JEE Main 2025 Jan 29 Shift 1, Mathematics Q4: Points and Straight Lines

JEE Main2025Jan 29, Shift 1Mathematics

Let $A B C$ be a triangle formed by the lines $7 x - 6 y + 3 = 0$ , $x + 2 y - 31 = 0$ and $9 x - 2 y - 19 = 0$ . Let the point $(h, k)$ be the image of the centroid of $△ A B C$ in the line $3 x + 6 y - 53 = 0$ . Then $h^{2} + k^{2} + hk$ is equal to

A
$37$
B
$47$
C
$40$
D
$36$

Solution

Solving the three line-pairs gives the vertices $A (9, 11)$ , $B (3, 4)$ , $C (5, 13)$ .

Centroid $G = (\frac{9 + 3 + 5}{3}, \frac{11 + 4 + 13}{3}) = (\frac{17}{3}, \frac{28}{3}) .$

The reflection $(h, k)$ of $G$ in the line $3 x + 6 y - 53 = 0$ satisfies (i) the segment $G (h, k)$ is perpendicular to the line (slope of line $= - \frac{1}{2}$ , so slope of $G I = 2$ ) and (ii) the midpoint of $G$ and $(h, k)$ lies on the line.

From the perpendicularity: $\frac{k - 28/3}{h - 17/3} = 2.$

From the midpoint on the line: $3 (\frac{h + 17/3}{2}) + 6 (\frac{k + 28/3}{2}) = 53.$

Solving the two equations gives $h = 3, k = 4.$

Therefore $h^{2} + k^{2} + hk = 9 + 16 + 12 = 37.$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q4: Points and Straight Lines

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