PROPERTIES OF PERPENDICULAR LINES

Property 1 :

Let m1 and m2 be the slopes of two lines.

If, the two lines are perpendicular, then the product of their slopes is equal to -1.

m1 x m2 = -1

Property 2 :

Let us consider the general form of equation of a straight line ax + by + c = 0.  

If the two lines are perpendicular, then their general form of equations will differ as shown in the figure below.

Property 3 :

Let us consider the slope intercept form of equation of a straight line  y = mx + b.

If the two lines are perpendicular, then their slope-intercept form equations will differ as given in the figure below

Property 4 :

If the two lines are perpendicular, the angle between them will be 90°.

The figure shown below illustrates the above property.

Solving Problems Using Properties of Perpendicular Lines

Problem 1 :

The slopes of the two lines are 7 and (3k + 2). If the two lines are perpendicular, find the value of k.

Solution :

If the given two lines are perpendicular, then the product of the slopes is equal to -1.

7(3k + 2) = -1

Use distributive property.

21k + 14 = -1

Subtract 14 from each side.

21k = -15

Divide each side by 21.

k = -15/21

k = -5/7

Problem 2 :

The equations of the two perpendicular lines are

3x + 2y - 8 = 0

(5k+3) - 3y + 1 = 0

Find the value of k.

Solution :

If the two lines are perpendicular, then the coefficient 'y' term in the first line is equal to the coefficient of 'x' term in the second line.

5k + 3 = 2

Subtract 3 from each side.

5k = -1

Divide each side by 5.

k = -1/5

Problem 3 :

Find the equation  of a straight line is passing through (2, 3) and perpendicular to the line 2x - y + 7 = 0.

Solution :

Required line is perpendicular to 2x - y + 7 = 0.

Then, the equation of the required line is

x  + 2y + k = 0 ----(1)

The required line is passing through (2, 3).

Substitute x = 2 and y = 3 in (1).

2 + 2(3) + k = 0

2 + 6 + k = 0

8 + k = 0

Subtract 8 from each side.

k = -8

Problem 4 :

Verify, whether the following two lines  re perpendicular.

3x - 2y - 7 = 0

y = -(2x/3) + 4

Solution :

In the equations of the given two lines, the equation of the second line is not in general form.

Let us write the equation of the second line in general form.

y = -(2x/3) + 4

Multiply each side by 3.

3y = -2x + 12

2x + 3y - 12 = 0

Compare the equations of two lines,

3x - 2y - 7 = 0

2x + 3y - 12 = 0

When we look at the general form of equations of the above two lines, we get the following points.

(i) The sign of y terms are different.

(ii) The coefficient of x term in the first equation is the coefficient of y term in the second equation.

(iii) The coefficient of y term in the first equation is the coefficient of x term in the second equation.

(iv)  The above equations differ in constant terms.

Considering the above points, it is clear that the given two lines are perpendicular.

Problem 5 :

Verify, whether the following two lines are perpendicular.

5x + 7y - 1 = 0

14x - 10y + 5 = 0

Solution :

In the equation of the second line 14x - 10y + 5 = 0, the coefficients of x and y have the common divisor 2.

Divide the second equation by 2.

7x - 5y + 2.5 = 0

Comparing the equations of two lines,

5x + 7y - 1 = 0

7x - 5y + 2.5 = 0

When we look at the general form of equations of the above two lines, we get the following points.

(i) The sign of y- terms are different.

(ii) The coefficient of x term in the first equation is the coefficient of y term in the second equation.

(iii) The coefficient of y term in the first equation is the coefficient of x term in the second equation.

(iv)  The above equations differ in constant terms.

Considering the above points, it is clear that the given two lines are perpendicular.

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