SOLUTION: the slope intercept form given (-6,5) and perpendicular to -5x-7y=-17 the standard form of the line parallel to the given line y-3x

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Question 1043745: the slope intercept form given (-6,5) and perpendicular to -5x-7y=-17
the standard form of the line parallel to the given line y-3x
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
-5x-7y=-17
change all signs; it makes it easier.
5x+7y=17
7y=-5x+17
y=(-5x/7)+(17/7)
The slope of the original line is -5/7
the slope of a perpendicular line is the negative reciprocal or +7/5
point slope formula
y-y1=m(x-x1)
the point is (-6,5), and the slope is 7/6
y-5=(7/5)(x+6)
y-5=(7x/5)+(42/5)
add 5 or 25/5 to both sides
y=(7x/5)+67/5

parallel to the line y=3x (edited)
y=3x+4 is one example. There are infinite numbers
Standard form is first -3x+y=4
Then change sign of everything to make x coefficient positive
3x-y=-4