Consider a triangle ABC having the vertices A(1, 2), and and angles and . If the points B and C lie on the line y = x + 4, then is equal to __________ [2024]

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Solution

Q.
Consider a triangle ABC having the vertices A(1, 2), $B (α, β)$ and $C (γ, δ)$ and angles $∠ A B C = \frac{π}{6}$ and $∠ B A C = \frac{2 π}{3}$ . If the points B and C lie on the line y = x + 4, then $α^{2} + γ^{2}$ is equal to __________ [2024]

Ans.
(14)

$P = \frac{| - 1 + 2 - 4 |}{\sqrt{1^{2} + 1^{2}}} = \frac{3}{\sqrt{2}}$

$\sin (\frac{π}{6}) = \frac{3 / \sqrt{2}}{A B} \Rightarrow \frac{3}{\sqrt{2} A B} = \frac{1}{2} \Rightarrow A B = \frac{6}{\sqrt{2}}$

$\Rightarrow {(α - 1)}^{2} + {(α + 4 - 2)}^{2} = {(\frac{6}{\sqrt{2}})}^{2}$

$\Rightarrow α^{2} + 1 - 2 α + α^{2} + 4 + 4 α = 18$

$\Rightarrow 2 α^{2} + 2 α - 13 = 0 \to α and γ$ satisfy same equation

$∴$ $α$ and $γ$ are the roots of the equation $2 x^{2} + 2 x - 13 = 0$

Sum of roots = $α + γ = \frac{- 2}{2} = - 1$

Product of roots = $α γ = \frac{- 13}{2}$

Now, $α^{2} + γ^{2} = {(α + γ)}^{2} - 2 α γ$

$= {(- 1)}^{2} - 2 (\frac{- 13}{2}) = 1 + 13 = 14$ .

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