question_answer

The pair of straight lines \[{{x}^{2}}-3{{y}^{2}}=0\]and the line \[x=1\]form a triangle which is

A)  right angled     

B)  isosceles

C)  scalene          

D)  equilateral

Correct Answer: D

Solution :

Given, lines are \[{{x}^{2}}-3{{y}^{2}}=0\] and \[x=1\] \[\Rightarrow \] \[(x-\sqrt{3}y)=0,\,(x+\sqrt{3}\,y)=0\] and \[x=1\] Here, \[AB=\sqrt{{{(0-1)}^{2}}+{{\left( 0-\frac{1}{\sqrt{3}} \right)}^{2}}}\] \[=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}\] \[BC=\sqrt{{{(1-1)}^{2}}+{{\left( \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}} \right)}^{2}}}\] \[=\sqrt{{{\left( \frac{2}{\sqrt{3}} \right)}^{2}}}=\frac{2}{\sqrt{3}}\] \[CA=\sqrt{{{(1-0)}^{2}}+{{\left( -\frac{1}{\sqrt{3}}-0 \right)}^{2}}}\] \[=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}\] Hence, all the three sides of \[\Delta \,ABC\] are equal. Therefore, the triangle is equilateral.