y2 – y1 -2 – -3 1

——— = ———- = ——

x2- x1 5 – 10 – 5

Now you have y = (1/-5) x +b

Substitute points in for x & y.

-2 = (1/-5) 5 +b

-2 = -1 +b

2 = b

The final answer is y= 1/-5 x + 2

Y2 – Y1 / X2 – X1 = Slope

(-2 – -3) / (5 – 10) = Slope

Slope = -1/5

You have two points on the line, so plug either one of those cordinates (sp?) into the equation along with the slope. I’m going to use (5,-2) because they’re smaller numbers. (Still works if you use (10,-3))

y=mx+b

-2 = (-1/5) (5) +b

-2 =- 1 +b

b = -1

So the equation is y= (-1/5)x – 1

Your points are (10,-3) and (5,-2).

The slope (m) is found by this formula m = (Y2-Y1) / (X2-X1)

where y2 and y1 are just the y’s in your points (x,y) and x2,x1 are the x’s in those points.

So do the math by installing your two points (10, -3) and (5,-2).

[(-3) – ( -2)] / [10- 5] = (-1) / 5

therefore m = -1/5. the slope is – 1/5

slope intercept form is: y = mx + b

now plug in one point (use (5, -2)) and your slope to get:

-2=(-1/5) 5 + b

Solve for b: (-2) = (-1) + b

(-1) = b

Now the answer in slope-intercept form

y=(-1/5) x + (-1) or simply y = (-1/5) x -1

There you have it!

What you do is y sub 1 minus y sub 2 divided by x sub 1 minus x sub 2

-3–2(-3+2)

10-5=

-1 over 5 (1/5) thats you slope. and then for intercept y=1/5x + b…hmm, I cant remeber how to get b…which is the intercept.

let P1 = (10, -3) and P2 = (5, -2)

P1 = (x1 =10, y1 = -3) and

P2 = (x2 =5 , y2 = -2)

1) Find the slope m

m = (y2 -y1)/(x2 -x1)

m = (-2 -(-3) ) /(5 -10)

m = (-2 +3)/(-5) = 1/(-5) = -1/5

the equation of the line passing through P1 and P2 is

y – y1 = m (x -x1)

y1 = -3 , x1 =10

y -(-3) = (-1/5) (x -10)

y +3 = -x/5 +2

y = -x/5 +2 -3 = -x/5 -1

-2–3/5-10=1/-5=-1/5

and then you set it upp….

y–3=1/5(x-10)

y+3=1/5x-2

y=1/5x+1 is your answer…(: