The triangle formed by \[{{x}^{2}}-9{{y}^{2}}=0\]and \[x=4\]is [Orissa JEE 2004]

Given lines are \[{{x}^{2}}-9{{y}^{2}}=0\] and \[x=4\].            We have \[{{x}^{2}}-9{{y}^{2}}=0\]            So equation of line is,                    \[x-3y=0\]                           .....(i)                    \[x+3y=0\]                         .....(ii)                            \[x=4\]                         .....(iii)            By solving (i), (ii) and (iii) we GET                      \[A(0,\,0),\,\,B\,\left( 4,\,\FRAC{-4}{3} \RIGHT),\,\,C\,\left( 4,\,\frac{4}{3} \right)\]            Now,           \[AB=\sqrt{{{(4-0)}^{2}}+{{\left( 0+\frac{4}{3} \right)}^{2}}}=\frac{4\sqrt{10}}{3}\]                     \[AC=\sqrt{{{(4-0)}^{2}}+{{\left( 0-\frac{4}{3} \right)}^{2}}}=\frac{4\sqrt{10}}{3}\]                    \[BC=\sqrt{{{(4-4)}^{2}}+{{\left( \frac{4}{3}+\frac{4}{3} \right)}^{2}}}=\frac{8}{3}\]            Hence ABC is an isosceles TRIANGLE.