MHT CET · Maths · Straight Lines
The equation of a line, whose perpendicular distance from the origin is 5 units and the angle, which the perpendicular to the line from the origin makes, is \(210^{\circ}\) with positive \(\mathrm{X}\)-axis, is
- A \(-x \sqrt{3}+y+10=0\)
- B \(x \sqrt{3}+y-10=0\)
- C \(x \sqrt{3}+y+10=0\)
- D \(x \sqrt{3}-y+10=0\)
Answer & Solution
Correct Answer
(C) \(x \sqrt{3}+y+10=0\)
Step-by-step Solution
Detailed explanation
Here, \(P=5\) and \(\alpha=210^{\circ}\)
Writing the equation of straight line in normal form
\(\begin{aligned} & \Rightarrow x \cos \left(210^{\circ}\right)+y \sin 30^{\circ}=5 \\ & \Rightarrow-x \cos 30^{\circ}-y \sin 30^{\circ}=5 \\ & \Rightarrow-x \cdot \frac{\sqrt{3}}{2}-y \cdot \frac{1}{2}=5 \\ & \Rightarrow-\sqrt{3} x-y=10 \\ & \Rightarrow \sqrt{3} x+y+10=0\end{aligned}\)
Writing the equation of straight line in normal form
\(\begin{aligned} & \Rightarrow x \cos \left(210^{\circ}\right)+y \sin 30^{\circ}=5 \\ & \Rightarrow-x \cos 30^{\circ}-y \sin 30^{\circ}=5 \\ & \Rightarrow-x \cdot \frac{\sqrt{3}}{2}-y \cdot \frac{1}{2}=5 \\ & \Rightarrow-\sqrt{3} x-y=10 \\ & \Rightarrow \sqrt{3} x+y+10=0\end{aligned}\)
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