Question

The slope of a line is $\sqrt{3}.$ What is the equation of the line perpendicular to this line that has its $y$-intercept as $–\sqrt{2}$?

Answer 1

Let the slope of the first line be $m_1$ and the slope of the line perpendicular to it be $m_2.$

$\text{Given:}{m}_1=\sqrt{3}$


If two lines are perpendicular to each other, then the product of their slopes is –1.

$⟹m_1\times m_2=–1$
$⟹m_2=\frac{–1}{m_1}=\frac{–1}{\sqrt{3}}$

We need to find the equation of the second line that has slope $–\frac{1}{\sqrt{3}}$ and $y$-intercept $–\sqrt{2}.$

We use the slope-intercept form of the equation for this.

$y=mx+c,$
where $m$ is the slope and $c$ is the $y$-intercept of the line.

$c=–\sqrt{2}\text{and}m=m_2=\frac{–1}{\sqrt{3}}$

Hence, the equation of the line will be:

$y=\frac{–1}{\sqrt{3}}\times x+(–\sqrt{2})$

$y=\frac{–x}{\sqrt{3}}–\sqrt{2}$