question_answer

The ends of the base of an isosceles triangle are at \[(2a,\ 0)\]and\[(0,\ a).\] The equation of one side is \[(lx+my)(a+b)=(l+m)\ ab\] The equation of the other side is

A)            \[x+2y-a=0\]                           

B)            \[x+2y=2a\]

C)            \[3x+4y-4a=0\]                      

D)            \[3x-4y+4a=0\]

Correct Answer: D

Solution :

           Obviously, other line AB will pass through (0, a) and \[(2a,k)\].                    But as we are given \[AB=AC\]                    \[\Rightarrow k=\sqrt{4{{a}^{2}}+{{(k-a)}^{2}}}\]Þ \[k=\frac{5a}{2}\]                    Hence the required equation is \[3x-4y+4a=0\].