Question

The vertices of triangle are
(1,k), (4,-3) and (-9, 1) and
area Triangle 159. units then
find Kvalue?​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:k=-26.53}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{given :}} \\ \tt: \implies vertices \: of \: triangle = (1.k). \: (4. - 3) \: and \: ( - 9.1) \\ \\ \tt: \implies area \: of \: triangle = 159 \: unit \\ \\ \red{\underline \bold{to \: find :}} \\ \tt: \implies value \: of \: k = ?$

• According to given question :

$\bold{as \: we \: know \: that} \\ \tt: \implies area \: of \: triangle = \frac{1}{2} (x_{1}( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2}) ) \\ \\ \tt: \implies 159 = \frac{1}{2} (1( - 3 - 1) + 4(1 - k) - 9(k - ( - 3)) )\\ \\ \tt: \implies 159 \times 2 = (1 \times - 4 + 4 - 4k - 9k - 27) \\ \\ \tt: \implies 318 = (- 4 + 4 - 4k - 9k - 27) \\ \\ \tt: \implies 318 = - 13k - 27 \\ \\ \tt: \implies 318 + 27 = - 13k \\ \\ \tt: \implies 345 = - 13k \\ \\ \tt: \implies k = - \frac{345}{ 13} \\ \\ \green{\tt: \implies k = -26.53}$

Answer 2

Heya!!

Here's your answer!!

kindly see in the attached picture.⬆️⬆️

glad help you,

it helps you,

thanks.