Answer:
case a) y=1/3x+8
case b) y=-3x+28
Step-by-step explanation:
case a)
parallel lines has the same slope, then, the searched line has the form:
[tex]y=\frac{1}{3} x +b[/tex]
Now, in order to find b, we must use the given point (6,10). By substituying this point into the last equation, one has
[tex]10=\frac{1}{3} 6+b\\10=\frac{6}{3} +b\\10=2+b\\b=10-2\\b=8[/tex]
Then, the line equation is y=1/3x+8
case b)
the slope of a perpendicular line is the "negative reciprocal" of the slope of the original line, that is
[tex]m=-\frac{1}{\frac{1}{3} } \\m=-3[/tex]
Hence, the searched line has the form y=-3x + b. Now, in order to find the y-intercept b, we must use the given point (6,10). Therefore,
[tex]10=-3(6)+b\\10=-18+b\\b=10+18\\b=28[/tex]
Hence, the searced equation is y=-3x+28
