AP EAMCET · Maths · Straight Lines
Find the equation of a straight line which passes through the point \((-1,-1)\) and makes an angle \(150^{\circ}\) with positive direction of \(X\)-axis.
- A \(\sqrt{3} x+y=1\)
- B \(\sqrt{3} y+x+(1+\sqrt{3})=0\)
- C \(x+\sqrt{3} y+(\sqrt{3}-1)=0\)
- D \(x+y=0\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3} y+x+(1+\sqrt{3})=0\)
Step-by-step Solution
Detailed explanation
Now, \(\tan \theta=\tan 150^{\circ}=\tan \left(\pi-30^{\circ}\right)\) \[ =-\tan 30^{\circ}=-\frac{1}{\sqrt{3}} \] \(\therefore\) Slope of given line \(=-\frac{1}{\sqrt{3}}\) Since, line passes through \((-1,-1)\). \(\therefore\) Equation of line will be…
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