Question

How do you find the equation of the line that goes through (3, -5) and (5, 4)?

Answer 1

See below

Explanation:

The method is called Two point form ( At least in my school )

The method is

let
3 = `x_1`
-5 = `y_1`
5 = `x_2`
4 = `y_2`

To find an equation passing through these two points,

`(y - y_1) = (y_2 - y_1)/(x_2 - x_1) * (x - x_1)`

Where y and x are all the points on the equation, that has to be found out.

Now plugging in the values,

`(y + 5) = ( 4 + 5)/(5 - 3) * (x - 3)`

`(y + 5) = ( 9)/(2) * (x - 3)`

`(y + 5) 2 = 9(x+3)`

`2y + 10 = 9x + 18`

`9x - 2y + 18 - 10 = 0`

`9x - 2y + 8 = 0`

Answer 2

`y=9/2x-37/2`

Explanation:

Begin by finding the slope via the slope formula: `m=(y_2-y_1)/(x_2-x_1)`

If we let,
`(3,-5)->(color(red)(x_1),color(blue)(y_1)) and (5,4) ->(color(red)(x_2),color(blue)(y_2))`

Then,

`m=color(blue)(4-(-5))/color(red)(5-3)=9/2`

Now that we have the slope, we can find the equation of the line by using the point-slope formula:

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

Where `m` is the slope and `(x_1,y_1)` is a point on the function. We can use any of the two coordinates given. I will use `(5,4)` as my `(x_1,y_1)`

Thus, the equation of the line is...

`y-4=9/2(x-5)larr` Equation in point-slope form

We can rewrite the equation above in `y=mx+b` if desired by solving t=for the variable `y`

`y-4=9/2x-45/2`

`ycancel(-4+4)=9/2x-45/2+4`

`y=9/2x-45/2+4(2/2)`

`y=9/2x-45/2+8/2`

`y=9/2x-37/2larr` Equation in `y=mx+b` form