Animated Solution for Mathematics - Straight Lines: The coordinates of $A,B,C$ are $(6,3)$, $(-3,5)$, $(4,-2)$ respectively, and $P$ is any point $(x,y)$. Show that the ratio of the area of the triangles $△PBC$ and $△ABC$ is $|\frac{x+y-2}{7}|$.

A Journey Through Coordinate Space

Welcome, fellow traveler of the JEE landscape. Today, we aren't just solving a coordinate geometry problem; we are uncovering the hidden relationship between a fixed foundation and a wandering point.

Imagine you are standing on a coordinate plane with three fixed anchors: $A(6,3)$, $B(-3,5)$, and $C(4,-2)$. These three points define a rigid triangle, $△ABC$.

As a wanderer point $P(x,y)$ moves, the triangle $△PBC$ changes shape, stretching and shrinking. Our goal is to find the ratio of the area of this shifting triangle to the area of our fixed anchor triangle.

Calculating the Fixed Area

Before we can understand the motion of $P$, we must understand the static nature of $△ABC$. We reach into our mathematical toolkit for the classic area formula for a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$:

Plugging in our coordinates for $A$, $B$, and $C$, we get:

As we simplify this, watch the numbers dance: $6(7)+(-3)(-5)+4(-2)$ becomes $42+15-8$, which sums to $49$. Thus, the area of our fixed triangle is $\frac{49}{2}$. This is our constant, our denominator, the bedrock upon which our ratio stands.

Defining the Variable Area

Now, let us turn our attention to $△PBC$. The vertices are $P(x,y)$, $B(-3,5)$, and $C(4,-2)$. Applying the same formula, we get:

I know, seeing variables $x$ and $y$ inside the absolute value bars can feel intimidating. But take a deep breath. Let's expand this systematically.

We have $x(7)-3(-2-y)+4(y-5)$. Distributing those terms gives us $7x+6+3y+4y-20$.

Combining the like terms, we arrive at $7x+7y-14$. If we factor out the $7$, we get $\frac{7}{2}|x+y-2|$. This expression is the heartbeat of our problem; it tells us exactly how the area of $△PBC$ depends on the position of $P$.

Now, we bring it all together. The ratio we seek is:

Look at the beauty of this moment! The $\frac{1}{2}$ terms cancel out instantly. We are left with $\frac{7|x+y-2|}{49}$.

Simplifying the fraction $\frac{7}{49}$ gives us the final, elegant result: $\frac{|x+y-2|}{7}$.

Why did this happen? Why is the ratio so clean? If you look at the line $BC$, its equation is $x+y-2=0$.

The perpendicular distance from any point $P(x,y)$ to this line is given by $d=\frac{|x+y-2|}{\sqrt{1^2+1^2}}$. Since both triangles share the base $BC$, the ratio of their areas is simply the ratio of their heights.

The height of $△ABC$ from $A$ to $BC$ is a constant, and the height of $△PBC$ from $P$ to $BC$ is a variable. By calculating the ratio, we have essentially normalized the distance of $P$ from the line $BC$ relative to the distance of $A$ from that same line.

You have just mastered the interplay between algebraic manipulation and geometric intuition. Keep this clarity with you—it is the key to conquering the JEE.