How is the area of a parallelogram given by |(c1-c2)(d1-d2)/ (m1-m2)|?

Question

My book gives this formula for the area of a parallelogram bounded by the lines $$y = m_1x + c_1,\ \ y = m_1x + c_2 \,\ \ y= m_2x + d_1, \text{ and } \ y = m_2x + d_2$$is given by $$\operatorname{abs}\left(\frac{(c_1-c_2)(d_1-d_2)} {m_1-m_2}\right).$$ I understood that $c_1-c_2$ and $d_1-d_2$ are the perpendicular distances between the two pairs of opposite sides of the parallelogram. Since area of a parallelogram equals base $\times$ height, taking $c_1-c_2$ as the height, the value of $$\frac{d_1-d_2}{m_1-m_2}$$ must necessarily supply the base. How does it do this?

Answer 1

This i because c1-c2 is not the height. If you want to determine the height between $y = m_1x + c_1$ and $y = m_1x + c_2$ you've to compute the minimal distance between them.The distance between the two lines is the distance between the two intersection points of these lines with the perpendicular line. So the height is: \begin{equation} h=\frac{|d_2-d_1|}{\sqrt{m_2²+1}} \end{equation} The base can be calcutated by considering the distance between two points. The first one is obtained by the intersection of between $y = m_1 x + c_1$ and $y = m_2 x + d_1$ while the second one is obtained by the intersection of between $y = m_1 x + c_2$ and $y = m_2 x + d_1$. The first point has coordinates $(\frac{c_1-d_1}{m_2-m_1},\frac{m_2 c_1-m_1 d_1}{m_2-m_1})$ while the second one is $(\frac{c_2-d_1}{m_2-m_1},\frac{m_2 c_2-m_1 d_1}{m_2-m_1})$. The length of the base is: \begin{equation} b=\frac{|c_1-c_2|\sqrt{m_2²+1}}{|m_2-m_1|} \end{equation}