The area of the triangle ABC with the vertices A(–5, 7), B(–4, –5) and C(4, 5) is
Question
The area of the triangle ABC with the vertices A(–5, 7), B(–4, –5) and C(4, 5) is
Solution
To find the area of a triangle with vertices at three points (x1, y1), (x2, y2), and (x3, y3), you can use the formula:
Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Here, the vertices of the triangle ABC are A(-5, 7), B(-4, -5), and C(4, 5). So, x1 = -5, y1 = 7, x2 = -4, y2 = -5, x3 = 4, and y3 = 5.
Substitute these values into the formula:
Area = 1/2 * |-5(-5 - 5) - 4(5 - 7) + 4(7 - (-5))| = 1/2 * |-5(-10) - 4(-2) + 4(12)| = 1/2 * |50 + 8 + 48| = 1/2 * 106 = 53 square units
So, the area of the triangle ABC is 53 square units.
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