The area of the triangle whose vertices are A(6,7), B(-4,11) and C(2,-9) is

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Solution

The correct option is A 68
The area of the triangle whose vertices are $A (X_{1}, Y_{1}), B (X_{2}, Y_{2})$ and $C (X_{3}, Y_{3})$ is $\frac{1}{2} [(X_{1} Y_{2} + X_{2} Y_{3} + X_{3} Y_{1}) - (X_{2} Y_{1} + X_{3} Y_{2} + X_{1} Y_{3})]$
Substituting the values given in the question,we get area = 68 sq. units.

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△ A B C

68

A (6, 7), B (- 4, 1)

C (a, - 9)

a

(a, a^{2}), (b, b^{2}), (c, c^{2})

(p, p^{2}), (q, q^{2})

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∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 + a p)^{2} & (1 + b p)^{2} & (1 + c p)^{2} (1 + a q)^{2} & (1 + b q)^{2} & (1 + c q)^{2} (1 + a r)^{2} & (1 + b r)^{2} & (1 + c r)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

△ A B C

A (2, 2), B (6, 3)

C (4, 11)

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