Key Concepts and Formulas

• Area of a Triangle (Coordinate Geometry): Given vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area of the triangle is $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
• Equation of a Line (Two-Point Form): The equation of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
• Intersection of Lines: To find the intersection point of two lines, solve their equations simultaneously.

Step-by-Step Solution

Step 1: Calculate the Area of $\Delta ABC$ ($A_1$)

• Objective: Determine the area of the triangle ABC using the given coordinates. This will be our reference area.
• Vertices: A(-1, 1), B(3, 4), C(2, 0)
• Applying the area formula: $A_1 = \frac{1}{2} |(-1)(4 - 0) + (3)(0 - 1) + (2)(1 - 4)|$
• Calculation: $A_1 = \frac{1}{2} |(-1)(4) + (3)(-1) + (2)(-3)|$ $A_1 = \frac{1}{2} |-4 - 3 - 6|$ $A_1 = \frac{1}{2} |-13|$ $A_1 = \frac{13}{2}$

Step 2: Find the Equation of Line AC and the Coordinates of Point P

• Objective: Find the equation of line AC and then find the intersection point P of line AC and y = mx.
• Equation of Line AC:
  • Points A(-1, 1) and C(2, 0)
  • Using the two-point form: $y - 1 = \frac{0 - 1}{2 - (-1)}(x - (-1))$ $y - 1 = -\frac{1}{3}(x + 1)$ $3y - 3 = -x - 1$ $x + 3y - 2 = 0$
• Intersection of AC and y = mx:
  • Substitute y = mx into the equation of AC: $x + 3(mx) - 2 = 0$ $x(1 + 3m) = 2$ $x = \frac{2}{1 + 3m}$
  • Find y: $y = m \left( \frac{2}{1 + 3m} \right) = \frac{2m}{1 + 3m}$
  • Coordinates of P: $\left( \frac{2}{1 + 3m}, \frac{2m}{1 + 3m} \right)$

Step 3: Find the Equation of Line BC and the Coordinates of Point Q

• Objective: Find the equation of line BC and then find the intersection point Q of line BC and y = mx.
• Equation of Line BC:
  • Points B(3, 4) and C(2, 0)
  • Using the two-point form: $y - 0 = \frac{4 - 0}{3 - 2}(x - 2)$ $y = 4(x - 2)$ $y = 4x - 8$ $4x - y - 8 = 0$
• Intersection of BC and y = mx:
  • Substitute y = mx into the equation of BC: $4x - mx - 8 = 0$ $x(4 - m) = 8$ $x = \frac{8}{4 - m}$
  • Find y: $y = m \left( \frac{8}{4 - m} \right) = \frac{8m}{4 - m}$
  • Coordinates of Q: $\left( \frac{8}{4 - m}, \frac{8m}{4 - m} \right)$

Step 4: Calculate the Area of $\Delta PQC$ ($A_2$)

• Objective: Calculate the area of triangle PQC using the coordinates of P, Q, and C.
• Vertices: P$\left( \frac{2}{1 + 3m}, \frac{2m}{1 + 3m} \right)$, Q$\left( \frac{8}{4 - m}, \frac{8m}{4 - m} \right)$, C(2, 0)
• Applying the area formula: $A_2 = \frac{1}{2} \left| \frac{2}{1 + 3m} ( \frac{8m}{4 - m} - 0) + \frac{8}{4 - m}(0 - \frac{2m}{1 + 3m}) + 2 (\frac{2m}{1 + 3m} - \frac{8m}{4 - m}) \right|$ $A_2 = \frac{1}{2} \left| \frac{16m}{(1 + 3m)(4 - m)} - \frac{16m}{(4 - m)(1 + 3m)} + 2 \left( \frac{2m(4 - m) - 8m(1 + 3m)}{(1 + 3m)(4 - m)} \right) \right|$ $A_2 = \frac{1}{2} \left| 2 \left( \frac{8m - 2m^2 - 8m - 24m^2}{(1 + 3m)(4 - m)} \right) \right|$ $A_2 = \frac{1}{2} \left| 2 \left( \frac{-26m^2}{(1 + 3m)(4 - m)} \right) \right|$ $A_2 = \left| \frac{-26m^2}{(1 + 3m)(4 - m)} \right|$ Since $m > 0$, we have $A_2 = \frac{26m^2}{|(1 + 3m)(4 - m)|}$

Step 5: Apply the Area Condition ($A_1 = 3A_2$) and Solve for $m$

• Objective: Use the given area relation to solve for the value of m.

• Given $A_1 = 3A_2$

• Substituting the calculated areas: $\frac{13}{2} = 3 \left( \frac{26m^2}{|(1 + 3m)(4 - m)|} \right)$ $\frac{13}{2} = \frac{78m^2}{|(1 + 3m)(4 - m)|}$ $13 |(1 + 3m)(4 - m)| = 156m^2$ $|(1 + 3m)(4 - m)| = 12m^2$ We consider two cases:

  • Case 1: $(1 + 3m)(4 - m) = 12m^2$ $4 - m + 12m - 3m^2 = 12m^2$ $15m^2 - 11m - 4 = 0$ $m = \frac{11 \pm \sqrt{121 - 4(15)(-4)}}{30} = \frac{11 \pm \sqrt{121 + 240}}{30} = \frac{11 \pm \sqrt{361}}{30} = \frac{11 \pm 19}{30}$ $m = \frac{30}{30} = 1 \quad \text{or} \quad m = \frac{-8}{30} = -\frac{4}{15}$ Since $m > 0$, we have $m = 1$. If $m = 1$, then $(1+3m)(4-m) = (4)(3) = 12 > 0$, so this case is consistent.

  • Case 2: $(1 + 3m)(4 - m) = -12m^2$ $4 - m + 12m - 3m^2 = -12m^2$ $9m^2 + 11m + 4 = 0$ The discriminant is $D = 11^2 - 4(9)(4) = 121 - 144 = -23 < 0$. Thus, there are no real solutions in this case.

Therefore, $m = 1$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when expanding determinants and simplifying algebraic expressions.
• Absolute Value: Remember to use the absolute value in area calculations. It's necessary to consider both positive and negative cases.
• Checking for Validity: Always check if your solutions satisfy any given conditions (e.g., $m > 0$) and any implied conditions (e.g., $m \ne 4$).

Summary

We calculated the areas of triangles ABC and PQC using coordinate geometry. We then used the given area relationship to form an equation and solved for m, considering both positive and negative cases arising from the absolute value in the area formula. The final solution is m = 1.

Final Answer

The value of m is \boxed{1}, which corresponds to option (A).