question_answer

One diagonal of a square is along the line \[8x-15y=0\] and one of its vertex is (1, 2). Then the equation of the sides of the square passing through this vertex, are                       [IIT 1962]

A)            \[23x+7y=9,\ 7x+23y=53\]

B)            \[23x-7y+9=0,\ 7x+23y+53=0\]

C)            \[23x-7y-9=0,\ 7x+23y-53=0\]

D)            None of these                          

Correct Answer: C

Solution :

           Slope of \[BD\] is \[\frac{8}{15}\] and angle made by \[BD\]with AD and DC is \[{{45}^{o}}\]. So let slope of DC be m, then \[\tan {{45}^{o}}=\pm \frac{m-\frac{8}{15}}{1+\frac{8}{15}m}\] \[\Rightarrow (15+8m)=\pm (15m-8)\]                    Þ \[m=\frac{23}{7}\]and \[-\frac{7}{23}\]                    Hence the equations of DC and AD are                           \[y-2=\frac{23}{7}(x-1)\]\[\Rightarrow 23x-7y-9=0\]                    and \[y-2=-\frac{7}{23}(x-1)\]\[\Rightarrow 7x+23y-53=0\].