JEE Main 2026 — Apr 5, Shift 1 — Mathematics solution

JEE Main2026Apr 5, Shift 1Mathematics

In an equilateral triangle $P QR$ , let the vertex $P$ be at $(3, 5)$ and the side $QR$ be along the line $x + y = 4$ . If the orthocentre of the triangle $P QR$ is $(α, β)$ , then $9 (α + β)$ is equal to :

A
16
B
27
C
36
D
48

Solution

For an equilateral triangle, the orthocentre, centroid, and circumcentre coincide, and the centroid divides the segment from any vertex to the midpoint of the opposite side in the ratio $2 : 1$ .

The foot of the perpendicular from $P (3, 5)$ to the line $x + y = 4$ (which is the midpoint $M$ of $QR$ ) is found by moving along the normal direction $(1, 1) / 2$ . Signed distance: $\frac{3 + 5 - 4}{2} = \frac{4}{2}$ . So $M = (3, 5) - \frac{4}{2} (1, 1) = (1, 3)$ .

The orthocentre is the centroid here: $(\frac{3 + 2 ( 1 )}{3}, \frac{5 + 2 ( 3 )}{3}) = (\frac{5}{3}, \frac{11}{3})$ .

Therefore $9 (α + β) = 9 \cdot \frac{16}{3} = 48$ .

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