Question:
If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is
A. 12
B. −2
C. −12, −2
D. 12, −2
Solution:
Answer: D
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1\end{array}\right|$
$=\frac{1}{2}[2(4-4)+6(5-k)+1(20-4 k)]$
$=\frac{1}{2}[30-6 k+20-4 k]$
$=\frac{1}{2}[50-10 k]$
$=25-5 k$
It is given that the area of the triangle is ±35.
Therefore, we have:
$\Rightarrow 25-5 k=\pm 35$
$\Rightarrow 5(5-k)=\pm 35$
$\Rightarrow 5-k=\pm 7$
When $5-k=-7, k=5+7=12$.
When $5-k=7, k=5-7=-2$.
Hence, $k=12,-2$.
The correct answer is D.