MHT CET · Maths · Straight Lines
The equation of a line, whose perpendicular distance from the origin is 7 units and the angle, which the perpendicular to the line from the origin makes, is \(120^{\circ}\) with positive \(\mathrm{X}\)-axis, is
- A \(x+\sqrt{3} y-14=0\)
- B \(x+\sqrt{3} y+14=0\)
- C \(x-\sqrt{3} y+14=0\)
- D \(x-\sqrt{3} y-14=0\)
Answer & Solution
Correct Answer
(C) \(x-\sqrt{3} y+14=0\)
Step-by-step Solution
Detailed explanation
Normal form of the equation of line is
\(x \cos \alpha+y \sin \alpha=\mathrm{p} \)
\( \text { Here, } \alpha=120^{\circ} \text { and } \mathrm{p}=7 \)
\( \therefore x \cos 120^{\circ}+y \sin 120^{\circ}=7 \)
\( \therefore x\left(\frac{-1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=7 \)
\( \therefore \frac{-x+\sqrt{3} y}{2}=7 \)
\( \therefore -x+\sqrt{3} y=14 \)
\( \therefore -x+\sqrt{3} y-14=0 \)
\( \therefore x-\sqrt{3} y+14=0\)
\(x \cos \alpha+y \sin \alpha=\mathrm{p} \)
\( \text { Here, } \alpha=120^{\circ} \text { and } \mathrm{p}=7 \)
\( \therefore x \cos 120^{\circ}+y \sin 120^{\circ}=7 \)
\( \therefore x\left(\frac{-1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=7 \)
\( \therefore \frac{-x+\sqrt{3} y}{2}=7 \)
\( \therefore -x+\sqrt{3} y=14 \)
\( \therefore -x+\sqrt{3} y-14=0 \)
\( \therefore x-\sqrt{3} y+14=0\)
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