Question

find the area of the triangle whose vertices are (8,-4) (-6,6) and (-3,9)

Answer 1

$\frak{Here} \begin{cases} & \sf{(x_1 , y_1) = \bf{(8, -4)}} \\ & \sf{(x_2 , y_2) = \bf{(-6,6)}} \\ & \sf{(x_3 , y_3) = \bf{(-3,9)}}\end{cases}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

We know that,

$\star\;{\boxed{\sf{\purple{Area_{ \triangle} = \dfrac{1}{2} \bigg[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg]}}}}$

Therefore,

⠀⠀⠀

$:\implies\sf \dfrac{1}{2} \bigg[ 8(6 - 9) - 6(9 - (-4)) + (-3) (-4 - 6) \bigg]$

$:\implies\sf \dfrac{1}{2} \bigg[ 8 (-3) -6 (13) -3 (8) \bigg]$

$:\implies\sf \dfrac{1}{2} \bigg[ -24 - 78 - 24 \bigg]$

$:\implies\sf \dfrac{1}{2} \bigg[ -114 \bigg]$

$:\implies\sf \bigg[ -57 \bigg]$

$\qquad \qquad\tiny \dag \: {\underline{\frak{As \;we\;know\;area\;of\;\triangle \;cannot\;be\;negative\;so,}}}$

$:\implies{\boxed{\frak{\pink{57}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;area\;of\;triangle\;is\; \bf{57}.}}}$