Find values of mathrm{k} if area of triangle | vedclass

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Question

We know that the area of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is the ab

Vedclass · Accepted Answer

$0,8$

Vedclass · Answer

We know that the area of a triangle whose vertices are $\left(x_{1}, y_{1}ight),\left(x_{2}, y_{2}ight)$ and $\left(x_{3}, y_{3}ight)$ is the absolute value of the determinant ( $\Delta$ ),  where

$\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}ight|$

It is given that the area of triangle is $4$ square units.

$	herefore \Delta=\pm 4$

The area of the triangle with vertices $(k, 0),(4,0),(0,2)$ is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{lll}k & 0 & 1 \ 4 & 0 & 1 \ 0 & 2 & 1\end{array}ight|$

$=\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)]$

$=\frac{1}{2}[-2 k+8]=k+4$

$	herefore-k+4=\pm 4$

When $-k+4=-4, k=8$

When $-k+4=-4, k=0$

Hence, $k=0,8$

Find values of $\mathrm{k}$ if area of triangle is $4$ square units and vertices are $(\mathrm{k}, 0),(4,0),(0,2)$

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