The points on the x -axis whose perpendicular distance from the line x 3 + y 4 =1 is 4 units are

Step-by-step solution

The given line is x 3 + y 4 = 1 , which can be rewritten as 4x + 3y - 12 = 0 . Let the point on the x -axis be (a, 0) . The perpendicular distance d from a point (x₁, y₁) to the line Ax + By + C = 0 is given by d = |Ax₁ + By₁ + C| A^2 + B^2 . Substituting the values, we have 4 = |4(a) + 3(0) - 12| 4^2 + 3^2 . 4 = |4a - 12| 16 + 9 = |4a - 12| 5 . |4a - 12| = 20 . This gives two cases: Case 1: 4a - 12 = 20 4a = 32 a = 8 . Case 2: 4a - 12 = -20 4a = -8 a = -2 . The points are (8, 0) and (-2, 0) . Answer: (8,0) and (-2