Question

If the area of a triangle with vertices \((k, 0)\), \((2, 0)\) and \((0, -2)\) is 2 square units, the value of \(k\) is \(\underline{\hspace{1cm}}\).

Hint:

Absolute value equations give two solutions—check both when solving coordinate geometry area problems.

Answer 1

Correct Answer: 0

Using the coordinate geometry 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 points: \((k,0)\), \((2,0)\), \((0,-2)\):
\[ \text{Area} = \frac{1}{2} \left| k(0 + 2) + 2(-2 - 0) + 0(0 - 0) \right| \]
\[ = \frac{1}{2} | 2k - 4 | \]
Given area = 2:
\[ \frac{1}{2}|2k - 4| = 2 \]
\[ |2k - 4| = 4 \]
Thus: 1) \( 2k - 4 = 4 $\Rightarrow$ k = 4 \)
2) \( 2k - 4 = -4 $\Rightarrow$ k = 0 \)