Step 
<u>Find the slope of the given line</u>
Let

slope mAB is equal to

Step 
<u>Find the slope of the line that is perpendicular to the given line</u>
Let
CD ------> the line that is perpendicular to the given line
we know that
If two lines are perpendicular, then the product of their slopes is equal to 
so

Step 
<u>Find the equation of the line with mCD and the point (3,0)</u>
we know that
the equation of the line in the form point-slope is equal to

Multiply by
both sides


therefore
the answer is
the equation of the line that is perpendicular to the given line is the equation 

Here's the solution ~
The given points are : (-7 , -4) and (1 , 1)
now, let's use distance formula to find the distance between them !





Answer: 0.6
Reason: 3 divide by 5 is 0.6
For a line to be parallel it has to have the same slope as the original line. so the slope is 1/2