Animated Solution for Mathematics - Straight Lines: Orthocentre of triangle with vertices $(0,0),(3,4)$ and $(4,0)$ is

Finding the Orthocentre

Welcome, future engineers. Today, we are going to dismantle a classic coordinate geometry problem. Often, when students see a problem asking for the 'orthocentre,' they immediately panic, searching their memory for complex formulas or heavy algebraic expressions.

But I want you to take a deep breath. In JEE Advanced, the most powerful tool in your arsenal is not a formula sheet; it is your ability to visualize the geometry. Let us look at our triangle with vertices $O(0,0)$, $A(3,4)$, and $B(4,0)$.

The Power of Observation

Before we touch a pen to paper, look at the coordinates. We have the origin $O(0,0)$ and the point $B(4,0)$. Both have a y-coordinate of zero.

This means the side $OB$ lies perfectly flat on the x-axis. This is not a coincidence; it is a gift from the paper-setter. In coordinate geometry, whenever you see a side lying on an axis, you have found a shortcut.

We are looking for the orthocentre, which is defined as the intersection of the altitudes. Remember, an altitude is just a line dropped from a vertex perpendicular to the opposite side. Because we have a horizontal base, our lives are about to get much easier.

The First Altitude

Let us drop an altitude from vertex $A(3,4)$ to the opposite side $OB$. Since $OB$ is horizontal (the x-axis), any line perpendicular to it must be vertical.

A vertical line is the simplest line in the coordinate plane—it has a constant x-coordinate. Since this line must pass through $A(3,4)$, its x-coordinate must be $3$ everywhere.

Therefore, our first altitude is simply the line $x=3$. See? No complex algebra, no scary slopes. Just pure geometric intuition.

The Second Altitude

Now, we need a second altitude to find the intersection. Let us drop an altitude from vertex $B(4,0)$ to the side $OA$.

To find the equation of this line, we first need its slope. We know that the slope of the line segment $OA$ is calculated by the change in y over the change in x:

Now, here is the crucial rule: the altitude is perpendicular to $OA$. When two lines are perpendicular, the product of their slopes is $-1$.

If the slope of $OA$ is $\frac{4}{3}$, then the slope of our altitude must be the negative reciprocal, which is $-\frac{3}{4}$. We have the slope ($m=-\frac{3}{4}$) and we have the point it passes through ($B(4,0)$).

Using the point-slope form, $y-y_1=m(x-x_1)$, we get:

The Intersection

We are at the finish line. The orthocentre $H$ is the intersection of our two altitudes: $x=3$ and $y=-\frac{3}{4}(x-4)$.

Since we already know $x=3$, we simply substitute this value into our second equation:

This simplifies to $y=-\frac{3}{4}(-1)$, which gives us $y=\frac{3}{4}$.

And there it is. The x-coordinate is $3$, and the y-coordinate is $\frac{3}{4}$. The orthocentre is $(3,\frac{3}{4})$.

Notice how we didn't need to memorize a single complex formula? We used the definition of the orthocentre, the properties of perpendicular lines, and the simplicity of vertical lines.

This is the essence of JEE Advanced mathematics: it is not about how much you memorize, but how clearly you can see the structure of the problem. Keep practicing this habit of observation, and you will find that even the most intimidating problems start to look like simple puzzles waiting to be solved.