Open Math›Year 10›Linear relationships›Perpendicular lines

Perpendicular lines

Find the gradient of a perpendicular line using

m_1 \times m_2 = -1

and write equations of perpendicular lines.

Worked examples

Finding the perpendicular gradient

Straightforward

Problem

A line has equation

y = 5x - 3

. Find the gradient of a line perpendicular to it.

Read the gradient of the given line.

y = 5x - 3

, the gradient is

m_1 = 5

Apply the perpendicular gradient rule

m_1 \times m_2 = -1

5 \times m_2 = -1 \implies m_2 = -\frac{1}{5}

State the perpendicular gradient as the negative reciprocal.

To find the negative reciprocal of

5

: flip it to

\frac{1}{5}

, then negate to get

-\frac{1}{5}

. Check:

5 \times \left(-\frac{1}{5}\right) = -1

✓

Answer

The gradient of any perpendicular line is

-\dfrac{1}{5}

Writing the equation of a perpendicular line with a given y-intercept

Moderate

Problem

Write the equation of a line perpendicular to

y = -2x + 5

with y-intercept

3

Identify the gradient of the given line.

The gradient of

y = -2x + 5

m_1 = -2

Find the perpendicular gradient using

m_1 \times m_2 = -1

-2 \times m_2 = -1 \implies m_2 = \frac{1}{2}

The negative reciprocal of

-2

\frac{1}{2}

Write the equation using the perpendicular gradient and the given y-intercept.

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

Verify the gradients satisfy

m_1 \times m_2 = -1

m_1 \times m_2 = (-2) \times \frac{1}{2} = -1

✓

Answer

y = \dfrac{1}{2}x + 3

Writing the equation of a perpendicular line through a given point

Challenging

Problem

Find the equation of a line perpendicular to

y = 3x + 1

and passing through the point

(6, 2)

Identify the gradient of the given line.

The gradient of

y = 3x + 1

m_1 = 3

Find the perpendicular gradient.

3 \times m_2 = -1 \implies m_2 = -\frac{1}{3}

Write the equation of the perpendicular line as

y = -\frac{1}{3}x + c

The gradient is

-\frac{1}{3}

, so the equation has the form

y = -\frac{1}{3}x + c

for some unknown

c

Substitute the given point

(6, 2)

to find

c

2 = -\frac{1}{3}(6) + c = -2 + c \implies c = 4

Write the final equation and verify.

Equation:

y = -\frac{1}{3}x + 4

. Check point:

y = -\frac{1}{3}(6) + 4 = -2 + 4 = 2

✓. Check perpendicularity:

3 \times \left(-\frac{1}{3}\right) = -1

✓

Answer

y = -\dfrac{1}{3}x + 4

Practise

Q1·Straightforward

What is the gradient of a line perpendicular to

y = 2x + 3

Explanation

The given line has gradient

m_1 = 2

. For perpendicular lines,

m_1 \times m_2 = -1

, so:

2 \times m_2 = -1 \implies m_2 = -\frac{1}{2}

The perpendicular gradient is the negative reciprocal of

2

Q2·Straightforward

What is the gradient of a line perpendicular to

y = -3x + 1

Explanation

The given line has gradient

m_1 = -3

. Applying

m_1 \times m_2 = -1

-3 \times m_2 = -1 \implies m_2 = \frac{1}{3}

The negative reciprocal of

-3

\frac{1}{3}

Q3·Straightforward

Which of the following lines is perpendicular to

y = 4x - 2

$y = -\dfrac{1}{4}x + 1$$y = 4x + 1$$y = \dfrac{1}{4}x + 1$$y = -4x + 1$

Explanation

The gradient of

y = 4x - 2

m_1 = 4

. The perpendicular gradient is

m_2 = -\frac{1}{4}

because

4 \times \left(-\frac{1}{4}\right) = -1

. Only

y = -\frac{1}{4}x + 1

has gradient

-\frac{1}{4}

. The option

y = -4x + 1

has gradient

-4

, which gives

4 \times (-4) = -16 \neq -1

, so it is not perpendicular.

Q4·Straightforward

What is the gradient of a line perpendicular to

y = \dfrac{1}{5}x - 4

Explanation

The given line has gradient

m_1 = \frac{1}{5}

. Using

m_1 \times m_2 = -1

\frac{1}{5} \times m_2 = -1 \implies m_2 = -5

The negative reciprocal of

\frac{1}{5}

-5

Q5·Moderate

Write the equation of a line perpendicular to

y = 2x + 1

and passing through the origin.

