Answer:
[tex]y= 3x+17; y= -\frac13x +\frac{11}3[/tex]
Step-by-step explanation:
First problem. If you want a parallel to a given line, you keep the slope.
Then we use the point-slope form of a line
[tex]y-y_0=m(x-x_0)[/tex] and we plug in there everything we need.
[tex]y-5=3(x+4) \rightarrow y=3x+12+5\\y=3x+17[/tex]
The second is quite similar. This time we want the perpendicular. It means that the product of the slopes has to be -1.
[tex]3\cdot m = -1 \rightarrow m=-\frac13[/tex]
At this point we have everything, let's replace and write down the line in a better looking form
[tex]y-5=-\frac13(x+4) \rightarrow y= -\frac13 x -\frac43 +5\\y= -\frac13x -\frac43 +\frac{15}3 \rightarrow y= -\frac13x +\frac{11}3[/tex]
