A) Isosceles
B) Equilateral
C) Right angled
D) None of these
Correct Answer: A
Solution :
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.You need to login to perform this action.
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