Prove that two perpendicular lines have slopes that are negative reciprocal of each other

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.
Questions on 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
More References and LinksEquations of Lines in Different Forms,
Slope of a Line
Tutorial on Equation of Line
Line Problems
Equations of Line Through Two Points And Parallel and Perpendicular. 