Question

what is the area of triangle ABC with vertices A(x¹,y¹), B(x²,y²)and C (x³,y³)??????????
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Answer 1

Answer:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

$Area = 2\ units^2$

Explanation:

Given

$A = (x_1,y_1)$

$B = (x_2,y_2)$

$C = (x_3,y_3)$

Required

Determine the area

The area of a triangle is :

$Area = \frac{1}{2}|A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)|$

By substituting values for the x and y coordinates of A, B and C;

We have:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

So:

For instance

$A = (0,3)$

$B= (2,1)$

$C = (2,3)$

The area is:

$Area = \frac{1}{2}|0(1-3) + 2(3-3) + 2(3-1)|$

$Area = \frac{1}{2}| 2*0 + 2*2|$

$Area = \frac{1}{2}| 0 + 4|$

$Area = \frac{1}{2}|4|$

$Area = \frac{1}{2} * 4$

$Area = 2\ units^2$