Can someone please show me how to do this? im confused and thank you :)write an equation of a line that passes through the given point and is parallel to graph of the given equation. (2,-1);y=5x-2
y = mx + b is the form for slope-intercept form m = slope and b = y-intercept the line you are looking for would have to have the same slope as the given equation in order for the two lines to be parallel so m = 5 from y = 5x - 2 then make an equation given that slope = m = 5 and the point (2, -1)
oh , ok I I have this one (0,-5);y=9x so the slope would be 9?
yep :)
but how would I find the line that passes through? that's what its asking and idk what it is :/
I'm not quite sure what you are asking, may I ask what the full question was?
yeah. write an equation of a line that passes through the given point and is parallel to graph of the given equation. this is what its asking and it give me equations
okay, so for given the point (0,-5); and the equation y=9x a line parallel to y = 9x would have to have a slope of 9 then y = 9x + b and temporarily plug in the points (0, -5) into y = 9x + b to find b. so (-5) = 9(0) + b then b = -5 so the equation that they want would be y = 9x - 5
Then how would I be able to do this one (2,-1);y=5x-2
would -2 replace B?
from y=5x-2, the slope would be 5 then with y = 5x + b you temporarily plug in the points (2, -1) into y = 5x + b to find b so (-1) = 5(2) + b but b would Not be -2
so the equation would be -1=10+b? sorry if i'm askig too much question
you would solve for b from -1 = 10 + b then plug in what you got from b back into the y = 5x + b
oh ok thank you so much. May I ask you on more quesio?
sure :)
ok, so this one asks us graph the equation but the equations are like y+3=2(x-1)
can show me example so I can do the rest by myself :)
ah yes, okay, so that would be in point-slope form: y - y1 = m(x - x1) when m = slope and given point on line (x1, y1) so for an equation in point-slope form like y + 3 = 2(x - 1) you would graph it knowing that the slope = 2 and a point is (1, -3)
cool thank you again.
you're welcome :)
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