines and the equation of a tangent

What you'll learn

1. Straight lines in $y = m x + c$ form

Parallel lines

2. Finding a line from two points

3. Perpendicular gradients

4. Perpendicular bisectors

5. Tangents to circles

Recap questions

You've reached the end

Perpendicular Lines and the equation of a tangent

What you'll learn

1. Straight lines in y=mx+cy=mx+cy=mx+c form

Parallel lines

2. Finding a line from two points

3. Perpendicular gradients

4. Perpendicular bisectors

5. Tangents to circles

Recap questions

You've reached the end

Perpendicular Lines and the equation of a tangent Revision Guide

Perpendicular Lines and the equation of a tangent

What you'll learn

1. Straight lines in y=mx+cy=mx+cy=mx+c form

Parallel lines

2. Finding a line from two points

3. Perpendicular gradients

4. Perpendicular bisectors

5. Tangents to circles

Recap questions

You've reached the end

Perpendicular Lines and the equation of a tangent Revision Guide

1. Straight lines in $y = m x + c$ form

1. Straight lines in $y = m x + c$ form