Consider a triangle \(ABC\) having the vertices \(A(1,2), B(\alpha,\beta)\) and \(C(\gamma,\delta)\) and angles \(\angle ABC=\dfrac{\pi}{6}\) and \(\angle BAC=\dfrac{2\pi}{3}\). If the points \(B\) and \(C\) lie on the line \(y=x+4\), then \(\alpha^{2}+\gamma^{2}\) is equal to ____.

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Straight Lines

Consider a triangle

A B C

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 ____.

Answer: 14

∠ B A C = \frac{2 π}{3}, ∠ A B C = \frac{π}{6} \Rightarrow ∠ A C B = π - \frac{2 π}{3} - \frac{π}{6} = \frac{π}{6}

, so triangle

A B C

is isosceles with base angles equal at B and C, and

B C

lies along

y = x + 4

(slope 1, inclination

4 5^{\circ}

).

Each of

A B, A C

makes a

3 0^{\circ}

angle with

B C

A

, so their slopes are

tan (4 5^{\circ} + 3 0^{\circ}) = tan 7 5^{\circ} = 2 + 3

and

tan (4 5^{\circ} - 3 0^{\circ}) = tan 1 5^{\circ} = 2 - 3

.

A line through

A (1, 2)

with slope

m

meets

y = x + 4

where

m (x - 1) + 2 = x + 4

, i.e.

x = \frac{m + 2}{m - 1}

.

For

m = 2 + 3

x = \frac{4 + 3}{1 + 3} = \frac{( 4 + 3 ) ( 3 - 1 )}{2} = \frac{3 3 - 1}{2}

.

For

m = 2 - 3

x = \frac{4 - 3}{1 - 3} = \frac{( 4 - 3 ) ( 1 + 3 )}{- 2} = - \frac{1 + 3 3}{2}

.

These are the

x

-coordinates

α, γ

of B and C:

α^{2} = \frac{( 3 3 - 1 ) ^{2}}{4} = \frac{28 - 6 3}{4}, γ^{2} = \frac{( 1 + 3 3 ) ^{2}}{4} = \frac{28 + 6 3}{4}

α^{2} + γ^{2} = \frac{56}{4} = 14

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Integer answer