Straight-Line Equations: Slope-Intercept Form

http://www.purplemath.com/modules/strtlneq.htm

• Find the equation of the line that passes through the points (–2, 4) and (1, 2).

Well, if I have two points on a straight line, I can always find the slope; that's what the slope formula is for.

The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x₁, y₁) and a slope m, and have you plug it into this formula:

y – y₁ = m(x – x₁)

• Find the equation of the line that passes through the points (–2, 4) and (1, 2).

I've already answered this one, but let's look at the process. I should get the same result (namely,  y = ( – ²/₃ ) x + ⁸/₃ ). Given two points, I can always find the slope:
Then I can use either point as my (x₁, y₁), along with this slope Ive just calculated, and plug in to the point-slope form. Using (–2, 4) as the (x₁, y₁), I get:

y – y₁ = m(x – x₁) y – (4) = ( – ²/₃ )(x – (–2)) y – 4 = ( – ²/₃ )(x + 2) y – 4 = ( – ²/₃ ) x – ⁴/₃y = ( – ²/₃ ) x – ⁴/₃ + 4 y = ( – ²/₃ ) x – ⁴/₃ + ¹²/₃ y = ( – ²/₃ ) x + ⁸/₃

y=(y2-y1/x2-x1)*X+(x2y1-x1y2/x2-x1)