Animated Solution for Mathematics - Straight Lines: The orthocentre of the triangle formed by the lines $xy=0$ and $x+y=1$ is

Welcome, future engineers! Today, we are going to peel back the layers of a seemingly simple coordinate geometry problem. We are tasked with finding the orthocenter of a triangle formed by the equations $xy=0$ and $x+y=1$.

At first glance, this might look like a standard problem, but it is actually a beautiful lesson in geometric intuition.

Let us start by deconstructing the first equation: $xy=0$. In algebra, if the product of two variables is zero, then at least one of them must be zero.

This means our equation $xy=0$ is actually the union of two distinct lines: $x=0$ and $y=0$. If you recall your basic coordinate geometry, $x=0$ is the equation of the y-axis, and $y=0$ is the equation of the x-axis.

Suddenly, the problem becomes much clearer: two of the sides of our triangle are the coordinate axes themselves!

Now, we turn our attention to the third side, defined by the linear equation $x+y=1$. To visualize this triangle, we need its vertices. We already have one vertex: the intersection of the x-axis and the y-axis, which is the origin $(0,0)$.

To find the other two vertices, we look for where the line $x+y=1$ intersects our axes:

1. For the y-axis ($x=0$), we substitute $x=0$ into $x+y=1$, giving us $y=1$. So, the first vertex is $(0,1)$.

2. For the x-axis ($y=0$), we substitute $y=0$ into $x+y=1$, giving us $x=1$. So, the second vertex is $(1,0)$.

We now have our three vertices: $(0,0)$, $(1,0)$, and $(0,1)$.

Here is where the magic happens. We are asked for the orthocenter, which is defined as the point of intersection of the three altitudes of a triangle. A brute-force approach would involve finding the slopes of the sides, determining the equations of the altitudes, and solving for their intersection.

But wait—look at the vertices again. The sides of our triangle lie along the x-axis and the y-axis. Since the x-axis and y-axis are perpendicular, the angle at the origin $(0,0)$ is exactly $90^{\circ }$.

This is a right-angled triangle!

In any right-angled triangle, the two legs (the sides forming the right angle) are themselves the altitudes. The altitude from $(1,0)$ to the y-axis is the x-axis itself, and the altitude from $(0,1)$ to the x-axis is the y-axis itself.

Since these two altitudes intersect at the vertex where the right angle is located, that vertex is, by definition, the orthocenter.

Because our right angle is at the origin $(0,0)$, the orthocenter of this triangle is simply $(0,0)$.

We have arrived at our answer through geometric reasoning rather than tedious algebra. This is the essence of JEE Advanced preparation: recognizing the underlying structure of a problem to find the most elegant path to the solution.

Keep practicing this kind of visualization, and you will find that even the most intimidating problems have a simple, beautiful core.