Animated Solution for Mathematics - Straight Lines: Let k be an integer such that the triangle with vertices $(k,-3k)$, $(5,k)$ and $(-k,2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point:

Unlocking the Orthocenter

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are embarking on a journey of visualization.

Often, when students see a problem involving variables like $k$ and vertices defined by algebraic expressions, they panic. They reach for the most complex formulas they can remember.

But in the world of JEE Advanced, the most elegant solutions are rarely the most complex ones. They are the ones that respect the geometry of the situation.

The Detective Work

We begin with a triangle defined by vertices $A(k,-3k)$, $B(5,k)$, and $C(-k,2)$. We are told the area is 28 square units.

Our first task is to find $k$. We invoke the determinant formula for the area of a triangle:

This implies the absolute value of the determinant is 56. When we expand this determinant along the first row, we are essentially calculating the signed area.

The expansion gives us a quadratic equation: $5k^2+13k-46=0$. Now, pause for a moment. This is where many students rush.

They see a quadratic and immediately reach for the quadratic formula. But look at the constraint: $k\in \mathbb{Z}$. This is a massive hint.

It tells us that our roots must be integers. When we factorize $5k^2+23k-10k-46=0$, we get $(k-2)(5k+23)=0$.

The roots are $k=2$ and $k=-\frac{23}{5}$. Since $k$ must be an integer, we discard the fraction. We have found our key: $k=2$.

The Geometric Reveal

With $k=2$, our vertices become concrete: $A(2,-6)$, $B(5,2)$, and $C(-2,2)$. Now, I want you to stop and look at these coordinates.

Do you see it? Look at the $y$-coordinates of $B$ and $C$. They are both 2.

This is not a coincidence; it is a gift from the problem setter. It means the side $BC$ is a horizontal line segment lying on the line $y=2$.

Why does this matter? Because the altitude from vertex $A$ to side $BC$ must be perpendicular to a horizontal line.

A line perpendicular to a horizontal line is a vertical line. Since the altitude must pass through $A(2,-6)$, its equation is simply $x=2$.

We have found our first altitude without doing any heavy lifting. This is the power of observation in coordinate geometry.

The Orthocenter Hunt

To find the orthocenter, we need the intersection of two altitudes. We have the first one: $x=2$. Now we need a second one.

Let's find the altitude from vertex $B$ to side $AC$. First, we calculate the slope of $AC$:

The altitude from $B$ must be perpendicular to $AC$. Therefore, its slope $m_{alt}$ must be the negative reciprocal of $m_{AC}$.

Thus, $m_{alt}=-\frac{1}{-2}=\frac{1}{2}$. Now, we use the point-slope form for the line passing through $B(5,2)$ with slope $\frac{1}{2}$:

This is our second altitude. We are now at the finish line. The orthocenter is the intersection of $x=2$ and $y-2=\frac{1}{2}(x-5)$.

Substituting $x=2$ into the second equation:

So, the orthocenter is at $(2,0.5)$.

Look at how we navigated this. We didn't use a single complex formula for the orthocenter.

We used the definition of an altitude, the property of perpendicular slopes, and the simple geometry of horizontal and vertical lines.

The lesson here is clear: before you start calculating, look at the geometry. Does the triangle have a horizontal side? Is there a vertical altitude?

These are the shortcuts that separate the good students from the great ones. Keep practicing this mindset, and you will find that even the most intimidating problems start to look like simple puzzles waiting to be solved.