JEE Mains · Maths · STD 11 - 9. straight line
Consider a triangle \(\mathrm{ABC}\) having the vertices \(\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)\) and \(\mathrm{C}(\gamma, \delta)\) and angles \(\angle \mathrm{ABC}=\frac{\pi}{6}\) and \(\angle \mathrm{BAC}=\frac{2 \pi}{3}\). If the points \(\mathrm{B}\) and \(\mathrm{C}\) lie on the line \(\mathrm{y}=\mathrm{x}+4\), then \(\alpha^2+\gamma^2\) is equal to ....................
- A \(46\)
- B \(13\)
- C \(15\)
- D \(14\)
Answer & Solution
Correct Answer
(D) \(14\)
Step-by-step Solution
Detailed explanation
Equation of line passes through point \(\mathrm{A}(1,2)\) which makes angle \(\frac{\pi}{6}\) from \(y=x+4\) is \( \mathrm{y}-2=\frac{1 \pm \tan \frac{\pi}{6}}{1 \mp \tan \frac{\pi}{6}}(\mathrm{x}-1) \) \( \mathrm{y}-2=\frac{\sqrt{3} \pm 1}{\sqrt{3} \mp 1}(\mathrm{x}-1)\)…
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