determining equations of straight lines?

Question

1. (2,0) (0.1/3)

2. (5,0) (0,2/5)

3. (0,-1/2) (0, 8)

Answer 1

Y = -x/6+(1/3)
Y = -2x/25+(2/5)
x = 0 (the equation of the Y axis)

Answer 2

y = mx + b

Where x and y are any points on the line. m is the slope. Think of the slope as how steep or straight the line is and b is shift.

so the slope is basically change in y divided by change of x.

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

I will do the first problem you do the rest

1. m = 0 - 1/3 / 2 - 0 which gives me - 1/6
y = mx + b
0 = -1/6 * 2 + b
b = -1/3
So the equation for this line would be
y = (-1/6)*x + (-1/3) or (-1/6)*x - (1/3)

Answer 3

2. Find the equation of the line thru the points (5,0) and (0,2/5).

First calculate the slope m.

m = ∆y/∆x = (2/5 - 0)/(0 - 5) = (2/5)/(-5) = -2/25

Now plug one of the points into the equation of the line.

(y - 2/5) = (-2/25)(x - 0)
y - 2/5 = (-2/25)x
y = (-2/25)x + 2/5

The other two questions work the same way.

Answer 4

1)Slope formula:
m=(y2-y1)/(x2-x1)
m=(1/3-0)/(0-2)
m=1/3/-2
m=-1/6

Point-slope formula:
(y-y1)=m(x-x1)
y-0=-1/6(x-2)
y=-1/6x+1/3

Check:
0=-1/6(2)+1/3
0=-2/6+1/3
0=-1/3+1/3
0=0

For the other two problems, just apply this example to them.