We have the equation of the line;

We will rewrite this equation as;

Inspecting this equation, we can see that the slope of this line is;

We can now find the equations of the lines needed.
1. A line parallel to the given line and passing through (1,4)
We know that parallel lines have equal slopes, so the slope of this line is still -1/2, lets use the equation of a line fiven slope and a point to obtain the line.

2. A line perpendicular to the given line and passing through (1,4)
The product of the slopes of perpendicular lines is -1, Therefore,the slope of this very line is;

lets use the equation of a line fiven slope and a point to obtain the line.
