Find the points on the x-axis, whose distances from the line x3+y4=1 are 4 units.
The given equation of line is x3+y4=1 or 4x+3y−12=0
Let (a,0) be the point on the x-axis whose distance from the given line is 4 units.
Thus, using distance formula,
4=|4a+3⋅0−12|√42+32
⇒4=|4a−12|5
⇒|4a−12|=20
⇒±(4a−12)=20
⇒(4a−12)=20or−(4a−12)=20
⇒4a=20+12or4a=−20+12
⇒a=8or−2
Thus, the required points on the x-axis are (−2,0) and (8,0)