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Hi,
I have a question that I got from my maths teacher that pretty much says this:
Find the equation of the line with the gradient 2/3 (two thirds), passing through (1 2/3, -6) (one and two-thirds minus six).
All I know is I am supposed to use the point-gradient formula, but how is it actually done? As there are fractions when I substitute the numbers accordingly, I end up with 3y + 15 = 2x.
I get the feeling I've got the wrong answer...can someone please explain how I've gone wrong, and provide an solutions with an explanation on how to do it properly?
Thank you.
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All lines have the form y = mx + b where m is the gradient and b is the y-intercept. You have the m (gradient) and you have a point (x,y). Substituting yields you the b and hence the equation of the line.
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The given point is (1 2/3, -6), that is (5/3, -6).
The equation of the straight line through a given point is given by the equation
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