Find the area of a triangle whose vertices are (3,8), (-4,2) and (5,-1).

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Solution

Let A(3,8), B(-4,2) and C(5,-1) be the vertices of the given $△$ ABC. Then,
$(x_{1} = 3, y_{1} = 8), (x_{2} = - 4, y_{2} = 2), (x_{3} = 5, y_{3} = - 1)$
Area of $△$ ABC = $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | 3 (3) - 4 (- 9) + 5 (6) |$

$= \frac{1}{2} | 9 + 36 + 30 | = \frac{75}{2} = 37.5 s q . u n i t s$

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Similar questions

Find the area of $△ A B C$ whose vertices are :

(i) A(1, 2), B (-2, 3) and C(-3, -4)

(ii) A (-5,7),B(-4, -5) and C(4,5)

(iii) A(3, 8), B(-4, 2) and C(5, -1)

(iv) A (10, -6),B (2,5) and C(-1, 3)

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