Question

a line passes through (3, –2) and (6, 2). a. write an equation for the line in point-slope form. b. rewrite the equation in standard form using integers. a. y minus 2 equals four thirds times the quantity x minus 3 end of quantity; negative 4 x plus 3 y equals 18. b. y plus 2 equals four thirds times the quantity x plus 3 end of quantity; negative 4 x plus 3 y equals negative eighteen. c. y minus 3 equals four thirds times the quantity x plus 2 end of quantity; negative 4 x plus 3 y equals 17. d. y plus 2 equals four thirds times the quantity x minus 3 end of quantity; negative 4 x plus 3 y equals negative eighteen.

Answer 1

Part A)
`y+2=(4)/(3)(x-3)` ----> y plus
`2` equals four thirds times the quantity x minus
`3` end of quantity

Part B)
`-4x+3y=-18` ---> negative
`4x` plus
`3y` equals negative eighteen

The answer is the option D

Step-by-step explanation:

Part A) we know that

The equation of a line into point slope for is equal to

`y-y1=m(x-x1)`

Find the slope

The formula to calculate the slope between two points is equal to

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

we have

`A(3,-2)\ B(6,2)`

Substitute the values

`m=(2+2)/(6-3)`

`m=(4)/(3)`

With the slope m and the point A find the equation of the line

`y+2=(4)/(3)(x-3)` -----> equation of the line into point slope form

Part B) we know that

The equation of the line into standard form is equal to

`Ax+By=C`

we have

`y-2=(4)/(3)(x-6)` ------> convert to standard form

`y=(4)/(3)x-8+2`

`y=(4)/(3)x-6`

Multiply by
`3` both sides

`3y=4x-18`

`-4x+3y=-18` ----> equation of the line in standard form

Answer 2

`(y-y_1)/(x-x_1) = (y_2-y_1)/(x_2-x_1) \\ (y-(-2))/(x-3) = (2-(-2))/(6-3) \\ (y+2)/(x-3) = (2+2)/(6-3) \\ (y+2)/(x-3) = (4)/(3) \\ y+2= (4)/(3) (x-3)`
In standard form

`y+2= (4)/(3) (x-3) \\ 3(y+2)=4(x-3) \\ 3y+6=4x-12 \\ -4x+3y=-18`