JEE Main 2026 — Straight Lines Question with Solution

Let the vertices of the triangle $ABC$ be $A(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$.

Let the given midpoints be of sides $BC$, $CA$, and $AB$ respectively:
Midpoint of $BC=\left(\frac{5}{2},7\right)\Rightarrow x_2+x_3=5$ and $y_2+y_3=14$
Midpoint of $CA=\left(\frac{5}{2},3\right)\Rightarrow x_3+x_1=5$ and $y_3+y_1=6$
Midpoint of $AB=(4,5)\Rightarrow x_1+x_2=8$ and $y_1+y_2=10$

Adding the three equations for the $x$-coordinates:
$2(x_1+x_2+x_3)=5+5+8=18$
$x_1+x_2+x_3=9$

Substituting the known sums:
$x_1=9-5=4$
$x_2=9-5=4$
$x_3=9-8=1$

Similarly, for $y$-coordinates:
$2(y_1+y_2+y_3)=14+6+10=30$
$y_1+y_2+y_3=15$

$y_1=15-14=1$
$y_2=15-6=9$
$y_3=15-10=5$

So, the vertices are $A(4,1)$, $B(4,9)$, and $C(1,5)$.

Lengths of the sides:
$a=BC=\sqrt{(4-1{)}^2+(9-5{)}^2}=\sqrt{9+16}=5$
$b=CA=\sqrt{(4-1{)}^2+(1-5{)}^2}=\sqrt{9+16}=5$
$c=AB=\sqrt{(4-4{)}^2+(9-1{)}^2}=\sqrt{0+64}=8$

Coordinates of the incentre $(h,k)$:
$I(x,y)=\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$

$h=\frac{5(4)+5(4)+8(1)}{5+5+8}=\frac{20+20+8}{18}=\frac{48}{18}=\frac{8}{3}$

$k=\frac{5(1)+5(9)+8(5)}{5+5+8}=\frac{5+45+40}{18}=\frac{90}{18}=5$

So, the incentre is $(h,k)=\left(\frac{8}{3},5\right)$.

Therefore:
$3h+k=3\cdot \frac{8}{3}+5=8+5=13$

Hence, the correct option is $(3)13$.