Question 366438
What is the equation of a line that is perpendicular to the line y = 3x + 9 and containing the point (6, –4)?
..
y=3x+9
slope = 3
slope of line perpendicular to this line will be negative reciprocal
= -1/3
Point(6,-4)
equation of line passing through point (x1,y1) 
=(y-y1)=m(x-x1)
=(y-(-4)=-1/3(x-6)
=y+4 = -x/3+2
y=-1/3*x -4+2
y= -1/3 *x -2
...
The graph will be as below

{{{graph(300,300,-5,5,-5,5,(3x+9),((-x/3)-2))}}}

and 

What is the equation of a line that is parallel to the line y=1/2x+7 equation and containing the point (–4, 7)? 
..
y= 1/2 x +7
slope = 1/2
since line required is parallel slope will be same = 1/2
Point ( -4,7)
..
y-y1=m(x-x1)
y-7=1/2(x+4)
y-7= x/2 +2
y=x/2 +9
...
{{{graph(300,300,-10,2,-2,10,((x/2)+7),((x/2)+9))}}}