Question

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

Answer 1

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


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


=>1/2[(-5)(-5-5)+(-4)(5+5)+(4)(7+5)]
=> 1/2 [50-40+48]
=>1/2 [58]
=>29.

so , Area of a triangle =29





hope it's help u!

Answer 2

Answer:

SOLUTION :-

Given that,

The points A ( -5 , 7 ), B ( -4 , -5 ) and C ( 4 , 5 ) are the vertices of ΔABC .

$\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ ]}$

Here,

$\bullet\sf \ x_1=-5 \ , \ y_1=7 \\\\ \bullet \ x_2= -4 \ , \ y_2= -5 \\\\\bullet \ x_3= 4 \ , \ y_3= 5$

Area of triangle ABC,

$\longrightarrow \sf \dfrac{1}{2} \times [ \ -5(-5-5)+-4(5-7)+4(7-(-5) \ ] \\\\\longrightarrow \dfrac{1}{2}\times [ \ -5(-5-5)+-4(5-7)+4(7+5) \ ] \\\\\longrightarrow \dfrac{1}{2}\times [ \ (-5\times -10)+(-4\times -2 )+(4\times 12) \ ] \\\\\longrightarrow \dfrac{1}{2}\times [ \ 50+8 +48 \ ] \\\\\longrightarrow \dfrac{1}{2}\times 106 \\\\\longrightarrow\bf 53 \ sq.units$

Area of ΔABC = 53 sq.units