Animated Solution for Mathematics - Straight Lines: If the line $3x+4y-24=0$ intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is

Coordinate geometry is not just about plugging numbers into formulas; it is about seeing the hidden architecture of space. When we look at the line $3x+4y-24=0$, we are looking at a boundary that carves a perfect right-angled triangle out of the Cartesian plane.

Let us embark on this journey to find the incentre of $△OAB$.

Every triangle needs vertices. To find where our line intersects the axes, we treat the axes as simple algebraic constraints.

For the $x$-axis, we set $y=0$. The equation $3x+4(0)-24=0$ simplifies to $3x=24$, giving us $x=8$. Thus, our first vertex is $A(8,0)$.

Similarly, for the $y$-axis, we set $x=0$, leading to $4y=24$, or $y=6$. Our second vertex is $B(0,6)$. With the origin $O(0,0)$ as our third vertex, we have successfully anchored our triangle.

Because the $x$ and $y$ axes are perpendicular, the angle at the origin $O$ is $90^{\circ }$. This is a gift, as it confirms we are dealing with a right-angled triangle.

The base $OA$ has length $8$, and the perpendicular $OB$ has length $6$. Using the Pythagorean theorem, the hypotenuse $AB$ is calculated as follows:

We have identified a classic $(6,8,10)$ Pythagorean triplet.

Now, we need the incentre. While the general formula for the incentre is powerful, for a right-angled triangle, we have a beautiful shortcut for the inradius $r$:

Here, $a$ and $b$ are the legs and $c$ is the hypotenuse. Substituting our values, we get:

This is the radius of the circle that touches all three sides of our triangle.

Finally, visualize the incircle nestled in the corner of the first quadrant. It touches the $x$-axis at $(2,0)$ and the $y$-axis at $(0,2)$.

For a circle of radius $r=2$ to be tangent to both positive axes, its center must be exactly $2$ units from the $x$-axis and $2$ units from the $y$-axis. Therefore, the coordinates of the incentre $I$ are $(2,2)$.

We have arrived at our destination. Remember, the beauty of JEE problems lies in spotting these geometric shortcuts. Keep practicing, and these patterns will become second nature!