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