JEE Main 2025 Jan 22 Shift 1, Mathematics Q3: Points and Straight Lines

JEE Main2025Jan 22, Shift 1Mathematics

Let the triangle PQR be the image of the triangle with vertices $(1, 3)$ , $(3, 1)$ and $(2, 4)$ in the line $x + 2 y = 2$ . If the centroid of $△$ PQR is the point $(α, β)$ , then $15 (α - β)$ is equal to :

A
24
B
19
C
21
D
22

Solution

The centroid of the original triangle is $G = (\frac{1 + 3 + 2}{3}, \frac{3 + 1 + 4}{3}) = (2, \frac{8}{3})$ .

Reflecting a point through the line $a x + b y + c = 0$ uses $\frac{x ^{'} - x}{a} = \frac{y ^{'} - y}{b} = - \frac{2 ( a x + b y + c )}{a ^{2} + b ^{2}}$ . For $x + 2 y - 2 = 0$ and $G = (2, 8/3)$ :

$a x + b y + c = 2 + \frac{16}{3} - 2 = \frac{16}{3}$ , and $a^{2} + b^{2} = 5$ .

$α - 2 = - \frac{2 \cdot 16/3}{5} \cdot 1 = - \frac{32}{15}$ , so $α = 2 - \frac{32}{15} = - \frac{2}{15}$ .

$β - \frac{8}{3} = - \frac{2 \cdot 16/3}{5} \cdot 2 = - \frac{64}{15}$ , so $β = \frac{8}{3} - \frac{64}{15} = - \frac{24}{15}$ .

$15 (α - β) = 15 \cdot \frac{- 2 + 24}{15} = 22$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q3: Points and Straight Lines

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