Animated Solution for Mathematics - Straight Lines: Let the area of a $△PQR$ with vertices $P(5,4)$, $Q(-2,4)$ and $R(a,b)$ be 35 square units. If its orthocenter and centroid are $O(2,\frac{14}{5})$ and $C(c,d)$ respectively, then $c+2d$ is equal to

Welcome, JEE warrior! Today, we are not just solving a coordinate geometry problem; we are embarking on a journey of visualization. When you see a problem involving vertices, orthocenters, and centroids, your first instinct might be to panic and reach for the distance formula.

Stop. Take a breath. Let us look at the geometry with fresh eyes.

We are given vertices $P(5,4)$ and $Q(-2,4)$. Look closely at the $y$-coordinates; they are both $4$. This is our first gift from the problem setter.

The line segment $PQ$ is perfectly horizontal. This means the base of our triangle is simply the difference in $x$-coordinates: $|5-(-2)|=7$.

We are told the area is $35$. Since the area formula is defined as:

Substituting our known values, we have:

Solving this gives us a height $h=10$. This height is the vertical distance from the third vertex $R(a,b)$ to the line $y=4$. Thus, $b$ must be $4+10=14$ or $4-10=-6$. We have two candidates for $R$, but only one is the true vertex.

Now, consider the orthocenter $O(2,\frac{14}{5})$. The orthocenter is the intersection of all altitudes.

The altitude from $R$ to $PQ$ must be perpendicular to $PQ$. Since $PQ$ is horizontal, this altitude is a vertical line with the equation $x=a$.

Because the orthocenter $O$ must lie on this altitude, the $x$-coordinate of $O$ must be the same as the $x$-coordinate of $R$. Therefore, $a=2$. We have successfully pinned down the $x$-coordinate of $R$!

We still need to decide between $b=14$ and $b=-6$. This is where we use the second altitude.

The altitude from $Q$ must be perpendicular to the side $PR$. The slope of $PR$ is:

The slope of the altitude from $Q$ is the negative reciprocal: $m_{alt}=\frac{3}{b-4}$. The equation of this altitude, passing through $Q(-2,4)$, is:

We know the orthocenter $O(2,\frac{14}{5})$ lies on this line. Substituting these coordinates, we get:

This simplifies to:

Cross-multiplying, we find $b-4=-10$, which means $b=-6$. Our vertex $R$ is confirmed at $(2,-6)$.

With the vertices $P(5,4)$, $Q(-2,4)$, and $R(2,-6)$ in hand, finding the centroid $C(c,d)$ is a victory lap. The centroid is the average of the coordinates:

The question asks for $c+2d$. Substituting our values, we get:

We have arrived at the answer with elegance and precision. Remember, in JEE Advanced, the most complex problems often yield to the simplest geometric insights. Keep practicing, and keep falling in love with the process! The final answer is 3.