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 $$y = -\frac{1}{2}x + \frac{1}{6}$$
$$y - (1) = -\frac{1}{2}(x-5)$$
$$Y = -\frac{1}{2}x - \frac{1}{10} - \frac{1}{10}$$
$$y = -\frac{1}{2}x$$

Notice, in general, the equation of the line passing through the point $(x_1, y_1)$ & having slope $m$ is given by the point-slope form: $$\color{blue}{y-y_1=m(x-x_!)}$$
We have, equation of line A: $3x+6y-1=0\iff \color{blue}{y=-\frac{1}{2}+\frac{1}{6}}$ having slope $-\frac{1}{2}$
1.) slope of the line passing through $(5, 1)$ & perpendicular to the line A $$=\frac{-1}{\text{slope of line A}}=\frac{-1}{-\frac{1}{2}}=2$$ Hence, the equation of the line: $$y-1=2(x-5)$$  $$y-1=2x-10$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{2x-y-9=0}}$$
2.) slope of the line passing through $(5, 1)$ & parallel to the line A $$=\text{slope of line A}=-\frac{1}{2}$$ Hence, the equation of the line: $$y-1=-\frac{1}{2}(x-5)$$  $$2y-2=-x+5$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x+2y-7=0}}$$
