JEE Main 2026 — Mathematics Coordinate Geometry

Let the coordinates of vertex $B$B be $(x_1, y_1)$(x1​,y1​) and vertex $C$C be $(x_2, y_2)$(x2​,y2​).

Since the mid-point of $AB$AB is $(5, -1)$(5,−1), we have:

$$\dfrac{1 + x_1}{2} = 5 \Rightarrow x_1 = 9$$21+x1​​=5⇒x1​=9

$$\dfrac{2 + y_1}{2} = -1 \Rightarrow y_1 = -4$$22+y1​​=−1⇒y1​=−4

Thus, $B \equiv (9, -4)$B≡(9,−4).

Given the centroid of $\triangle ABC$△ABC is $(3, 4)$(3,4), we get:

$$\dfrac{1 + 9 + x_2}{3} = 3 \Rightarrow x_2 = -1$$31+9+x2​​=3⇒x2​=−1

$$\dfrac{2 - 4 + y_2}{3} = 4 \Rightarrow y_2 = 14$$32−4+y2​​=4⇒y2​=14

Thus, $C \equiv (-1, 14)$C≡(−1,14).

The circumcenter $(\alpha, \beta)$(α,β) is the intersection of the perpendicular bisectors of the sides of the triangle.

For side $AB$AB, the mid-point is $(5, -1)$(5,−1) and the slope is $\dfrac{-4 - 2}{9 - 1} = -\dfrac{3}{4}$9−1−4−2​=−43​.

The slope of its perpendicular bisector is $\dfrac{4}{3}$34​.

Equation of the perpendicular bisector of $AB$AB is:

$$y - (-1) = \dfrac{4}{3}(x - 5) \Rightarrow 4x - 3y = 23$$y−(−1)=34​(x−5)⇒4x−3y=23

For side $AC$AC, the mid-point is $\left(\dfrac{1 - 1}{2}, \dfrac{2 + 14}{2}\right) = (0, 8)$(21−1​,22+14​)=(0,8) and the slope is $\dfrac{14 - 2}{-1 - 1} = -6$−1−114−2​=−6.

The slope of its perpendicular bisector is $\dfrac{1}{6}$61​.

Equation of the perpendicular bisector of $AC$AC is:

$$y - 8 = \dfrac{1}{6}(x - 0) \Rightarrow x - 6y = -48$$y−8=61​(x−0)⇒x−6y=−48

Solving the equations $4x - 3y = 23$4x−3y=23 and $x - 6y = -48$x−6y=−48:

Multiplying the first equation by $2$2 gives $8x - 6y = 46$8x−6y=46.

Subtracting the second equation from this gives:

$$7x = 94 \Rightarrow x = \dfrac{94}{7}$$7x=94⇒x=794​

Substituting $x$x into the second equation:

$$\dfrac{94}{7} - 6y = -48 \Rightarrow 6y = \dfrac{430}{7} \Rightarrow y = \dfrac{215}{21}$$794​−6y=−48⇒6y=7430​⇒y=21215​

Therefore, the circumcenter is $(\alpha, \beta) = \left(\dfrac{94}{7}, \dfrac{215}{21}\right)$(α,β)=(794​,21215​).

Finally, calculating $21(\alpha + \beta)$21(α+β):

$21(\alpha + \beta) = 21\left(\dfrac{94}{7} + \dfrac{215}{21}\right) = 3(94) + 215 = 282 + 215 = 497$21(α+β)=21(794​+21215​)=3(94)+215=282+215=497.

Answer: $497$497