JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

Let the angles made with the positive x-axis by two straight lines drawn from the point $P (2, 3)$ and meeting the line $x + y = 6$ at a distance $\frac{2}{3}$ from the point P be $θ_{1}$ and $θ_{2}$ . Then the value of $(θ_{1} + θ_{2})$ is :

A
$\frac{π}{12}$
B
$\frac{π}{6}$
C
$\frac{π}{2}$
D
$\frac{π}{3}$

Solution

Using the parametric form of a line through $P (2, 3)$ at angle $θ$ , a point at distance $2/3$ from P is

$Q = (2 + 2/3 cos θ, 3 + 2/3 sin θ)$ .

Q lies on $x + y = 6$ :

$\frac{2}{3} (cos θ + sin θ) + 5 = 6 \Rightarrow cos θ + sin θ = \frac{3}{2}$ .

Squaring: $1 + sin 2 θ = \frac{3}{2} \Rightarrow sin 2 θ = \frac{1}{2}$ .

In $[0, 2 π)$ , $2 θ = \frac{π}{6}$ or $\frac{5 π}{6}$ , giving $θ = \frac{π}{12}$ or $\frac{5 π}{12}$ .

$∴ θ_{1} + θ_{2} = \frac{π}{12} + \frac{5 π}{12} = \frac{π}{2}$ .

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