Question

Question #26d13

Answer 1

I'm going to say `y = (-5/2)x + 8`

Explanation:

The general form for the point-slope version of the equation of a line is:

`y=mx+b`

Here you are given the slope:

`-(5/2)`,

and the point of intersection with the y-axis, which is:

`(0,8)`

Let's calculate some values for `y = (-5/2)x + 8`, plug them in and check:

x | y

---

0 | 8
3 | 1/2
4 | -2

If I graph these points, I get:

graph{y=(-5/2)*x+8 [-20.75, 19.25, -8.08, 11.92]}

This is the graph the software gives me. It's a little confusing because 10 units on the graph are divided into only 4 parts instead of five.

Click on the graph to see the points of intersection with the x- and y-axes.

Answer 2

`color(red)(y-8=-5/2(x+8) larr" Point slope form")`

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

Explanation:

Slop (gradient) is given as `-5/2`

Given point `P_1->(x_1,y_1)=(-8,8)`

Gradient `->m=("change in y")/("change in x")`

So picking any hypothetical point `P_2-(x_2,y_2)`

We have:

`m=-5/2=("change in y")/("change in x") ->(y_2-y_1)/(x_2-x_1) = (y_2-8)/(x_2-(-8))`

We do not know the values of `y_2 and x_2` so write this as just:

`m=-5/2=(y-8)/(x+8)`

Cross multiply

`-5/2(x+8)=y-8`

Turn it round the other way:

`color(red)(y-8=-5/2(x+8) larr" Point slope form")`

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

Answer 3

`y-8=-5/2(x+8)`

Explanation:

`"the equation of a line in "color(blue)"point-slope form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))`

`"where m is the slope and "(x_1,y_1)" a point on the line"`

`"here "m=-5/2" and "(x_1,y_1)=(-8,8)`

`rArry-8=-5/2(x+8)larrcolor(red)"in point-slope form"`