Give the equation of a line that passes through the point (5,1) that is perpendicular and parallel to line A. The equation of line $A$ is $3x + 6y - 1 = 0$. Give the equation of a line that passes through the point $(5,1)$ that is


*

*Perpendicular to line $A$.

*Parallel to line $A$.
Attempting to find the parallel,
I tried $$y = -\frac{1}{2}x + \frac{1}{6}$$
$$y - (1) = -\frac{1}{2}(x-5)$$
$$Y = -\frac{1}{2}x - \frac{1}{10} - \frac{1}{10}$$
$$y = -\frac{1}{2}x$$

For the second case:
the line you want to find its equation is parallel to $A$, so it means they have the same slope$-\frac{1}{2}$, which yields:
$$
{y}_{1}  = -\frac{1}{2}x+p
$$
Now you'll find $p$ from another condition you gave(the line passes from $(5,1)$), so we get :
$$
1 = \frac{-5}{2}+p\\
p = 1+\frac{5}{2} = \frac{7}{2}
$$
We finally get:
$$
{y}_{1}  = -\frac{1}{2}x+\frac{7}{2}
$$
For the first question you should know that if $m$ is the slope of line $A$ and $m'$ is the slopeof line $B$, and $A$ and $B$ are perpendicular then we have:
$$
mm' = -1
$$
For our case we have $m = 1/2$, so $m' = 2$ and we get:
$$
{y}_{2} = 2x+p'
$$
Can you continue from here?
