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

Answer 1

Answer: (D)

12,-2

Answer 2

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
$\triangle$=\frac{1}{2}$$\begin{vmatrix}2&-6&1\\\5&4&1\\\k&4&1\end{vmatrix}
=$\frac{1}{2}$[2(4-4)+6(5-k)+1(20-4k)]
=$\frac{1}{2}$[30-6k+20-4k]
=$\frac{1}{2}$[50-10k]
=25-5k
It is given that the area of the triangle is ±35.
Therefore, we have:
$\Rightarrow$ 25-5k=±35
$\Rightarrow$ 5(5-k)=±35
$\Rightarrow$ 5-k=±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.