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Step-by-Step Solution
Step 1: Understand the Problem
We have three points:
A($-1$, $1$), B($3$, $4$), and C($2$, $0$).
A line defined by $y = m\,x$, with $m > 0$, intersects the lines AC and BC at points P and Q, respectively.
Let $A_{1}$ be the area of $\triangle ABC$ and $A_{2}$ be the area of $\triangle PQC$. We know that $A_{1} = 3 A_{2}$.
We aim to find the value of $m$ that satisfies this area condition.
Step 2: Find the Slope of AC and BC
• Segment AC: slope is
$$
m_{AC} = \frac{y_{C} - y_{A}}{x_{C} - x_{A}}
= \frac{0 - 1}{2 - (-1)}
= \frac{-1}{3}
= -\frac{1}{3}.
$$
• Segment BC: slope is
$$
m_{BC} = \frac{y_{C} - y_{B}}{x_{C} - x_{B}}
= \frac{0 - 4}{2 - 3}
= \frac{-4}{-1}
= 4.
$$
Step 3: Equation of Lines AC and BC
We write the equations of lines AC and BC in slope-point form:
Line AC passing through A($-1$, $1$) with slope $-\frac{1}{3}$:
$$
y - 1 = -\frac{1}{3}(x + 1).
$$
Line BC passing through B($3$, $4$) with slope $4$:
$$
y - 4 = 4(x - 3).
$$
Step 4: Find Coordinates of P and Q
The line $y = m\,x$ intersects AC at P and BC at Q:
Point P lies on both $y = m\,x$ and AC:
$$
m\,x - 1 = -\frac{1}{3}(x + 1).
$$
Solve this for $x$, then use $y = m\,x$.
Point Q lies on both $y = m\,x$ and BC:
$$
m\,x - 4 = 4(x - 3).
$$
Solve for $x$, then use $y = m\,x$.
(These steps will yield the exact coordinates for P and Q in terms of $m$.)
Step 5: Express Area of Triangles and Use the Relation $A_{1} = 3A_{2}$
• The area of a triangle given by coordinates $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ can be found using the determinant formula
$$
\text{Area} = \frac{1}{2} \Big|\, x_1(y_2 - y_3) \;+\; x_2(y_3 - y_1) \;+\; x_3(y_1 - y_2) \Big|.
$$
• Compute $A_{1} = \text{Area}(\triangle ABC)$ using A, B, C.
• Compute $A_{2} = \text{Area}(\triangle PQC)$ in terms of $m$.
• Set $A_{1} = 3 A_{2}$ and solve for $m$.
Step 6: Conclude the Value of $m$
After simplification and satisfying $A_{1} = 3 A_{2}$, the positive value of $m$ turns out to be $m = 1$.
Hence, the line $y = x$ satisfies the required condition where the area of $\triangle ABC$ is thrice the area of $\triangle PQC$.
Final Answer
The value of $m$ is $1$.
