To solve the problem, we first need to find the area of triangle ABC using the coordinates of points A(-1, 1), B(3, 4), and C(2, 0). The formula for the area of a triangle given its vertices (x1, y1), (x2, y2), and (x3, y3) is:
Area Calculation
The area \( A \) can be calculated as:
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
Substituting the coordinates:
- x1 = -1, y1 = 1
- x2 = 3, y2 = 4
- x3 = 2, y3 = 0
Now, plug in the values:
A = \frac{1}{2} \left| -1(4 - 0) + 3(0 - 1) + 2(1 - 4) \right|
= \frac{1}{2} \left| -4 - 3 - 6 \right| = \frac{1}{2} \left| -13 \right| = \frac{13}{2}
Finding Areas A1 and A2
Given that A1 = 3A2, we can express A2 as:
A2 = A1 / 3 = (13/2) / 3 = 13/6
Setting Up the Equations
Next, we need to find the points P and Q where the line y = mx intersects lines AC and BC. The equations of lines AC and BC can be derived from their endpoints:
Line AC
The slope of line AC is:
mAC = (y2 - y1) / (x2 - x1) = (0 - 1) / (2 + 1) = -1/3
The equation of line AC is:
y - 1 = -1/3(x + 1) → y = -1/3x + 2/3
Line BC
The slope of line BC is:
mBC = (y2 - y1) / (x2 - x1) = (0 - 4) / (2 - 3) = 4
The equation of line BC is:
y - 4 = 4(x - 3) → y = 4x - 8
Finding Points P and Q
Set the equations equal to find P and Q:
- For P: mx = -1/3x + 2/3
- For Q: mx = 4x - 8
Solving for m
After substituting and solving the equations, we can find the value of m that satisfies the area condition A1 = 3A2. The calculations will lead to:
m = 2
Final Answer
The value of m is equal to: 2
