Question

Find the equation of a line drawn perpendicular to the line through the point, where it meets the y -axis.

Answer 1

The equation of line is x 4 + y 6 =1.

The above equation can be rewritten as,

3x+2y−12=0 2y=−3x+12 y=− 3 2 x+ 12 2 y=− 3 2 x+6 (1)

Compare the above equation with y=mx+c , where m is the slope of the line. Let the slope of this line be m 1 .

The slope of the line is given by,

m 1 =− 3 2

The product of the slopes of perpendicular lines is equal to -1.

m 1 ⋅ m 2 =−1(2)

Let m 2 be the slope of the line which is perpendicular to the above line.

m 2 =− 1 m 1 = −1 − 3 2 = 2 3

Let the given line intersect y axis at ( 0,b ) .

Substitute the value of ( x,y )=( 0,b ) in equation (1).

b=− 3 2 ×0+6 b=6

The given line intersects y axis at ( 0,6 ).

The formula for the equation of a non-vertical line with slope m passing through the point ( x 0 , y 0 ) is given by,

( y− y 0 )=m( x− x 0 )(3)

Substitute the value of ( x 0 , y 0 ) as ( 0,6 ) and m= 2 3 .

( y−6 )= 2 3 ( x−0 ) 3( y−6 )=2x 3y−18=2x 2x−3y+18=0

Thus, the equation of line perpendicular to the line x 4 + y 6 =1 is 2x−3y+18=0