write an equation of the line containing the given point and parallel to the given line. (5,-6); 3x-8y=5 if some one could please explain to me how to do this. In my mind i would turn the equation into y=mx+b form and get the slope and then go from there to get the slope intercept for the other line is this correct? y=(-3/8)x+(-5/8) then from there i would have -6=(-3/8)*5+b y=(-3/8)x+(33/8) is this correct?
yes this is correct another useful general form of equation of straight line is y-y1 = m(x - x1) where m = slope and (x1,y1) is a point on the line
for some reason my book says that the answer was 63/8 not 33/8 and I hate that form it is confusing to me.
now I have the equation (4,-6); 8x-5y=7
oops sorry that was my daughter
ok - lets do it again using the other equation 3x - 8y = 5 8y = -3x - 5 y = (-3/8)x - 5/8 plug in (5,-6) to general form y + 6 = (-3/8) (x - 5) y = (-3/8) x +15/8 - 48/8 y = (-3/8) x - 33/8 the book's wrong!
I agree lol but it don't think so its an ebook and just tells me that is the answer not how it got it gotta love electronics
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