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

Correct answer: 0 Equation of line passes through point A(1,2) which makes angle (π)/(6) from y=x+4 is aligned & y-2=(1 ± (π)/(6))/(1 (π)/(6))(x-1) & y-2= 3 ± 1 3 1 (x-1) & array c|c & y-2=(2+ 3 )(x-1) & y-2=(2- 3 )(x-1) array & solve with y=x+4 solve with y=x+4 & x+2=(2+ 3 ) x-2- 3 x+2=(2- 3 ) x-2+ 3 & x= 4+ 3 1+ 3 x= 4- 3 1- 3 & aligned aligned & α^2+γ^2=( 4+ 3 1+ 3 )^2+( 4- 3 1- 3 )^2 & α^2+γ^2=14 aligned

JEE Main 20244 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Straight Lines

Consider a triangle ABC having the vertices A(1,2), B(α, β) and...

Consider a triangle

\mathrm{ABC}

having the vertices

\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)

and

\mathrm{C}(\gamma, \delta)

and angles

\angle A B C=\frac{\pi}{6}

and

\angle B A C=\frac{2 \pi}{3}

. If the points

\mathrm{B}

and

\mathrm{C}

lie on the line

y=x+4

, then

\alpha^2+\gamma^2

is equal to ________

Correct answer

Step-by-step solution

Equation of line passes through point

\mathrm{A}(1,2)

which makes angle

\frac{\pi}{6}

from

y=x+4

\begin{aligned} & 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) \\ & \begin{array}{c|c} \oplus & \oplus \\ \mathrm{y}-2=(2+\sqrt{3})(\mathrm{x}-1) & \mathrm{y}-2=(2-\sqrt{3})(\mathrm{x}-1) \end{array} \\ & \text{ solve\text{ with } } \mathrm{y}=\mathrm{x}+4 \quad \text{ solve\text{ with } } \mathrm{y}=\mathrm{x}+4 \\ & x+2=(2+\sqrt{3}) x-2-\sqrt{3} \quad x+2=(2-\sqrt{3}) x-2+\sqrt{3} \\ & x=\frac{4+\sqrt{3}}{1+\sqrt{3}} \quad x=\frac{4-\sqrt{3}}{1-\sqrt{3}} \\ & \end{aligned}

\begin{aligned} & \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\end{aligned}

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