Straight Lines Question 1
Question 1 - 2024 (01 Feb Shift 2)
Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, AB=AC$ and $B$ is on the positive $x$ axis. If $BC=4 \sqrt{3}$ and the line $BC$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^{4}}{\alpha^{2}}$ is :
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Answer (36)
Solution
$\frac{c}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}$ [By sine rule]
$2 c=8 \Rightarrow c=4$
$AB=|(b+1)|=4$
$b=3, m _{AB}=0$
$m _{BC}=\frac{-1}{\sqrt{3}}$
$BC:-y=\frac{-1}{\sqrt{3}}(x-3)$
$\sqrt{3} y+x=3$
Point of intersection : $y=x+3, \sqrt{3} y+x=3$
$(\sqrt{3+1}) y=6$
$y=\frac{6}{\sqrt{3}+1}$
$x=\frac{6}{\sqrt{3}+1}-3$
$=\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1}$
$=3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3})^{2}}$
$\frac{\beta^{4}}{\alpha^{2}}=36$
