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

Answer (a) $\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{aq -bp}\right|$

$\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{aq -bp}\right|$

$\frac{|\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)|}{2|aq -bp|}$

$\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{ab -pq}\right|$

$\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{ab -bq}\right|$

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

$\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{aq -bp}\right|$

$\frac{|\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)|}{2|aq -bp|}$

$\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{ab -pq}\right|$

$\left|\frac{\left(c_{1}-c_{2}\right)\left(r_{1}-r_{2}\right)}{ab -bq}\right|$

The area of the parallelogram formed by the lines

Notes on Horizontal and vertical lines

Concepts Used:

Horizontal and vertical lines

Horizontal Lines:

A horizontal line is a sleeping line that means "side-to-side".
These are the lines drawn from left to right or right to left and are parallel to the x-axis.

Equation of the horizontal line:

In all cases, horizontal lines remain parallel to the x-axis. It never intersects the x-axis but only intersects the y-axis. The value of x can change, but y always tends to be constant for horizontal lines.