Find the value of x for which the area of the triangle with vertices at (-1, -4), (x, 1) and (x, -4) is 12¹/₂ sq. units.

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-3

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-6

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Solution

The correct options are
B 4
D -6
Area of the triangle $= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

The area of the triangle with vertices at (-1, -4), (x, 1) and (x, -4) is

$= \frac{1}{2} | (- 1 - 4 x - 4 x) - (- 4 x + x + 4) |$

$= \frac{1}{2} | - 1 - 8 x + 3 x - 41 = \frac{1}{2} | - 5 x - 5 | s q . u n i t s .$

By problem,
$\frac{1}{2} | 5 x - 5 | = 12 \frac{1}{2} = \frac{25}{2}$

$\Rightarrow$ 5x + 5 = ± 25
$\Rightarrow$ x + 1 = ± 5
$\Rightarrow$ x = 5 - 1 or x = -5 - 1

Therefore, x = 4 or - 6.

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