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Mathematics Question on Horizontal and vertical lines

The area of the parallelogram formed by the lines ax+by+c1=0, ax+by+c2=0, px+qy+r1=0, px+qy+r2=0ax+by+c_1=0, \: ax+by+c_2=0, \: px+qy+r_1=0, \: px+qy+r_2=0

A

∣(c1−c2)(r1−r2)aq−bp∣\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{aq -bp}\right|

B

∣(c1−c2)(r1−r2)∣2∣aq−bp∣\frac{|\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)|}{2|aq -bp|}

C

∣(c1−c2)(r1−r2)ab−pq∣\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{ab -pq}\right|

D

∣(c1−c2)(r1−r2)ab−bq∣\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{ab -bq}\right|

Answer

∣(c1−c2)(r1−r2)aq−bp∣\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{aq -bp}\right|

Explanation

Solution

Answer (a) ∣(c1−c2)(r1−r2)aq−bp∣\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{aq -bp}\right|