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

Given equation of line

4x + 3y = 12

4x + 3y – 12 = 0

Comparing above equation with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = -12

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

|4a – 12| = 4 × 5

(4a – 12) = 20

4a – 12 = 20 or - (4a – 12) = 20

4a = 20 + 12 or 4a = -20 + 12

a = 32/4 or a = -8/4

a = 8 or a = -2

The required points on the x – axis are (-2, 0) and (8, 0)

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