Question

What is the function of the line that passes through the points `(-8.3, -5.2)` and `(6.4, 9.5)`?

Answer 1

`y=mx+c" "->" "y=x+3.1`

Solution provided in a lot of detail taking you through it 1 step at a time.

Explanation:

Set point 1 as `P_1->(x_1,y_1) = (-8.3,-5.2)`
Set point 1 as `P_2->(x_2,y_2)=(6.4,9.5)`

Consider the standard straight line equation form of `y=mx+c` where `m` is the gradient.

Gradient (slope) is the change in the up or down for the change in along reading left to right. So we are traveling from `P_1" to "P_2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the gradient (slope)")`

Change in up or down:

change in `y -> y_2-y_1 = 9.5-(-5.2) = 14.7`

Change in along:

change in `x->x_2-x_1= 6.4-(-8.3) = 14.7`

So `("change in up or down")/("change in along")->color(red)(m=14.7/14.7 = 1)`

so `color(green)(y=color(red)(m)x+c" "->" "y=color(red)(1)x+c)`

It is bad practice to show the 1 so we write:

`y=x+c`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the value of the constant c")`

Picking any point. I chose `P_2 ->(x_2,y_2)=(6.4,9.5)`

So by substitution:

`y=x+c" "->" "9.5=6.4+c`

Subtract `6.4` from both sides

`9.5-6.4" "=" "6.4-6.4+c`

`3.1=0+c`

`c=3.1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

So our equation becomes:

`y=mx+c" "->" "y=x+3.1`

Answer 2

Showing you trick

Explanation:

Lets make determining the gradient easier:

I do not like decimals so lets get rid of them.

Multiply everything by 10.

Changing the scale should not change the slope

`(-8.3,-5.2) ->(-83,-52)`

`(6.4,9.5)->(64,95)`

so the gradient `m=(95-(-52))/(64-(-83)) = 147/147 = 1`as in the other solution