Question

5: use the points (20,10) and (70,110) and the slope formula to generate the point-slope equation of the trend line

6: apply the properties of equality to rewrite the equation you found in question 5 in slope-intercept and standard form

Answer 1

(5) use the points (20,10) and (70,110)

slope =
`(y_2-y_1)/(x_2-x_1)`

(20,10) is (x1,y1) and (70,110) is (x2,y2)

slope =
`(110-10)/(70-20)=2`

so slope = 2

the point-slope equation of the trend line is

`y - y_1 = m(x - x_1)`

m is the slope. m=2, x1= 20 and y1= 10

`y - 10 = 2(x - 20)`

The point slope equation is y - 10 = 2(x - 20)

(6) rewrite the equation you found in question 5 in slope-intercept and standard form

y - 10 = 2(x - 20)

To find slope intercept form we need to get y alone,

distribute 2 inside the parenthesis

y - 10 = 2x - 40

Add 10 on both sides

Slope intercept form is y = 2 x - 30

Standard form is Ax + By = C

y = 2x - 30

Subtract 2x from both sides

-2 x + y = -30