Consider a triangle ABC having the vertices A (1,2), B (α, β) and C (γ, δ) and angles angle ABC =π/6 and angle BAC =2 π/3 . If the points B and C lie on the line y=x+4 , then α 2 +γ 2 is equal to ldots . .

Mathematics · Straight lines

JEE Main 2024 — 4 April, Shift 2 — Question 28

Consider a triangle 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 B and C lie on the line $y=x+4$ , then $\alpha^{2}+\gamma^{2}$ is equal to $\ldots .$ .

Answer: 14

Numerical answer — enter this value.

Step-by-step solution

Equation of line passes through point $\mathrm{A}(1,2)$

which makes angle $\frac{\pi}{6}$ from $y=x+4$ is

$y-2=\frac{1 \pm \tan \frac{\pi}{6}}{1 \mp \tan \frac{\pi}{6}}(x-1)$

$y-2=\frac{\sqrt{3} \pm 1}{\sqrt{3} \mp 1}(x-1)$

$y-2=(2+\sqrt{3})(x-1)$

Solve with $y=x+4$

then $x+2=(2+\sqrt{3}) x-2-\sqrt{3}$

$x=\frac{4+\sqrt{3}}{1+\sqrt{3}}$

$\alpha^{2}+\gamma^{2}=\left(\frac{4+\sqrt{3}}{1+\sqrt{3}}\right)^{2}+\left(\frac{4-\sqrt{3}}{1-\sqrt{3}}\right)^{2}$

$\alpha^{2}+\gamma^{2}=14$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Mathematics
Chapter: Straight lines
Topic: Angle between lines, perpendicular distance & distance between parallel lines, foot, image