Question

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Answer 1

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:

$y=\frac{1}{3} x +b$

Now, in order to find b, we must use the given point (6,10). By substituying this point into the last equation, one has

$10=\frac{1}{3} 6+b\\10=\frac{6}{3} +b\\10=2+b\\b=10-2\\b=8$

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

$m=-\frac{1}{\frac{1}{3} } \\m=-3$

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,

$10=-3(6)+b\\10=-18+b\\b=10+18\\b=28$

Hence, the searced equation is y=-3x+28