Question

What is the equation of a line that is parallel to −x+3y=6 and passes through the point (3, 5) ?



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Answer 1

First find the slope of the given line -x+3y=6.

3y=x+6

y=(1/3)x+2

Hence slope is 1/3.

Now the equation of a line whose slope is m and passing through point (x1, y1) is y-y1=m(x-x1)

y-5=1/3(x-3)

3y-15=x-3

x-3y+12=0

Answer 2

slope will will 1/3 and the equation will
y-5=1/3 (x-3) ... 3y-x=12