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Straight Lines — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsStraight Lines2 Marks
Question
What are the points on the y-axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.
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Answer

Let point on y-axis be (0, y).
The given equation of line is $\frac{x}{3} + \frac{y}{4} = 1$.
$ \Rightarrow 4x + 3y = 12$ $ \Rightarrow 4x + 3y - 12 = 0$
Now perpendicular distance from point (0, y) to line 4x + 3y - 12 = 0 is
$\left| {\frac{{4 \times 0 + 3y - 12}}{{\sqrt {{{(4)}^2} + {{(3)}^2}} }}} \right| = \left| {\frac{{3y - 12}}{{\sqrt {25} }}} \right| = \left| {\frac{{3y - 12}}{5}} \right|$
It is given that
$\left| {\frac{{3y - 12}}{5}} \right| = 4 \Rightarrow \left| {\frac{{3y - 12}}{5}} \right| = \pm 4$
When $\frac{{3y - 12}}{5} = 4$ $ \Rightarrow 3y - 12 = 20 \Rightarrow y = \frac{{32}}{3}$
When $\frac{{3y - 12}}{5} = - 4$$ \Rightarrow 3y - 12 = - 20 \Rightarrow y = \frac{{ - 8}}{3}$
Thus required points are $\left( {0,\frac{{32}}{3}} \right)$ and $\left( {0,\frac{{ - 8}}{3}} \right)$.

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