Question

The triangle formed by the lines $x+\sqrt{3}y-2=0,\sqrt{3}x-y+1=0,y=0$ is

• A. Equilateral
• B. Right angled
• C. Isosceles
• D. None of these

Answer 1

Answer: (B)

Right angled

Given:

$L_1:x+\sqrt{3}y-2=0$ $[m=-\frac{1}{\sqrt{3}}]$

$L_2:\sqrt{3}x-y+1=0$ $[m_2=-\frac{\sqrt{3}}{-1}]$

$L_3:y=0$

Since product of slope of line $1$ and line $2$ is

$=\sqrt{3}\times -\frac{1}{\sqrt{3}}=-1$

Hence $L_1$ and $L_2$ are perpendicular.

So triangle formed by these lines are right angled triangle.