Question

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

Answer 1

Answer: (D)

equilateral

Answer 2

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.