The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If third vertex is $(\frac{7}{2}, y)$ , find y.

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Solution

Area of triangle = $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = 5$

$\Rightarrow | 2 (- 2 - y) + 3 (y - 1) + \frac{7}{2} (1 + 2) | = 10$

$\Rightarrow | - 4 - 2 y + 3 y - 3 + \frac{21}{2} | = 10$

$\Rightarrow y + \frac{7}{2} = 10$ or $y + \frac{7}{2} = - 10$

$∴ y = \frac{13}{2}$ or $y = \frac{- 27}{2}$

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