Let ABC be an isosceles triangle in which A is at (–1, 0), , AB = AC and B is on the positive x-axis. If and the line BC intersects the line y = x + 3 at , then is __________. [2024]

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Solution

Q.
Let ABC be an isosceles triangle in which A is at (–1, 0), $∠ A = \frac{2 π}{3}$ , AB = AC and B is on the positive x-axis. If $B C = 4 \sqrt{3}$ and the line BC intersects the line y = x + 3 at $(α, β)$ , then $\frac{β^{4}}{α^{2}}$ is __________. [2024]

Ans.
(36)

We have, $\frac{\sin 30 °}{A C} = \frac{\sin 120 °}{B C}$

$\Rightarrow A C = 4 \sqrt{3} \times \frac{1}{2} \times \frac{2}{\sqrt{3}}$

$\Rightarrow A C = 4 = A B$

$\Rightarrow A B^{2} = {(x + 1)}^{2} \Rightarrow x + 1 = \pm 4$

But B is on positive x-axis

$∴$ Coordinates of B = (3, 0)

Now, $A C^{2} = 16$

$\Rightarrow {(p + 1)}^{2} + q^{2} = 16$ ... (i)

Also, $B C^{2} = 48$

$\Rightarrow {(p - 3)}^{2} + q^{2} = 48$ ... (ii)

Solving (i) and (ii), we get

$p = - 3, q = 2 \sqrt{3}$

So, coordinates of $C = (- 3, 2 \sqrt{3})$

$∴$ Equation of line BC is, $y - 0 = \frac{2 \sqrt{3} - 0}{- 3 - 3} (x - 3)$

$\Rightarrow x + \sqrt{3} y = 3$ ... (iii)

Now, point of intersection of line (iii) and x + 3 = y is

$(3 (\sqrt{3} - 2), 3 (\sqrt{3} - 1))$

$\Rightarrow α = 3 (\sqrt{3} - 2), β = 3 (\sqrt{3} - 1)$

$∴$ $\frac{β^{4}}{α^{2}} = \frac{3^{4} {(3 + 1 - 2 \sqrt{3})}^{2}}{3^{2} {(\sqrt{3} - 2)}^{2}} = \frac{9 (2 {(2 - \sqrt{3}))}^{2}}{{(2 - \sqrt{3})}^{2}} = 9 \times 4 = 36$ .

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