If ad-bc neq 0,then the area (in sq. units) o | vedclass

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(D) The area of a parallelogram formed by the 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 formul

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$\frac{12}{|ad-bc|}$

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(D) The area of a parallelogram formed by the 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 lines are $ax+by+2=0$,$ax+by+5=0$,$cx+dy+3=0$,and $cx+dy+7=0$.Comparing with the formula,we have $c_1=2, c_2=5, d_1=3, d_2=7, a_1=a, b_1=b, a_2=c, b_2=d$.Substituting these values into the formula:$	ext{Area} = \left| \frac{(2-5)(3-7)}{ad-bc} ight|$$	ext{Area} = \left| \frac{(-3)(-4)}{ad-bc} ight|$$	ext{Area} = \frac{12}{|ad-bc|}$ sq. units.

If $ad-bc \neq 0$,then the area (in sq. units) of the parallelogram formed by the lines $ax+by+2=0$,$ax+by+5=0$,$cx+dy+3=0$ and $cx+dy+7=0$ is

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