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Maths Question 11 – JEE-MAIN 2026

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In an equilateral triangle PQR, 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 PQR is (α,β), then 9(α+β) is equal to:

In an equilateral triangle, the orthocenter, centroid, circumcenter, and incenter all coincide at a single point.

Step 1: Determine the properties of the orthocenter in an equilateral triangle.✦ Active

In an equilateral triangle, the orthocenter, centroid, circumcenter, and incenter all coincide. Let the orthocenter be O(α,β). The altitude from vertex P to side QR will pass through O. This altitude is also the median.

Step 2: Find the foot of the altitude from P to QR and the orthocenter's coordinates.○ Expand

The vertex P is (3,5) and the side QR lies on the line L:x+y=4. The slope of L is −1. The altitude from P to QR is perpendicular to L, so its slope is 1. The equation of the altitude passing through P(3,5) is y−5=1(x−3)⇒x−y+2=0.

The foot of the altitude, M, is the intersection of x+y=4 and x−y+2=0. Solving these two equations simultaneously:

(x+y)+(x−y)=4+2⇒2x=6⇒x=3

Substituting x=3 into x+y=4 gives 3+y=4⇒y=1. So, M=(3,1).

The orthocenter O(α,β) (which is also the centroid) divides the median PM in the ratio 2:1 (from P). Using the section formula:

α=2(3)+1(3)2+1=6+33=3
β=2(1)+1(5)2+1=2+53=73

Thus, the orthocenter is (α,β)=(3,73).

Step 3: Calculate the required expression.○ Expand

We need to find 9(α+β).

α+β=3+73=93+73=163
9(α+β)=9×163=3×16=48
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