Maths Q12 – JEE-MAIN 2026

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.

Step 1: Determine Vertices B and C✦ Active

Let the vertices be $A (1, 2)$ , $B (x_{B}, y_{B})$ , and $C (x_{C}, y_{C})$ . The midpoint of AB is $(5, - 1)$ . Using the midpoint formula:

5 = \frac{1 + x_{B}}{2} \Rightarrow x_{B} = 9

- 1 = \frac{2 + y_{B}}{2} \Rightarrow y_{B} = - 4

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

3 = \frac{1 + 9 + x_{C}}{3} \Rightarrow 9 = 10 + x_{C} \Rightarrow x_{C} = - 1

4 = \frac{2 - 4 + y_{C}}{3} \Rightarrow 12 = - 2 + y_{C} \Rightarrow y_{C} = 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 $O A^{2} = O B^{2}$ :

(α - 1)^{2} + (β - 2)^{2} = (α - 9)^{2} + (β + 4)^{2}

α^{2} - 2 α + 1 + β^{2} - 4 β + 4 = α^{2} - 18 α + 81 + β^{2} + 8 β + 16

16 α - 12 β = 92 \Rightarrow 4 α - 3 β = 23 (1)

Equating $O A^{2} = O C^{2}$ :

(α - 1)^{2} + (β - 2)^{2} = (α + 1)^{2} + (β - 14)^{2}

α^{2} - 2 α + 1 + β^{2} - 4 β + 4 = α^{2} + 2 α + 1 + β^{2} - 28 β + 196

- 4 α + 24 β = 192 \Rightarrow - α + 6 β = 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), $α = 6 β - 48$ . Substitute this into equation (1):

4 (6 β - 48) - 3 β = 23

24 β - 192 - 3 β = 23

21 β = 215 \Rightarrow β = \frac{215}{21}

Substitute $β$ back into the expression for $α$ :

α = 6 (\frac{215}{21}) - 48 = \frac{2 \times 215}{7} - 48 = \frac{430}{7} - \frac{336}{7} = \frac{94}{7}

Now, calculate $21 (α + β)$ :

21 (\frac{94}{7} + \frac{215}{21}) = 21 (\frac{3 \times 94 + 215}{21}) = 21 (\frac{282 + 215}{21}) = 21 (\frac{497}{21}) = 497

💡 Teacher's Secret Hint

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

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Maths Question 12 – JEE-MAIN 2026