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Maths Question 12 – JEE-MAIN 2026

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Let the vertex A of a triangle ABC be (1,2), and the mid-point of the side AB be (5,−1). If the centroid of this triangle is (3,4) and its circumcenter is (α,β), then 21(α+β) is equal to:

Recall the definitions and formulas for the midpoint of a line segment, the centroid of a triangle, and the circumcenter of a triangle.

● Video Walkthrough
Step 1: Determine Vertices B and C✦ Active

Let the vertices be A(1,2), B(xB,yB), and C(xC,yC). The midpoint of AB is (5,−1). Using the midpoint formula:

5=1+xB2⇒xB=9 −1=2+yB2⇒yB=−4

So, B(9,−4). The centroid G(3,4) is given by:

3=1+9+xC3⇒9=10+xC⇒xC=−1 4=2−4+yC3⇒12=−2+yC⇒yC=14

So, C(−1,14). The vertices are A(1,2), B(9,−4), and C(−1,14).

Step 2: Formulate Equations for Circumcenter○ Expand

Let the circumcenter be O(α,β). It is equidistant from A, B, and C. Equating OA2=OB2:

(α−1)2+(β−2)2=(α−9)2+(β+4)2 α2−2α+1+β2−4β+4=α2−18α+81+β2+8β+16 16α−12β=92⇒4α−3β=23(1)

Equating OA2=OC2:

(α−1)2+(β−2)2=(α+1)2+(β−14)2 α2−2α+1+β2−4β+4=α2+2α+1+β2−28β+196 −4α+24β=192⇒−α+6β=48(2)
💡 Teacher's Secret Hint

Remember to expand and simplify the squared terms carefully to avoid algebraic errors.

Step 3: Solve for Circumcenter and Final Expression○ Expand

From equation (2), α=6β−48. Substitute this into equation (1):

4(6β−48)−3β=23 24β−192−3β=23 21β=215⇒β=21521

Substitute β back into the expression for α:

α=6(21521)−48=2×2157−48=4307−3367=947

Now, calculate 21(α+β):

21(947+21521)=21(3×94+21521)=21(282+21521)=21(49721)=497
💡 Teacher's Secret Hint

Double-check your arithmetic, especially when dealing with fractions and solving the system of equations.

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