- #1

- 3

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mogilem
- Start date

- #1

- 3

- 0

- #2

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 4,429

- 513

[tex] \frac{ y_2-y_1}{x_2-x_1} = m [/tex]

where m is the slope of the line. In particular if I tell you the slope of the line and a point on it, let's say the slope is 3 and (1,5) is a point on the line, then if (x,y) is on the line it must satisfy

[tex] \frac{y-5}{x-1} = 3 [/tex]

where 3 is the m above, (1,5) is the [itex] (x_1,y_1)[/itex] above and (x,y) is the [itex] (x_2,y_2)[/itex] above. This equation is the point slope form of the line.

- #3

Mark44

Mentor

- 34,896

- 6,637

The slope-intercept form uses the slope and y-intercept.

From (y - 5)/(x - 1) = 3, multiply both sides by x - 1 to get y - 5 = 3(x - 1) = 3x - 3. Add 5 to both sides to get y = 3x - 3 + 5, or

y = 3x + 2

Here the slope is 3 (as before) and the y-intercept is 2, which means that the line goes through (0, 2).

- #4

- 3

- 0

- #5

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 4,429

- 513

What Office_Shredder described is actually thepoint-slopeform of the equation of a line

OK, I obviously need to get more sleep

- #6

- 3

- 0

Share: