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, \[D=\left( \frac{9}{2},\,\frac{1}{2} \right)\].You need to login to perform this action.
You will be redirected in
3 sec