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If a straight line $L$ passing through the point $(5,-3)$ is inclined at an angle of $60^{\circ}$ to the line $\sqrt{3} x+y-9=0$ and $\mathrm{L}$ intersects $\mathrm{X}$-axis then the equation of $\mathrm{L}$ is
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Verified Answer
The correct answer is:
$\sqrt{3} x-y-3-5 \sqrt{3}=0$
Let $L_1=\sqrt{3} x+y-9=0 \Rightarrow m_1=-\sqrt{3}$
$$
\mathrm{L}_2=\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \Rightarrow \mathrm{m}_2=-\mathrm{a} / \mathrm{b}
$$
given, angle between $\mathrm{L}_1$ and $\mathrm{L}_2$ is $60^{\circ}$
$$
\begin{aligned}
& \Rightarrow \tan 60^{\circ}=\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right| \\
& \Rightarrow \sqrt{3}=\left|\frac{-\sqrt{3}-\mathrm{m}_2}{1-\sqrt{3} \mathrm{~m}_2}\right|
\end{aligned}
$$
On solving we get $\mathrm{m}_2=\sqrt{3}$
Hence equation of $l_2$ e passing through $(5,-3)$ is $\sqrt{3} x-y-3-5 \sqrt{3}=0$
$$
\mathrm{L}_2=\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \Rightarrow \mathrm{m}_2=-\mathrm{a} / \mathrm{b}
$$
given, angle between $\mathrm{L}_1$ and $\mathrm{L}_2$ is $60^{\circ}$
$$
\begin{aligned}
& \Rightarrow \tan 60^{\circ}=\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right| \\
& \Rightarrow \sqrt{3}=\left|\frac{-\sqrt{3}-\mathrm{m}_2}{1-\sqrt{3} \mathrm{~m}_2}\right|
\end{aligned}
$$
On solving we get $\mathrm{m}_2=\sqrt{3}$
Hence equation of $l_2$ e passing through $(5,-3)$ is $\sqrt{3} x-y-3-5 \sqrt{3}=0$
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