Question

Find the equations to the diagonals of the rectangle the equations of whose sides are x = a, x = a', y= b and y = b'.

Answer 1

The rectangles formed by the lines x = a, x = a', y = b and y = b' is shown below:

Clearly, the vertices of the rectangle are

\[A \left( a, b \right), B \left( a^{\prime},b\right), C \left( a^{\prime} , b^{\prime} \right) \text { and } D \left( a, b^{\prime} \right)\].

The diagonal passing through

\[A \left( a, b \right) \text { and} C \left( a^{\prime} , b^{\prime}\right)\] is 

\[y - b = \frac{b^{\prime} - b}{a^{\prime} - a}\left( x - a \right)\]

\[ \Rightarrow \left( a^{\prime} - a \right)y - b\left( a^{\prime}- a \right) = \left( b^{\prime}- b \right)x - a\left( b^{\prime} - b \right)\]

\[ \Rightarrow \left( a^{\prime} - a \right)y - \left( b^{\prime} - b \right)x = - a\left( b^{\prime} - b \right) + b\left( a^{\prime} - a \right)\]

\[ \Rightarrow \left( a^{\prime}- a \right)y - \left( b^{\prime} - b \right)x = b a^{\prime} - a b^{\prime}\]

And, the diagonal passing through

\[B \left( a^{\prime} , b \right) \text { and } D \left( a, b^{\prime} \right)\] is 

\[y - b = \frac{b^{\prime} - b}{a - a^{\prime}}\left( x - a^{\prime} \right)\]

\[ \Rightarrow \left( a - a^{\prime} \right)y - b\left( a - a^{\prime} \right) = \left( b^{\prime} - b \right)x - a^{\prime} \left( b^{\prime}- b \right)\]

\[ \Rightarrow \left( a - a^{\prime} \right)y - \left( b^{\prime} - b \right)x = - a^{\prime}\left( b^{\prime} - b \right) + b\left( a - a^{\prime} \right)\]

\[ \Rightarrow \left( a^{\prime} - a \right)y + \left( b^{\prime}- b \right)x = a^{\prime} b^{\prime} - ab\]

Hence, the equations of the diagonals are

\[\left( a^{\prime} - a \right)y - \left( b^{\prime}- b \right)x = b a^{\prime} - a b^{\prime}\] and 

\[\left( a{\prime} - a \right)y + \left( b{\prime} - b \right)x = a{\prime}b{\prime} - ab\].