Straight Line And Pair Of Straight Lines Ques 2
- A straight line $L$ at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^{\circ}$ with the line $x+y=0$. Then, an equation of the line $L$ is
(2019 Main, 12 April II)
(a) $x+\sqrt{3} y=8$
(b) $(\sqrt{3}+1) x+(\sqrt{3}-1) y=8 \sqrt{2}$
(c) $\sqrt{3} x+y=8$
(d) $(\sqrt{3}-1) x+(\sqrt{3}+1) y=8 \sqrt{2}$
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Solution:
Formula:
Equation Of A Straight Line In Various Forms:
- According to the question, we have the following figure.
Let $\theta$ be the inclination of the line $x+y=0$. Then,
$$ \begin{array}{cc} & \tan \theta=-1=\tan \left(180^{\circ}-45^{\circ}\right) \\ \Rightarrow & \tan \theta=\tan 135^{\circ} \\ \Rightarrow & \theta=135^{\circ} \\ \Rightarrow & \alpha+60^{\circ}=135^{\circ} \\ \Rightarrow & \alpha=75^{\circ} \end{array} $$
Since, line $L$ having perpendicular distance $O M=4$.
So, equation of the line ’ $L$ ’ is
$$ \begin{array}{cc} & x \cos \alpha+y \sin \alpha=4 \\ \Rightarrow & x \cos 75^{\circ}+y \sin 75^{\circ}=4 \\ \Rightarrow \quad x \cos \left(45^{\circ}+30^{\circ}\right)+y \sin \left(45^{\circ}+30^{\circ}\right)=4 \end{array} $$
$$ \begin{aligned} & \Rightarrow x \frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}+y \frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=4 \\ & \Rightarrow \quad(\sqrt{3}-1) x+y(\sqrt{3}+1)=8 \sqrt{2} \end{aligned} $$
BETA
