Question

What are the points on y-axis whose distance from the line \[\frac{x}{3} + \frac{y}{4} = 1\]  is 4 units?

 

Answer 1

Let (0, t) be a point on the y-axis.
It is given that the perpendicular distance of the line \[\frac{x}{3} + \frac{y}{4} = 1\] from the point (0, t) is 4 units.

\[\therefore \left| \frac{0 + \frac{t}{4} - 1}{\sqrt{\frac{1}{3^2} + \frac{1}{4^2}}} \right| = 4\]
\[ \Rightarrow \left| t - 4 \right| = 4 \times 4 \times \frac{5}{3 \times 4}\]
\[ \Rightarrow \left| t - 4 \right| = \frac{20}{3}\]

\[\Rightarrow t - 4 = \pm \frac{20}{3}\]
\[ \Rightarrow t = 4 \pm \frac{20}{3}\]
\[ \Rightarrow t = 4 + \frac{20}{3}, 4 - \frac{20}{3}\]
\[ \Rightarrow t = \frac{32}{3}, - \frac{8}{3}\]

Hence, the required points on the y-axis are

\[\left( 0, \frac{32}{3} \right) \text{ and  } \left( 0, - \frac{8}{3} \right)\]