Animated Solution for Mathematics - Straight Lines: Let two straight lines drawn from the origin $O$ intersect the line $3x+4y=12$ at the points $P$ and $Q$ such that $△OPQ$ is an isosceles triangle and $\angle POQ=90^{\circ }$. If $l=O{P}^2+P{Q}^2+Q{O}^2$, then the greatest integer less than or equal to $l$ is :

A Journey into Coordinate Geometry

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey to uncover the hidden symmetry within a simple line.

Imagine you are standing at the origin $O$ of a Cartesian plane. Before you lies a line, defined by the equation $3x+4y=12$.

You are tasked with finding two points, $P$ and $Q$, on this line such that the triangle formed by the origin and these two points, $△OPQ$, is an isosceles right-angled triangle with the right angle at the origin. This is a beautiful setup, and I want you to feel the elegance of the solution before we even touch the algebra.

The Parametric Power

When we see a line and a distance from the origin, our first instinct should be to use parametric coordinates. Why? Because they allow us to collapse the complexity of $(x,y)$ into a single distance variable $r$ and an angle $\theta$.

Let the distance $OP=OQ=r$. If the line segment $OP$ makes an angle $\theta$ with the positive x-axis, we can define the coordinates of $P$ as $(r\cos \theta ,r\sin \theta )$.

This is our anchor. It connects the geometry of the triangle directly to the algebra of the line.

The Rotation

Now, here is where the magic happens. We are told $\angle POQ=90^{\circ }$. This is not just a condition; it is a transformation.

If $P$ is at $(r\cos \theta ,r\sin \theta )$, then $Q$ is simply $P$ rotated by $90^{\circ }$ counterclockwise about the origin. If you recall your rotation matrices or complex numbers, rotating a point $(x,y)$ by $90^{\circ }$ transforms it into $(-y,x)$.

Applying this to our point $P$, the coordinates of $Q$ become $(-r\sin \theta ,r\cos \theta )$. We have now described both points using only $r$ and $\theta$. The geometry is now fully captured in our algebraic net.

The Algebraic Dance

Since both $P$ and $Q$ lie on the line $3x+4y=12$, they must satisfy its equation. Let's substitute $P$ first:

Now, let's substitute $Q$:

We have two equations, but we have two unknowns: $r$ and $\theta$. However, we only need $r^2$. This is a classic JEE moment. Whenever you see $a\cos \theta +b\sin \theta$ and $b\cos \theta -a\sin \theta$, squaring and adding them is the key to unlocking the door.

Let's square both equations:

When we add these, the cross-terms $24\sin \theta \cos \theta$ and $-24\sin \theta \cos \theta$ will vanish into thin air. This is the beauty of the symmetry! We are left with:

Since ${\sin }^2\theta +{\cos }^2\theta =1$, we get $25r^2=288$, or $r^2=\frac{288}{25}=11.52$.

We are almost there. The question asks for $l=O{P}^2+P{Q}^2+Q{O}^2$. We know $O{P}^2=r^2$ and $O{Q}^2=r^2$.

Since $△OPQ$ is a right-angled triangle, $P{Q}^2=O{P}^2+O{Q}^2=2r^2$. Thus, $l=r^2+2r^2+r^2=4r^2$.

Substituting our value of $r^2$:

The greatest integer less than or equal to $46.08$ is 46. You have navigated the geometry, mastered the rotation, and conquered the algebra. Take a moment to appreciate how the symmetry of the problem made the path clear. Keep this intuition, and you will solve any problem the JEE throws at you.