Write the general forms of the equations of the lines that pass through the point and are (a) parallel to the given line and (b) perpendicular to the given line. Point (−1, 0) Line y=-3

(a) we have to write the equation of the line that is parallel to the line $y=-3$ and is passing through the point (-1,0).
as we know that if we have an equation of the line $ax+by+c=0$, then the slope m of the line is given by:
$m=\frac{-coefficientofx}{coefficientofy}=\frac{-a}{b}$
the equation of the line $y=-3$ can be written as:
$0x+y+3=0$
therefore the slope m1 of the line $y=-3$ is:
$m=\frac{-coefficientofx}{coefficientofy}=\frac{-0}{1}$
=0
and we know that if the two lines are parallel then the slopes of the two lines are equal.
therefore the slope m2 of the line parallel to the line $y=-3$ is 0.
now we want to find the equation of the line having slope 0 and passing through the point (-1,0).
as we know that the equation of the line passing through the point (x1,y1) and having slope m is given by:
$y-y_1=m(x-x_1)$
therefore the equation of the line having slope 0 and passing through the point (-1,0) is:
$y-0=0(x-(-1))$
y=0
therefore the slope of the line parallel to the line $y=-3$ and passing through the point (-1,0) is y=0
Step 4
(b)we have to write the equation of the line that is perpendicular to the line y=-3 and is passing through the point (-1,0).
as we know that if the two lines are perpendicular then the product of the slopes of the two lines is -1
therefore,
the slope $m_3$ of the line that is perpendicular to the line $y=-3$ is given by:
slope $m_3\ast slope{m}_1=-1$
$m_3\times 0=-1$
$m_3=\frac{-1}{0}$
now we have to find the equation of the line having slope $\frac{-1}{0}$ and passing through the point (-1,0).
therefore the equation is:
$y-0=\frac{-1}{0(x-(-1))}$
$0=-1(x+1)$
$0=x+1$
$x=-1$
therefore the equation of the line that is perpendicular to the line $y=-3$ and is passing through the point (-1,0) is $x=-1$