MHT CET · Maths · Three Dimensional Geometry
A straight line L through the point \((3,-2)\) is inclined at an angle of \(60^{\circ}\) to the line \(\sqrt{3} x+y=1\). If L also intersects the X -axis, then the equation of \(L\) is
- A \(y+\sqrt{3} x+2-3 \sqrt{3}=0\)
- B \(y-\sqrt{3} x+2+3 \sqrt{3}=0\)
- C \(\sqrt{3} y-x+3+2 \sqrt{3}=0\)
- D \(\sqrt{3} y+x-3+2 \sqrt{3}=0\)
Answer & Solution
Correct Answer
(B) \(y-\sqrt{3} x+2+3 \sqrt{3}=0\)
Step-by-step Solution
Detailed explanation
The equation of a straight line passing through \((3,-2)\) is
\(y+2=m(x-3)\)
The slope of the line \(\sqrt{3} x+y=1\) is \(-\sqrt{3}\)
So, \(\tan 60^{\circ}= \pm \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})} \Rightarrow \sqrt{3}= \pm \frac{m+\sqrt{3}}{1-\sqrt{3} m}\)
On solving, we get
\(\mathrm{m}=0 \text { or } \sqrt{3}\)
Putting the values of m in (i), the required equation of lines are \(y+2=0\) and
\(y-\sqrt{3} x+2+3 \sqrt{3}=0\)
\(y+2=m(x-3)\)
The slope of the line \(\sqrt{3} x+y=1\) is \(-\sqrt{3}\)
So, \(\tan 60^{\circ}= \pm \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})} \Rightarrow \sqrt{3}= \pm \frac{m+\sqrt{3}}{1-\sqrt{3} m}\)
On solving, we get
\(\mathrm{m}=0 \text { or } \sqrt{3}\)
Putting the values of m in (i), the required equation of lines are \(y+2=0\) and
\(y-\sqrt{3} x+2+3 \sqrt{3}=0\)
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