The area of the parallelogram formed by the l | vedclass

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(D) The given lines are $y - mx = 0$,$y - mx - 1 = 0$,$y - nx = 0$,and $y - nx - 1 = 0$.These lines form a parallelogram.The area of a parallelogram f

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$\frac{1}{|m - n|}$

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(D) The given lines are $y - mx = 0$,$y - mx - 1 = 0$,$y - nx = 0$,and $y - nx - 1 = 0$.These lines form a parallelogram.The area of a parallelogram formed by lines $a_1x + b_1y + c_1 = 0$,$a_1x + b_1y + c_2 = 0$,$a_2x + b_2y + d_1 = 0$,and $a_2x + b_2y + d_2 = 0$ is given by the formula:$	ext{Area} = \left| \frac{(c_1 - c_2)(d_1 - d_2)}{a_1b_2 - a_2b_1} ight|$Here,the equations are:$mx - y + 0 = 0$$mx - y + 1 = 0$$nx - y + 0 = 0$$nx - y + 1 = 0$Comparing with the formula,we have $c_1 = 0, c_2 = 1, d_1 = 0, d_2 = 1, a_1 = m, b_1 = -1, a_2 = n, b_2 = -1$.$	ext{Area} = \left| \frac{(0 - 1)(0 - 1)}{m(-1) - n(-1)} ight| = \left| \frac{(-1)(-1)}{-m + n} ight| = \left| \frac{1}{n - m} ight| = \frac{1}{|m - n|}$.

The area of the parallelogram formed by the lines $y = mx$,$y = mx + 1$,$y = nx$,and $y = nx + 1$ is equal to

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