Find the area of the triangle whose vertices are A(-2,-3) ,B(3,2) , C(-1,-8)
Answer (1)
Student Expert
27th Jul, 2020
First, let, x1 and y1 stands for the two points of A i.e. (-2 , -3), similarly, x2 and y2 stands for B(3 , 2) and x3 and y3 stand for C(-1 , -8).
The formula to find out the area of a triangle from its given vertices are=
(1/2) * [ x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * ( y3 - y2) ].
Already you know the value of x1,y1, x2, y2, x3 and y3.
x1 = -2 , y1 = -3
x2 = 3 , y2 = 2
x3 = -1 , y3 = -8.
Now putting all these values in the formula to find out the area of a triangle and solving it,
(1/2) * [ (-2) * (2- (-8)) + 3 * ((-8) - (-3)) + (-1) * ((-8) - 2) ]
It comes around (-12.5) square unit.
The formula to find out the area of a triangle from its given vertices are=
(1/2) * [ x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * ( y3 - y2) ].
Already you know the value of x1,y1, x2, y2, x3 and y3.
x1 = -2 , y1 = -3
x2 = 3 , y2 = 2
x3 = -1 , y3 = -8.
Now putting all these values in the formula to find out the area of a triangle and solving it,
(1/2) * [ (-2) * (2- (-8)) + 3 * ((-8) - (-3)) + (-1) * ((-8) - 2) ]
It comes around (-12.5) square unit.
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