Given a line written in the slope/intercept form:
[tex]y=mx+q[/tex]
The slope is the coefficient [tex]m[/tex]
If a line is parallel to the given line, then it has the same slope
If a line is perpendicular to the given slope, then their slope give -1 when multiplied.
In the first case, our line has slope 3. We want a parallel line (so it must have slope 3 as well). When we impose the passing through (-10,2.5), we have
[tex]y-2.5 = 3(x+10) \iff y = 3x+30+2.5 = 3x+32.5[/tex]
The second exercise is similar: the given slope is -4, so the perpendicular slope is 1/4.
Use again the formula [tex]y-y_0=m(x-x_0)[/tex] to get the desired line:
[tex]y+11 = \dfrac{1}{4}(x+16) \iff y = \dfrac{1}{4}x+4-11 =\dfrac{1}{4}x-7[/tex]
