If area of triangle is 35 mathrm{sq} mathrm{units} with vertices (2,-6),(5,4) and (mathrm{k}, 4) . T

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3 and 4 .Determinants and Matrices

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If area of triangle is $35$ $\mathrm{sq}$ $\mathrm{units}$ with vertices $(2,-6),(5,4)$ and $(\mathrm{k}, 4) .$ Then $\mathrm{k}$ is

A

$12$

B

$-2$

C

$12,-2$

D

$-12,-2$

Solution

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 id $\pm 35$.

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 $C$.

Standard 12

Mathematics

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