Straight LinesmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Orthocentre of Equilateral Triangle: 9(α+β) = 48 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
In an equilateral triangle PQRPQR, let the vertex PP be at (3,5)(3,5) and the side QRQR be along the line x+y=4x+y=4. If the orthocentre of the triangle is (α,β)(\alpha,\beta), then 9(α+β)9(\alpha+\beta) is equal to
A1616
B2727
C3636
D4848correct
Solution
Step 1: Foot MM of perpendicular from P(3,5)P(3,5) to x+y−4=0x+y-4=0: M=P−xP+yP−412+12(1,1)M=P-\dfrac{x_P+y_P-4}{1^2+1^2}(1,1). Here xP+yP−4=4x_P+y_P-4=4:
M=(3,5)−42(1,1)=(3−2, 5−2)=(1,3).M=(3,5)-\frac{4}{2}(1,1)=(3-2,\ 5-2)=(1,3).
 Step 2: Orthocentre == centroid G=P+2M3G=\dfrac{P+2M}{3} (using Q+R=2MQ+R=2M):
G=(3+2(1)3, 5+2(3)3)=(53, 113).G=\left(\frac{3+2(1)}{3},\ \frac{5+2(3)}{3}\right)=\left(\frac{5}{3},\ \frac{11}{3}\right).
∴ (α,β)=(53,113).\therefore\ (\alpha,\beta)=\left(\tfrac53,\tfrac{11}{3}\right).
 Step 3: 9(α+β)=9⋅5+113=9⋅163=489(\alpha+\beta)=9\cdot\dfrac{5+11}{3}=9\cdot\dfrac{16}{3}=48. Correct answer: (4)
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