Vertices of a parallelogram are $(0, 0),$ $(\frac{1}{m - n}, \frac{m}{m - n})$ , $(\frac{- 1}{m - n}, \frac{m}{m - n})$ ,and $(0, 1)$ where $m, n$ are the roots of the equation $441 x^{2} + 42 x - 8 = 0$ If area of the parallelogram is $A$ then $2 A$ is equal to:

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Solution

Given m,n are roots of equation $441 x^{2} + 42 x - 8 = 0$
$(21 x - 2) (21 x + 4) = 0$
$∴ m = \frac{2}{21}, n = \frac{- 4}{21}$
$m - n = \frac{6}{21}$
$\frac{1}{m - n} = \frac{21}{6}, = \frac{7}{2}, \frac{m}{m - n} = \frac{1}{3}$

(Ref diagram) Vertices of parallelogram are $(0, 0), (\frac{7}{2}, \frac{1}{3}), (- \frac{7}{2}, \frac{1}{3}), (0, 1)$
Diagonals are perpendicular to each other
$∴$ Area of parallelogram $= A = \frac{1}{2} \times \frac{7}{2} \times 1 + \frac{1}{2} \times \frac{7}{2} \times 1$
$= \frac{7}{2}$
$∴$ Area of Parallelogram= $A = \frac{7}{2}$
$2 A = 7$

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