Question

The slope of a line is $\sqrt{3}$. Find the equation of a line perpendicular to this line having $y$-intercept as $-\sqrt{2}.$

• A. $y=\frac{-x}{\sqrt{3}}-\sqrt{2}$
• B. $y=\frac{-x}{\sqrt{3}}-\sqrt{2}$
• C. $y=\frac{-x}{\sqrt{3}}+\sqrt{2}$
• D. $y=-x\times \sqrt{3}-\sqrt{2}$
• E. $y=-x\times \sqrt{3}-\sqrt{2}$

Answer 1

The correct option is A $y=\frac{-x}{\sqrt{3}}-\sqrt{2}$

Let the slope of the line be $m_1$ and the slope of perpendicular line be $m_2.$
Given, $m_1=\sqrt{3}$
Concept: If two line are perpendicular to each other, then product of the slope of both line is $-1.$
$m_1\times m_2=-1$
$m_2=\frac{-1}{m_1}$

The slope of one line $(m_1=\sqrt{3})$ is given
$So,m_2=\frac{-1}{m_1}=\frac{-1}{\sqrt{3}}$

Hence, $m_1=\sqrt{3}$
$m_2=-\frac{1}{\sqrt{3}}$

Now, find the equation of a line
Given, the intercept of the line is $-\sqrt{2}.$

Let the slope intercept form of equation of a line.
$y=mx+c$
Where ${}^{\prime }{m}^{\prime }$ is the slope of the line
${}^{\prime }{c}^{\prime }$ is the $y-$intercept of the line.

Given, $c=-\sqrt{2}$
$m=m_2=\frac{-1}{\sqrt{3}}$

Hence, Equation of the line: $y=mx+cy=\frac{-1}{\sqrt{3}}\times x+(-\sqrt{2})$
$y=\frac{-x}{\sqrt{3}}-\sqrt{2}$
Option (a) is correct.