Question

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

Answer 1

Let (0, b) be the point on the y-axis whose distance from line \(\frac{ x}{3} +\frac{ y}{4} = 1\) is 4 units. 
The given line can be written as

\(4x + 3y - 12 = 0 … (1) \)

On comparing equation (1) to the general equation of line \(Ax + By + C = 0,\)  we obtain \(A = 4, B = 3, and\space C = -12.\)
It is known that the perpendicular distance (d) of a line \(Ax + By + C = 0\)  from a point \((x_1, y_1)\) is given by

\(d=\frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^2+B^2}}\)

Therefore, if (0, b) is the point on the y-axis whose distance from line \(\frac{ x}{3} +\frac{ y}{4} = 1\) is 4 units, then:

\(4=\frac{\left|4(0)+3(0)-12\right|}{\sqrt{4^2+3^2}}\)

\(⇒ 4=\frac{\left|3b-12\right|}{5}\)

\(⇒ 20=|3b-12|\)

\(⇒ 20=±(3b-12)\)
\(⇒ 20=(3b-12) \space or 20=-(3b-12)\)
\(⇒ 3b=20+12 \space or\space  3b=-20+12\)

\(⇒ b=\frac{32}{3} \) or \(b=\frac{-8}{3}\)

Thus, the required points are\( (0, \frac{32}{3})\) and \((0, \frac{-8}{3})\).