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 \( C(k, 8) \) is calculated using the coordinate area formula:

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

Substituting the given points, with the area set to 5:

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

Simplifying the expression within the absolute value:

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

Further simplification yields:

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

Multiplying both sides by 2:

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

This absolute value equation leads to two possible linear equations:

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

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

Solving the first equation:

\[22-10=4k\]

\[12=4k\]

\[k=3\]

Solving the second equation:

\[22+10=4k\]

\[32=4k\]

\[k=8\]