MHT CET · Maths · Pair of Lines
The measure of the acute angle between the lines given by the equation
\(3 x^{2}-4 \sqrt{3} x y+3 y^{2}=0\) is
- A \(45^{\circ}\)
- B \(60^{\circ}\)
- C \(70^{\circ}\)
- D \(30^{\circ}\)
Answer & Solution
Correct Answer
(D) \(30^{\circ}\)
Step-by-step Solution
Detailed explanation
Comparing \(3 x^{2}-4 \sqrt{3} x y+3 y^{2}=0\), with \(a x^{2}+2 h x y+b y^{2}=0\), we get
\(\mathrm{a}=3, \mathrm{~h}=-2 \sqrt{3}, \mathrm{~b}=3\)
We know that, \(\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)
\(\therefore \tan \theta=\left|\frac{2 \sqrt{12-9}}{3+3}\right|=\left|\frac{2 \sqrt{3}}{6}\right|=\left|\frac{1}{\sqrt{3}}\right|\)
\(\therefore \quad \theta=30^{\circ}\)
\(\mathrm{a}=3, \mathrm{~h}=-2 \sqrt{3}, \mathrm{~b}=3\)
We know that, \(\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)
\(\therefore \tan \theta=\left|\frac{2 \sqrt{12-9}}{3+3}\right|=\left|\frac{2 \sqrt{3}}{6}\right|=\left|\frac{1}{\sqrt{3}}\right|\)
\(\therefore \quad \theta=30^{\circ}\)
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