How do you find the area of a triangle whose vertices are (0, 5), (2, -2), and (5, 1)?

Question

How do you find the area of a triangle whose vertices are (0, 5), (2, -2), and (5, 1)?

1 Answer

Impact of this question

10013 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-28T07:18:30

13.50

Explanation:

if the three vertices of a triangle are $(x_{1}, y_{1}), (x_{2}, y_{2}) & (x_{3}, y_{3})$ $(x_1,y_1),(x_2,y_2) & (x_3,y_3)$
then the area of triangle is
$a = \frac{1}{2} \cdot (x_{1} \cdot (y_{2} - y_{3}) + x_{2} \cdot (y_{3} - y_{1}) + x_{3} \cdot (y_{1} - y_{2}))$ $a = 1/2*(x_1*(y_2-y_3)+x_2*(y_3-y_1)+x_3*(y_1-y_2))$

so

$a = \frac{1}{2} \cdot (0 \cdot (- 2 - 1) + 2 \cdot (1 - 5) + 5 \cdot (5 - (- 2))) = 13.50$ $a = 1/2*(0*(-2-1)+2*(1-5)+5*(5-(-2))) = 13.50$

Science

Math

Humanities

... and beyond

How do you find the area of a triangle whose vertices are (0, 5), (2, -2), and (5, 1)?

1 Answer

Explanation:

Related questions

Impact of this question