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

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(D) The lines $ax+by+2=0$ and $ax+by+5=0$ are parallel. The distance $d_1$ between them is given by $d_1 = \frac{|5-2|}{\sqrt{a^2+b^2}} = \frac{3}{\sq

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

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(D) The lines $ax+by+2=0$ and $ax+by+5=0$ are parallel. The distance $d_1$ between them is given by $d_1 = \frac{|5-2|}{\sqrt{a^2+b^2}} = \frac{3}{\sqrt{a^2+b^2}}$.The lines $cx+dy+3=0$ and $cx+dy+7=0$ are parallel. The distance $d_2$ between them is given by $d_2 = \frac{|7-3|}{\sqrt{c^2+d^2}} = \frac{4}{\sqrt{c^2+d^2}}$.The area of a parallelogram formed by two pairs of parallel lines $a_1x+b_1y+c_1=0, a_1x+b_1y+c_2=0$ and $a_2x+b_2y+d_1=0, a_2x+b_2y+d_2=0$ is given by the formula:Area $= \frac{|(c_1-c_2)(d_1-d_2)|}{|a_1d_2 - a_2b_1|}$.Here,$a_1=a, b_1=b, a_2=c, b_2=d$. The constants are $c_1=2, c_2=5$ and $d_1=3, d_2=7$.Area $= \frac{|(2-5)(3-7)|}{|ad-bc|} = \frac{|(-3)(-4)|}{|ad-bc|} = \frac{12}{|ad-bc|}$.

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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