JEE Main 2026 — Straight Lines Question with Solution

Let the coordinates of vertex $B$ be $(x_1,y_1)$ and vertex $C$ be $(x_2,y_2)$.

Since the mid-point of $AB$ is $(5,-1)$, we have:
$\frac{1+x_1}{2}=5\Rightarrow x_1=9$
$\frac{2+y_1}{2}=-1\Rightarrow y_1=-4$
Thus, $B\equiv (9,-4)$.

Given the centroid of $△ABC$ is $(3,4)$, we get:
$\frac{1+9+x_2}{3}=3\Rightarrow x_2=-1$
$\frac{2-4+y_2}{3}=4\Rightarrow y_2=14$
Thus, $C\equiv (-1,14)$.

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

For side $AB$, the mid-point is $(5,-1)$ and the slope is $\frac{-4-2}{9-1}=-\frac{3}{4}$.
The slope of its perpendicular bisector is $\frac{4}{3}$.
Equation of the perpendicular bisector of $AB$ is:
$y-(-1)=\frac{4}{3}(x-5)\Rightarrow 4x-3y=23$

For side $AC$, the mid-point is $\left(\frac{1-1}{2},\frac{2+14}{2}\right)=(0,8)$ and the slope is $\frac{14-2}{-1-1}=-6$.
The slope of its perpendicular bisector is $\frac{1}{6}$.
Equation of the perpendicular bisector of $AC$ is:
$y-8=\frac{1}{6}(x-0)\Rightarrow x-6y=-48$

Solving the equations $4x-3y=23$ and $x-6y=-48$:
Multiplying the first equation by $2$ gives $8x-6y=46$.
Subtracting the second equation from this gives:
$7x=94\Rightarrow x=\frac{94}{7}$
Substituting $x$ into the second equation:
$\frac{94}{7}-6y=-48\Rightarrow 6y=\frac{430}{7}\Rightarrow y=\frac{215}{21}$

Therefore, the circumcenter is $(\alpha ,\beta )=\left(\frac{94}{7},\frac{215}{21}\right)$.

Finally, calculating $21(\alpha +\beta )$:
$21(\alpha +\beta )=21\left(\frac{94}{7}+\frac{215}{21}\right)=3(94)+215=282+215=497$.

Answer: $497$