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If the area of a triangle is $68$ sq. unit and the vertices are $(6, 7), (- 4, 1)$ and $(a, - 9)$ then the value of $a$ is

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answer is B.

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Detailed Solution

It is given that the area of a triangle is

68

sq. unit and the vertices are

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

and

(a, - 9)

.
Area of a triangle with vertices

(x_{1}, y_{1}); (x_{2}, y_{2})

and

(x_{3}, y_{3})

is ,

A = |\frac{x_{1} ({y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - {y_{2})}_{}}_{}}{2}|

Hence, substituting the points

(x_{1}, y_{1}) = (6, 7); (x_{2}, y_{2}) = (- 4, 1)

and

(x_{3}, y_{3}) = (a, - 9)

in the formula for area we get,

Δ = |\frac{6 (1 + 9) + (- 4) (- 9 - 7) + a (7 - 1)}{2}|

so we get the equation,

|\frac{6 (10) + (- 4) (- 16) + a (6)}{2}| = 68

\Rightarrow

| \frac{60 + 64 + 6 a}{2} | = 68

\Rightarrow

\frac{124 + 6 a}{2} = \pm 68

\Rightarrow

124 + 6 a = 136 or - 136

Here we must avoid the

- ve

quantity, as area is always positive.

\Rightarrow

6 a = 12

\Rightarrow a = 2

So , the correct option is 2.

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