Question

Find the coordinates of the foot of perpendicular from the point ( – 1, 3) to the line 3 x – 4 y – 16 = 0.

Answer 1

The perpendicular is drawn from the point ( −1,3 ) to the line 3x−4y−16=0 .

Let ( p,q ) be the coordinate of the foot of the perpendicular from point ( −1,3 )

The equation of line having slope m and making an intercept c with the y axis is given by

y=mx+c (1)

Let the slope of the line 3x−4y−16=0 is m 1 .

Rearrange the terms of the equation

4y=3x−16 y= 3 4 x− 16 4 y= 3 4 x−4

Compare the above expression with equation (1).

m 1 = 3 4 ,c=−4 (2)

The slope of the line is obtained as m 1 = 3 4 .

The formula for the slope of a line passes through points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by,

m= y 2 − y 1 x 2 − x 1 (3)

Let m 2 be the slope of the line segment which passes through the points ( −1,3 ) and ( p,q ) .

Substitute the value for ( x 1 , y 1 ) and ( x 2 , y 2 ) as ( −1,3 ) and ( p,q ) respectively in equation (3).

m 2 = q−3 p+1 (4)

The product of the slope of two lines perpendicular to each other is −1 .

m 1 ⋅ m 2 =−1 (5)

Substitute the value of m 1 and m 2 from equation (2) and equation (4) to equation (5) respectively.

3 4 ⋅( q−3 p+1 )=−1 3q−9 4p+4 =−1 3q−9=−1×( 4p+4 ) 3q−9=−4p−4

Further simplify the above expression.

4p+3q=9−4 4p+3q=5 (6)

As the point ( p,q ) lies on the line 3x−4y−16=0 so it satisfies the equation of the line.

3p−4q−16=0 3p−4q=16 (7)

Solve the equation (6) and equation (7) to obtain the values of ( p,q ) .

p= 68 25    q=− 49 25

Thus, the coordinates of the foot of the perpendicular drawn from the point ( −1,3 ) to the line 3x−4y−16=0 is ( 68 25 , 49 25 )