The area of the triangle whose vertices are (3,8), (-4,2) and (5,-1) is

75 s q . u n i t s

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37.5 s q . u n i t s

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45 s q . u n i t s

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22.5 s q . u n i t s

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Solution

The correct option is B $37.5 s q . u n i t s$
Let $A (3, 8), B (- 4, 2), C (5, - 1)$ be the vertices of the given $△ A B C$ .

Then,

$(x_{1} = 3, y_{1} = 8), (x_{2} = - 4, y_{2} = 2), (x_{3} = 5, y_{3} = - 1)$

Area of $△ A B C$ = $\frac{1}{2} | [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] |$

$= \frac{1}{2} | 3 [2 - (- 1)] - 4 (- 1 - 8) + 5 (8 - 2) |$

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

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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)

(1, 1),

(4, 2)

(3, 5) .

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