A detailed tutorial on how to prove that the slopes of two perpendicular lines are the negative reciprocal of each other. Questions on perpendicular lines are also included with their detailed solutions and explanations.

- The figure below shows two perpendicular lines L1 and L2.

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- Let the equation of line L1 be

y = m_{1}x + b_{1}

- We now select two points A and B on line L1 whose x coordinates are 6 and 8 respectively and find their y coordinates.

A(6 , 6 m_{1}+ b_{1})

B(7 , 7 m_{1}+ b_{1})

- We now find the components of vector AB.

vector(AB) =**<**7 - 6 , (7 m_{1}+ b_{1}) - (6 m_{1}+ b_{1})**>**

=**<**1 , m_{1}**>**

- Let y = m
_{2}x + b_{2}be the equation of line L2. We now select two points C and D on line L2 whose x coordinates are 4 and 5 respectively and find their y coordinates.

A(4 , 4 m_{2}+ b_{2})

B(5 , 5 m_{2}+ b_{2})

vector(CD) =**<**5 - 4 , (5 m_{2}+ b_{2}) - (4 m_{2}+ b_{2})**>**

=**<**1 , m_{2}**>**

- If the two lines are perpendicular then the two vectors AB and CD are orthogonal and their scalar product must be zero. Hence

**<**1 , m_{1}**>**.**<**1 , m_{2}**>**= 0

- Which gives

1 + m_{1}× m_{2}= 0

- The above may be written as

m_{1}× m_{2}= - 1 , the product of the slopes of two perpendicular lines is equal to - 1.

m_{1}= - 1 / m_{2}, m_{1}is equal to the negative reciprocal of m_{2}

m_{2}= - 1 / m_{1}, m_{2}is equal to the negative reciprocal of m_{1}

- NOTE: Points A, B, C and D and their x coordinates can chosen anywhere on the two lines. As an exercise choose other values for the x coordinates of points A, B, C and D and redo the calculation above. You should be able to make the same conclusion concerning the slopes of two perpendicular lines.

__Question 1__

Which of the lines given by the equations,

a) y = - 2 x - 3 b) - 4 y + 5 x = 4 c) (1 / 2) x - y = 2 d) - 6 y - 3 x = 9

are perpendicular?

Solution to Question 1

We first need to write each line in slope intercept form,
y = m x + b and find its slope m.

a) y = - 2 x - 3, slope = - 2

b) - 4 y + 5 x = 4, solve for y, y = (5 / 4) x - 1, slope = 5 / 4

c) (1 / 2) x - y = 2, solve for y, y = (1 / 2) x - 2, slope = 1 / 2

d) - 6 y - 3 x = 9, solve for y, y = - (1 / 2) x - 3 / 2, slope = - 1 / 2

The product of the slopes of the line in part a) and c) is

(-2)(1 / 2) = - 1

No other pair of lines have slopes with product equal to - 1.

Therefore the lines in part a) and c) are perpendicular.

__Question 2__

Find k so that the line through the points (1 , - 2) and (0 , 3) and the line through the points (3 , - k ) and ( 2 , - 5) are perpendicular.

Solution to Question 2

Find the slope m_{1} through the points (1 , - 2) and (0 , 3)

m_{1} = (3 - (- 2)) /(0 - 1) = - 5

Find the slope m_{2} through the points (3 , - k ) and ( 2 , - 5)

m_{2} = (- 5 - (- k)) / (2 - 3) = (- 5 + k) / (- 1) = 5 - k

The two lines are perpendicular, therefore the product of their slopes m_{1} and m_{2} must be equal to - 1

- 5 × (5 - k) = - 1

Solve for k

k = 24 / 5

__Question 3__

Find k so that the lines with equations - k x + 2 x - y = -3 and - 2 x + y = 7 are perpendicular.

Solution to Question 3

Solve for y each of the given equations and write it in slope intercept form y = m x + b then find its slope m.

- k x + 2 x - y = - 3 , solve for y, y = (- k + 2) x + 3 , slope = - k + 2

- 2 x + y = 7 , solve for y , y = 2 x + 7 , slope = 2

The two lines are perpendicular and therefore the product of their slopes is equal to - 1; hence

2 (- k + 2) = - 1

Solve the above equation for k to obtain

k = 5 / 2

__Question 4__

Find the coordinate of point C so that the points A(0 , 2) , B(2 , 6) and C(2 k , k) are the vertices of a right angled triangle with hypotenuse AC.

Solution to Question 4

For the points A, B and C to be the vertices of a right angled triangle with hypotenuse AC, segment AB must be perpendicular to segment BC.

slope of segment AB = (6 - 2) / (2 - 0) = 2

slope of segment BC = (k - 6) / (2 k - 2)

for the segments AB and BC to be perpendicular, the product of their slopes must be equal to - 1. Hence

2 × (k - 6) / (2 k - 2) = - 1

cross multiply to obtain

2 × (k - 6) = - 2 k + 2

We now solve the above equation for k to obtain.

k = 7 / 2

Coordinates of point C are given by

(2 k , k ) = (7 , 7 / 2)

__Question 5__

Find the equation of a line L_{1} that passes through the point (3 , 1) and is perpendicular to the line L_{2} with equation - x + 2 y - 6 = 0.

Solution to Question 5

In order to find the slope m_{2} of the line L_{2}, we need to write its equation in slope intercept form y = m x + b. Hence solve - x + 2 y - 6 = 0 for y to obtain

y = (1 / 2) x + 3

Hence

m_{2} = 1 / 2

Let m_{1} be the slope of the line L_{1}. L_{1} and L_{2} are perpendicular and therefore the product of their slopes is equal to -1. Hence

m_{1} × (1 / 2) = - 1

Solve the above to obtain

m_{1} = - 2

We now write the equation of the line L_{1} in point ( 3 , 1) slope m_{1} form is given by

y - 1 = - 2(x - 3)

which simplifies to

y = - 2 x + 7

__Question 6__

Find the equation of a line L_{1} that passes through the point (- 2 , - 1) and is perpendicular to the line L_{2} through the points (0 , - 2) and ( 2 , 4).

Solution to Question 6

The slope m_{2} of the line L_{2} through the points (0 , - 2) and ( 2 , - 4) is given by

m_{2} = ( 4 - (-2)) / (2 - 0) = 3

Let m_{1} be the slope of the line L_{1}. L_{1} and L_{2} being perpendicular the product of their slopes is equal to -1. Hence

m_{1} × (3) = - 1

Solve the above to obtain
m_{1} = - 1 / 3

The equation of line L_{1} in point ( - 2 , - 1) slope m_{1} form is given by

y - (-1) = ( - 1 / 3)(x - (-2))

which simplifies to

y = - (1 / 3) x - 5 / 3

Slope of a Line

Tutorial on Equation of Line

Line Problems

Equations of Line Through Two Points And Parallel and Perpendicular.