$y = -\dfrac{1}{2}x$$y = 2x$$y = \dfrac{1}{2}x$$y = -2x$

Explanation

The gradient of

y = 2x + 1

m_1 = 2

. The perpendicular gradient is

m_2 = -\frac{1}{2}

. Since the line passes through the origin,

c = 0

, so the equation is:

y = -\frac{1}{2}x

Note:

y = -2x

has gradient

-2

, but

2 \times (-2) = -4 \neq -1

, so it is not perpendicular to

y = 2x + 1

Q6·Moderate

Write the equation of a line perpendicular to

y = 3x - 5

with y-intercept

4

$y = -\dfrac{1}{3}x + 4$$y = 3x + 4$$y = \dfrac{1}{3}x + 4$$y = -3x + 4$

Explanation

The gradient of

y = 3x - 5

m_1 = 3

. The perpendicular gradient is

m_2 = -\frac{1}{3}

because

3 \times \left(-\frac{1}{3}\right) = -1

. With y-intercept

c = 4

y = -\frac{1}{3}x + 4

Q7·Moderate

Write the equation of a line perpendicular to

y = -4x + 2

with y-intercept

-3

$y = \dfrac{1}{4}x - 3$$y = -4x - 3$$y = 4x - 3$$y = -\dfrac{1}{4}x - 3$

Explanation

The gradient of

y = -4x + 2

m_1 = -4

. The perpendicular gradient:

-4 \times m_2 = -1 \implies m_2 = \frac{1}{4}

With y-intercept

c = -3

y = \frac{1}{4}x - 3

Q8·Moderate

Line

A

has equation

3x - y + 1 = 0

. A line perpendicular to

A

passes through the origin. What is its equation?

$y = -\dfrac{1}{3}x$$y = 3x$$y = \dfrac{1}{3}x$$y = -3x$

Explanation

Rearranging

3x - y + 1 = 0

y = 3x + 1

The gradient is

m_1 = 3

. The perpendicular gradient:

3 \times m_2 = -1 \implies m_2 = -\frac{1}{3}

Since the line passes through the origin,

c = 0

y = -\frac{1}{3}x

Q9·Challenging

A line is perpendicular to

y = 2x + 3

and passes through the point

(4, 1)

. What is the y-intercept of this line?

Explanation

The gradient of

y = 2x + 3

m_1 = 2

. Perpendicular gradient:

m_2 = -\frac{1}{2}

. Write

y = -\frac{1}{2}x + c

and substitute

(4, 1)

1 = -\frac{1}{2}(4) + c = -2 + c \implies c = 3

The perpendicular line is

y = -\frac{1}{2}x + 3

. Verify:

m_1 \times m_2 = 2 \times \left(-\frac{1}{2}\right) = -1

✓

Q10·Challenging

A line is perpendicular to

y = -3x + 1

and passes through the point

(3, 2)

. What is the y-intercept of this line?

Explanation

The gradient of

y = -3x + 1

m_1 = -3

. Perpendicular gradient:

-3 \times m_2 = -1 \implies m_2 = \frac{1}{3}

. Write

y = \frac{1}{3}x + c

and substitute

(3, 2)

2 = \frac{1}{3}(3) + c = 1 + c \implies c = 1

The perpendicular line is

y = \frac{1}{3}x + 1

. Verify:

(-3) \times \frac{1}{3} = -1

✓

Q11·Challenging

A line is perpendicular to

y = 4x - 1

and passes through the point

(2, -3)

. What is the y-intercept of this line?

Explanation

The gradient of

y = 4x - 1

m_1 = 4

. Perpendicular gradient:

4 \times m_2 = -1 \implies m_2 = -\frac{1}{4}

. Write

y = -\frac{1}{4}x + c

and substitute

(2, -3)

-3 = -\frac{1}{4}(2) + c = -\frac{1}{2} + c \implies c = -3 + \frac{1}{2} = -\frac{5}{2}

The perpendicular line is

y = -\frac{1}{4}x - \frac{5}{2}

. Verify:

4 \times \left(-\frac{1}{4}\right) = -1

✓

Q12·Challenging

A line is perpendicular to

y = \dfrac{1}{2}x - 1

and passes through the point

(2, 0)

. Which equation is correct?

$y = -2x + 4$$y = 2x - 4$$y = -2x - 4$$y = \dfrac{1}{2}x + 4$

Explanation

The gradient of

y = \frac{1}{2}x - 1

m_1 = \frac{1}{2}

. Perpendicular gradient:

\frac{1}{2} \times m_2 = -1 \implies m_2 = -2

. Write

y = -2x + c

and substitute

(2, 0)

0 = -2(2) + c = -4 + c \implies c = 4

The equation is

y = -2x + 4

. Verify:

\frac{1}{2} \times (-2) = -1

✓