Question
What are the points on y-axis whose distance from the line $\frac{\text{x}}{3}+\frac{\text{y}}{4}=1$ is 4 units?
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Answer
Let the required point be (0, a) Given, distnace of (0, a) from line 4x + 3y - 12 = 0 is 4units. $\text{D}=\Big|\frac{\text{ax}_1+\text{by}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$ $4=\Big|\frac{4(0)+3(\text{a})-12}{\sqrt{4^2+3^2}}\Big|$ $4=\Big|\frac{-3\text{a}+12}{5}\Big|$ $\Rightarrow4=\frac{-3\text{a}+12}{5}$ $\Rightarrow-3\text{a}=20-12$ $\text{a}=-\frac{8}{3}$ And $4=\frac{3\text{a}-12}{5}$ $\Rightarrow3\text{a}=20+12$ $\Rightarrow\text{a}=\frac{32}{3}$ So, Required points are $\Big(0,\frac{32}{3}\Big),\Big(0,\frac{-8}{3}\Big)$
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