Can you explain why an equation passing through the origin have variable constant? and its implication? if a line passes through the origin you cannot represent it by y=mx+b?


[quote="Anurag@Gurome"][quote="yellowho"]The points (a, -6) and (-3, b) lie on line m, which has a slope of 3 and passes through the origin. What is the value of a + b?


I'm doing something wrong here. Can someone comment.

y=3x+b. since it passes thru the origin => 0=3(0)+b thus B=0

-6=3a+0 => a= -2. a+b=-2+0 = -2. That's not the right answer. Also, I didn't get to use the info from the second point. What I might be doing wrong here is misusing the info. about the origin.[/quote]

Equation of a line which passes through the origin and has slope 3 is y = 3x.
Note that the equation will not have a constant term since it is passing through the origin.
So -6 = 3a and b = -9.
Or a = -2 and b = -9.
So a+b = -11.[/quote]

aaah i get it. i screwed up. defining the line y=mx+b was the downfall here because i re-used "b" as if it's the same as the constant.

[quote="Anurag@Gurome"]Let us assume that the equation of the line is y = mx+c, where m and c are constants.
Since the line is passing through (0, 0), y = 0 and x =0 should satisfy the equation y = mx+c.
Or 0 = m*0+c
Or 0 = 0 + c.
Or c = 0.
Hence, ultimately we will get the equation as y = mx.[/quote]