Question

find the area of triangle ABC with vertices A(-5, 7), B(-4, -5) and C(4, 5)
​

Answer 1

Step-by-step explanation:

A = 1/2 [ x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) ]

x1 = -5, y1 = 7, x2 = -4, y2 = -5, x3 = 4, y3 = 5

A = 1/2 [ -5(-5 - 5) + {-4(5 - 7)} + 4(7 - (-5) ]

A = 1/2 [ -5 (-10) - 4(-2) + 4(12) ]

A = 1/2 [ 50 + 8 + 48 ]

A = 1/2 [ 106 ]

A = 53 sq units

Answer 2

ANSWER:-

Area of ΔABC whose vertices are is

$(x_1,y_1),(x_2,y_2) \: and \: (x_3,y_3)are$

$Area \: of ΔABC = \frac{1}{2} [x_1(y_2,y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

In ΔABC, vertices are A(−5,7),B(−4,−5) and C(4,5)

$Area \: of \: triangle \: = \frac{1}{2} (- 5 (- 5 - 5 ) - 4(5 - 7) + 4(7 + 5))$

$= \frac{1}{2} ( - 50 + 8 + 48)$

$= 5 \: sq. \: units$

HOPE IT'S HELPS YOU ❣️