જો (-2,0),(0,4),(0, mathrm{k}) શિરો | 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 $(-2,0),(0,4),(0, k)$ is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}-2 & 0 & 1 \ 0 & 4 & 1 \ 0 & k & 1\end{array}ight|$

$=\frac{1}{2}[-2(4-k)]$

$=k-4$

$	herefore k-4=\pm 4$

When $k-4=-4, k=0$

When $k-4=4, k=8$

Hence, $k=0,8$

જો $(-2,0),(0,4),(0, \mathrm{k})$ શિરોબિંદુવાળા ત્રિકોણનું ક્ષેત્રફળ $4$ ચોરસ એકમ હોય, તો $\mathrm{k}$ નું મૂલ્ય શોધો.

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