91Ó°ÊÓ

Line $y=3$,

Point $(-1,-3)$.

We have to find the equation of line perpendicular to the line $y=3$and also passing through the point $(-1,-3)$.

Slopes of perpendicular lines are negative reciprocals to each other.

That is $m_2=-\frac{1}{m_1}$, where, $m_1,m_2$are slopes of the two perpendicular lines.

The given line $y=3$is a horizontal line.

So the slope of the line is $0$.

The given line is a horizontal line with slope zero.

A line perpendicular to a horizontal line will be a vertical line.

The slope of a vertical line is not defined.

The general equation of a vertical line is of the form $x=k$.

Where $k$is constant.

The required line contains the point $(-1,-3)$, so the line equation is $x=-1$.

Therefore the slope-intercept form of the line that is perpendicular to $y=3$and contains the point$(-1,-3)$is$x=-1$.