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

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 the above

Correct Answer: B

Solution :

Slop 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                 \[\tan {{45}^{o}}=\pm \frac{m-\frac{8}{15}}{1+\frac{8}{15}m}\] \[\Rightarrow \]               \[(15+8m)=\pm (15m-8)\] \[\Rightarrow \]               \[m=\frac{23}{7}\]and\[-\frac{7}{23}\] Hence, the equation of\[DC\]and\[AD\]are                 \[y-2=\frac{23}{7}(x-1)\] \[\Rightarrow \]               \[23x-7y-9=0\] and        \[y-2=-\frac{7}{23}(x-1)\] \[\Rightarrow \]               \[7x+23y-53=0\]