Question

Find the area of a triangle whose vertices are A( 6 2 ) B(- 3 - 5) C(- 1 2 ).​

Answer 1

Answer:

58

Step-by-step explanation:

it hope brainaly help

Answer 2

$\huge{ \pink{ \underline{ \underline{ \mathsf{solution}}}}}$

Vertices of the given triangle are A(6,3)B(−3,5),C(4,−2)

Therefore the Area of △ABC

${ = 21∣y1(x2−x3)+y2(x3−x1)+y3(x1−x2)∣sq.unit}</strong></p><p></p><p></p><p><strong>[tex] \bold{ = 21∣y1(x2−x3)+y2(x3−x1)+y3(x1−x2)∣sq.unit}$

$\bold{Here, x1=6,y1=3, x2=−3,y2=5, x3=4,y3=−2}$

$\bold{</strong></p><p><strong>[tex] \bold{ = 21∣3(−3−4)+5(4−6)+(−2)(6+3)∣sq.unit}$

$\tiny{</strong></p><p><strong>[tex] \tiny{ = 21∣−21−10−18∣sq.unit}$

$\tiny{</strong></p><p><strong>[tex] \tiny{ = 21×49 sq.units}$

${\bold{ \boxed{</strong></p><p><strong>[tex] {\bold{ \boxed{ = 249 sq.units</strong></p><p></p><p></p><p><strong>[tex] { \boxed{ = 249 sq.units}}}$