Animated Solution for Mathematics - Straight Lines: Let A(-1,1), B(3, 4) and C(2,0) be given three points. A line $y=mx,m>0$, intersects lines AC and BC at point P and Q respectively. Let $A_1$ and $A_2$ be the areas of $△ABC$ and $△PQC$ respectively, such that $A_1=3A_2$, then the value of m is equal to :

Welcome, student. Today, we are not just solving a coordinate geometry problem; we are choreographing a dance between a static triangle and a dynamic line.

Imagine you are standing on the Cartesian plane. You see the vertices $A(-1,1)$, $B(3,4)$, and $C(2,0)$ anchored in place, forming a rigid structure, $△ABC$.

Now, imagine a line $y=mx$ originating from the origin, sweeping across the plane like a radar beam. As it rotates, it slices through the sides $AC$ and $BC$, creating two new points, $P$ and $Q$. Our goal is to find the exact slope $m$ where the area of the smaller triangle $△PQC$ is exactly one-third of the area of the original $△ABC$.

Before we can slice the triangle, we must know its total size. We use the classic coordinate area formula:

Substituting our vertices $A(-1,1)$, $B(3,4)$, and $C(2,0)$, we calculate:

This simplifies to:

This is our target. We know that $A_1=3A_2$, which means $A_2$ must be $\frac{13}{6}$. Keep this number in your heart; it is our destination.

To find the area of $△PQC$, we need the coordinates of $P$ and $Q$. These points are born from the intersection of our line $y=mx$ with the lines $AC$ and $BC$.

First, we find the equations of these lines. The slope of $AC$ is $\frac{0-1}{2-(-1)}=-\frac{1}{3}$, leading to the equation $x+3y=2$. The slope of $BC$ is $\frac{0-4}{2-3}=4$, leading to $y=4x-8$.

Now, we solve for $P$ and $Q$:

For $P$, we substitute $y=mx$ into $x+3y=2$, yielding $x(1+3m)=2$, so $P=\left(\frac{2}{1+3m},\frac{2m}{1+3m}\right)$.

For $Q$, we substitute $y=mx$ into $y=4x-8$, yielding $mx=4x-8$, so $x(4-m)=8$, giving $Q=\left(\frac{8}{4-m},\frac{8m}{4-m}\right)$.

Now, we calculate the area of $△PQC$. With vertices $C(2,0)$, $P(x_P,y_P)$, and $Q(x_Q,y_Q)$, the area formula is:

Because $C$ is at $(2,0)$, the formula simplifies beautifully. When we plug in $y_P=m{x}_P$ and $y_Q=m{x}_Q$, the terms involving $x_P{x}_Q$ cancel out entirely.

We are left with $A_2=m|x_P-x_Q|$. We calculate the difference:

Thus, the area expression becomes:

We are at the finish line. We set $A_1=3A_2$, which means:

Dividing by 13, we get:

Cross-multiplying, we arrive at $|(1+3m)(4-m)|=12m^2$. Since $m>0$ and $Q$ lies on the segment $BC$, we know $4-m>0$, allowing us to remove the modulus.

Expanding this, we get $4+11m-3m^2=12m^2$, or $15m^2-11m-4=0$. Factoring this quadratic, we find:

Since $m$ must be positive, we reject $-\frac{4}{15}$ and embrace $m=1$. You have successfully navigated the geometry, the algebra, and the constraints. Well done.