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

$\text{The rectangle ABCD will have diagonals AC and BD and passes through A(a, b) and C(a', b').}$

$\text{Thus equation of Ac is :}\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\Rightarrow \frac{y-b}{b^{\prime }-b}=\frac{x-a}{a^{\prime }-a}\Rightarrow (y-b)(a^{\prime }-a)=(x-a)(b^{\prime }-b)\Rightarrow y(a^{\prime }-a)-a^{\prime }b+ab=x(b^{\prime }-b)-a{b}^{\prime }+ab\Rightarrow y(a^{\prime }-a)-x(b^{\prime }-b)-a{b}^{\prime }+a^{\prime }b\Rightarrow y(a^{\prime }-a)-x(b^{\prime }-b)=a^{\prime }b-a{b}^{\prime }\text{BD passes through B(a',b) and D(a,b')}\text{Thus equation of BD is :}\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\Rightarrow \frac{y-b}{b^{\prime }-b}=\frac{x-a^{\prime }}{a-a^{\prime }}\Rightarrow (y-b)(a-a^{\prime })=(x-a^{\prime })(b^{\prime }-b)\Rightarrow -y(a^{\prime }-a)-ab+a^{\prime }b=x(b^{\prime }-b)-a^{\prime }{b}^{\prime }+a^{\prime }b\Rightarrow a^{\prime }{b}^{\prime }-ab=x(b^{\prime }-b)+y(a^{\prime }-a^{\prime })\Rightarrow x(b^{\prime }-b)+y(a^{\prime }-a)=a^{\prime }{b}^{\prime }-ab$