JEE Main 2025 — 23 January, Morning Shift — Question 13

$P(5,4),;Q(-2,4)\Rightarrow$ base $PQ=5-(-2)=7.$ $\text{Area\;}=\tfrac12\cdot 7\cdot |b-4|=35\Rightarrow |b-4|=10.$ $\therefore b=14\ \text{or}\ b=-6.$

$PQ$ is horizontal so the altitude from $R(a,b)$ is $x=a.$ $\text{Given\; orthocenter\; }O(2,\tfrac{14}{5})$ so $a=2.$

$\text{Slope\; of\; }QR=\dfrac{b-4}{4}$. $\text{Slope\; of\; altitude\; through\; }P(5,4)=-\dfrac{4}{b-4}.$ $\text{Altitude\; through\; }P:y-4=-\dfrac{4}{b-4}(x-5).$

$O_x=2\Rightarrow O_y=4-3\left(-\dfrac{4}{b-4}\right)=4+\dfrac{12}{b-4}.$ $4+\dfrac{12}{b-4}=\dfrac{14}{5}\Rightarrow \dfrac{12}{b-4}=-\dfrac{6}{5}\Rightarrow b=-6.$

$\therefore R(2,-6).$ $\text{Centroid\; }C\big(c,d\big)=\Big(\dfrac{5+(-2)+2}{3},\dfrac{4+4+(-6)}{3}\Big)=\Big(\dfrac{5}{3},\dfrac{2}{3}\Big).$

$c+2d=\dfrac{5}{3}+2\cdot\dfrac{2}{3}=\dfrac{9}{3}=3.$

$\boxed{3}$