question_answer

The opposite angular points of a square are \[(3,\ 4)\] and \[(1,\ -\ 1)\]. Then the co-ordinates of other two points are  [Roorkee 1985]

A)            \[D\,\left( \frac{1}{2},\,\,\frac{9}{2} \right)\,,\,\,B\,\left( -\frac{1}{2},\,\,\frac{5}{2} \right)\]              

B)            \[D\,\left( \frac{1}{2},\,\,\frac{9}{2} \right)\,,\,\,B\,\left( \frac{1}{2},\,\,\frac{5}{2} \right)\]

C)            \[D\,\left( \frac{9}{2},\,\,\frac{1}{2} \right)\,,\,\,B\,\left( -\frac{1}{2},\,\,\frac{5}{2} \right)\]              

D)            None of these

Correct Answer: C

Solution :

           Obviously, slope of \[AC=5/2\].                    Let m be the slope of a line inclined at an angle of \[{{45}^{o}}\]to AC, then  \[\tan {{45}^{o}}=\pm \frac{m-\frac{5}{2}}{1+m.\frac{5}{2}}\Rightarrow m=-\frac{7}{3},\frac{3}{7}\].            Thus, let the slope of AB or DC be 3/7and that of AD or BC be \[-\frac{7}{3}\] . Then equation of AB is \[3x-7y+19=0\].                    Also the equation of BC is \[7x+3y-4=0\].                    On solving these equations, we get, \[B\,\,\,\left( -\frac{1}{2},\frac{5}{2} \right)\].                    Now let the coordinates of the vertex D  be (h, k). Since the middle points of AC and BD are same, therefore \[\frac{1}{2}\left( h-\frac{1}{2} \right)\,=\frac{1}{2}(3+1)\Rightarrow h=\frac{9}{2}\], \[\frac{1}{2}\left( k+\frac{5}{2} \right)=\frac{1}{2}(4-1)\]                    Þ \[k=\frac{1}{2}\]. Hence, \[D=\left( \frac{9}{2},\,\frac{1}{2} \right)\].