Question

Find the values of k for which the line is (a) Parallel to the x -axis, (b) Parallel to the y -axis, (c) Passing through the origin.

Answer 1

The equation of line is ( k−3 )x−( 4− k 2 )y+ k 2 −7k+6=0.

(a)

If the line is parallel to the x axis, then

Slope of the given line = slope of x axis(1)

Rearrange the terms of the equation of line.

( 4− k 2 )y=( k−3 )x+ k 2 −7k+6=0 y= ( k−3 ) ( 4− k 2 ) x+ k 2 −7k+6 ( 4− k 2 ) (2)

The above equation of line is in the form of equation y=mx+c, where m is the slope of the line.

Compare equation (2) with y=mx+c.

m= ( k−3 ) ( 4− k 2 )

Slope of x axis is 0.

Substitute the values in equation (1).

( k−3 ) ( 4− k 2 ) =0 k−3=0 k=3

Thus, if the line ( k−3 )x−( 4− k 2 )y+ k 2 −7k+6=0 is parallel to the x axis, then the value of k is 3.

(b)

If the line is parallel to the y axis, then

Slope of the given line = slope of y axis(1)

Rearrange the terms of the equation of line.

( 4− k 2 )y=( k−3 )x+ k 2 −7k+6=0 y= ( k−3 ) ( 4− k 2 ) x+ k 2 −7k+6 ( 4− k 2 ) (2)

The above equation of line is in the form of equation y=mx+c, where m is the slope of the line.

Compare equation (2) with y=mx+c.

m= ( k−3 ) ( 4− k 2 )

Slope of y axis is 1 0 = undefined.

Substitute the values in equation (1).

( k−3 ) ( 4− k 2 ) = 1 0 ( 4− k 2 )=0 k 2 =4 k=±2

Thus, if the line ( k−3 )x−( 4− k 2 )y+ k 2 −7k+6=0 is parallel to the y axis, then the value of k is ±2.

(c)

If the given line is passing through the origin, then point ( 0,0 ) satisfies the given equation of the line.

Substitute the value of ( x,y ) as ( 0,0 ) in equation of line.

( k−3 )×0−( 4− k 2 )×0+ k 2 −7k+6=0 k 2 −7k+6=0 k 2 −6k−k+6=0 k( k−6 )−1( k−6 )=0

Further simplify the above expression.

( k−1 )( k−6 )=0 k=1 or k=6

Thus, if the line ( k−3 )x−( 4− k 2 )y+ k 2 −7k+6=0 is passing through the origin, then the value of k is 1 or 6.