The area of the parallelogram formed by the l | vedclass

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(D) The lines are $L_1: mx - y = 0$,$L_2: mx - y + 1 = 0$,$L_3: nx - y = 0$,and $L_4: nx - y + 1 = 0$.The area of a parallelogram formed by lines $a_1

Vedclass · Accepted Answer

$1 / |m - n|$

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(D) The lines are $L_1: mx - y = 0$,$L_2: mx - y + 1 = 0$,$L_3: nx - y = 0$,and $L_4: nx - y + 1 = 0$.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|$Here,$a_1 = m, b_1 = -1, c_1 = 0, c_2 = 1$ and $a_2 = n, b_2 = -1, d_1 = 0, d_2 = 1$.Substituting these values into the formula:$	ext{Area} = \left| \frac{(0 - 1)(0 - 1)}{(m)(-1) - (n)(-1)} ight|$$	ext{Area} = \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 $y = mx, y = mx + 1, y = nx, y = nx + 1$ is:

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