Question

Find the points on the x -axis, whose distances from the line are 4 units.

Answer 1

The equation of line is x 3 + y 4 =1 . The distance of the line from a point on x axis is 4 units.

The general form of the equation of line is given by,

Ax+By+C=0 (1)

Rearrange the terms of equation of line.

4x+3y 12 =1 4x+3y=12 4x+3y−12=0

Compare the above expression with the general form of equation of line from equation (1).

A=4, B=3, C=−12 (2)

The given point is lie on x axis. Let the coordinate of that point is ( p,0 ) whose distance from x axis is 4 units.

The formula for the perpendicular distance d of a line Ax+By+C=0 from a point ( x 1 , y 1 ) is given by,

d= | A x 1 +B y 1 +C | A 2 +B 2 (3)

Substitute the value of ( x 1 , y 1 ) as ( p,0 ) , d as 4 , and the values of A , B ,and C from equation (2) to equation (3).

4= | 4×p+3×0+12 | 4 2 +3 2 4= | 4p+0+12 | 16+9 4= | 4p+12 | 25 4= | 4p+12 | 5

Further simplify the above equation.

5×4=| 4p+12 | 20=| 4p+12 | ±( 4p+12 )=20

If consider the positive sign.

4p−12=20 4p=32 p= 32 4 p=8

The coordinate of the point is ( 8,0 ) .

If consider the negative sign.

−( 4p−12 )=20 −4p+12=20 −4p=8 p=−2

The coordinate of the point is ( −2,0 ) .

Thus, the points on x axis whose distance from the line x 3 + y 4 =1 is 4 units.is ( 8,0 ) or ( −2,0 ) .