The equation of the lines on which the perpen | vedclass

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Question

(b) Let $p$ be the length of the perpendicular from the origin on the given line. Then its equation in normal form is $x\cos {30^o} + y\sin {30^o} = p

Vedclass · Accepted Answer

$\sqrt 3 x + y \pm 10 = 0$

Vedclass · Answer

(b) Let $p$ be the length of the perpendicular from the origin on the given line. Then its equation in normal form is $x\cos {30^o} + y\sin {30^o} = p$ or $\sqrt 3 x + y = 2p$
This meets the coordinate axes at $A\left( {\frac{{2p}}{{\sqrt 3 }},0} \right)$ and $B(0,\,2p)$.

Area of $\Delta OAB = \frac{1}{2}\left( {\frac{{2p}}{{\sqrt 3 }}} \right){\rm{ }}2p = \frac{{2{p^2}}}{{\sqrt 3 }}$

By hypothesis $\frac{{2{p^2}}}{{\sqrt 3 }} = \frac{{50}}{{\sqrt 3 }} \Rightarrow p = \pm 5$.
Hence the lines are $\sqrt 3 x + y \pm 10 = 0$.

The equation of the lines on which the perpendiculars from the origin make ${30^o}$ angle with $x$-axis and which form a triangle of area $\frac{{50}}{{\sqrt 3 }}$ with axes, are

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