Animated Solution for Mathematics - Straight Lines: The equations of the sides $AB,BC$ and $CA$ of a triangle $ABC$ are $2x+y=0$, $x+py=15a$ and $x-y=3$ respectively. If its orthocentre is $(2,a)$, $-\frac{1}{2}<a<2$, then $p$ is equal to

Welcome, future engineer! Today, we are not just solving a coordinate geometry problem; we are uncovering the hidden architecture of a triangle. When you look at a problem involving an orthocenter, I want you to stop seeing just lines and points. I want you to see a system of perpendicular relationships.

We are given the equations of the sides of a triangle $ABC$ and the coordinates of its orthocenter $H(2,a)$. The sides are defined as: $AB:2x+y=0$ $BC:x+py=15a$ $CA:x-y=3$

Our mission is to find the value of $p$. Let us embark on this journey.

First, we must find our bearings. Vertex $A$ is the intersection of $AB$ and $AC$. By solving the system $2x+y=0$ and $x-y=3$, we find that $y=-2x$.

Substituting this into the second equation, we get $x-(-2x)=3$, which simplifies to $3x=3$. Thus, $x=1$ and $y=-2$. Our vertex $A$ is at $(1,-2)$.

The orthocenter $H(2,a)$ is the intersection of the altitudes. This means the line segment $AH$ is an altitude, and by definition, it must be perpendicular to the side $BC$.

The slope of $AH$ is given by:

The equation of side $BC$ is $x+py=15a$, which can be rewritten as $y=-\frac{1}{p}x+\frac{15a}{p}$. Thus, the slope of $BC$ is $m_{BC}=-\frac{1}{p}$.

Because $AH\perp BC$, the product of their slopes must be $-1$:

Keep this in your back pocket; it is the key to the entire problem.

Now, we need to find vertex $B$, the intersection of $AB$ and $BC$. We know $y=-2x$ from $AB$. Substituting this into $BC$ ($x+py=15a$), we get $x+p(-2x)=15a$.

This simplifies to $x(1-2p)=15a$. Thus, the coordinates of $B$ are:

Since $BH$ is an altitude, it must be perpendicular to $AC$. The slope of $AC$ ($x-y=3$) is $1$. Therefore, the slope of $BH$ must be $-1$. Using the coordinates of $B$ and $H(2,a)$, we set the slope:

After cross-multiplying and simplifying the algebra, we arrive at the expression:

Now, we substitute our earlier finding, $p=a+2$, into this equation. This transforms the expression into a quadratic:

Factoring this, we get $(a-1)(a-3)=0$. We have two candidates: $a=1$ or $a=3$. Given the geometric constraints of the triangle, we identify $a=1$.

Finally, using $p=a+2$, we find: $p=1+2=3$

We have conquered the problem. The value of $p$ is $3$. Remember, in JEE, the path to the answer is often paved with algebraic rigor, but the geometric intuition is your compass.