Question

The area of a triangle with vertices at points $ A(1,2) $, $ B(4,6) $, and $ C(k, 8) $ is 5. Find the value of $ k $.

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Use the coordinate geometry formula for area of a triangle and solve absolute value equations carefully.

Answer 1

The Correct Option is C

The area of a triangle with vertices at points \( A(1,2) \), \( B(4,6) \), and \( C(k, 8) \) can be calculated using the formula for the area of a triangle given coordinates:

\[\text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\]

Substitute the given points into the formula:

\[\text{Area}=\frac{1}{2}\left|1(6-8)+4(8-2)+k(2-6)\right|=5\]

Compute inside the absolute value:

\[\frac{1}{2}\left|-2+24+k(-4)\right|=5\]

Simplify:

\[\frac{1}{2}\left|22-4k\right|=5\]

Multiply both sides by 2 to eliminate the fraction:

\[\left|22-4k\right|=10\]

This absolute value equation implies two scenarios:

1. \(22-4k=10\)

2. \(22-4k=-10\)

Solve the equations:

For \(22-4k=10\):

\[22-10=4k\]

\[12=4k\]

\[k=3\]

For \(22-4k=-10\):

\[22+10=4k\]

\[32=4k\]

\[k=8\]

Answer 2

Step 1: Use the area formula for a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \): \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substitute: \( A = (1, 2), B = (4, 6), C = (k, 8) \) \[ \text{Area} = \frac{1}{2} \left| 1(6 - 8) + 4(8 - 2) + k(2 - 6) \right| = 5 \] \[ \Rightarrow \frac{1}{2} \left| -2 + 24 - 4k \right| = 5 \Rightarrow \frac{1}{2} \left| 22 - 4k \right| = 5 \Rightarrow |22 - 4k| = 10 \] Step 2: Solve the absolute value equation: \[ 22 - 4k = 10 \quad \text{or} \quad 22 - 4k = -10 \] First case: \[ 22 - 4k = 10 \Rightarrow 4k = 12 \Rightarrow k = 3 \] Second case: \[ 22 - 4k = -10 \Rightarrow 4k = 32 \Rightarrow k = 8 \]