Math Is Fun Forum

linlum

Member

Registered: 2009-03-22

Posts: 1

Point Gradient Form of a Straight Line

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.

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Point Gradient Form of a Straight Line

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.

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,045

Re: Point Gradient Form of a Straight Line

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

where m is the slope or gradient, x[sub]1[/sub] and y[sub]1[/sub] are the x and y coordinates of the given point.
Hence,

Hence, the equation of the staright line is

.

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