Question

PLLSS HELP THANK YOU:) What is the equation of the line in standard form? A function graph of a line with two points (-3,2) and (2,-1) with an x axis of negative five to five and a y axis of negative five to five

A.) x + 3y = 5
B.0 3x + 5y = 1
C.) 5x + 3y = 1
D.) x + 5y = 3?

Answer 1

B. 3x + 5y = 1

Explanation:

Since, the equation of a line passing through
`(x_1, y_1)` and
`(x_2, y_2)` is,

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

Thus, the equation of the line passes through the points (-3,2) and (2,-1) is,

`y - 2 =(-1-2)/(2+3)(x+3)`

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

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

`5y - 10 = -3x - 9`

`3x + 5y = -9 + 10`

`3x + 5y = 1`

∵ Standard form of a line is ax + by = c, where, a, b and c are any constants,

Hence, the required equation is,

3x + 5y = 1

i.e. OPTION B is correct.

Answer 2

A function graph of a line with two points (-3,2) and (2,-1)

Now we have to find the equation of line passing thorough these points

So let's find slope first using formula:

`m=(y_2-y_1)/(x_2-x_1)=(2-\left(-1\right))/(-3-2)=(2+1)/(-5)=-(3)/(5)`

Now plug the value of slope m and any point say (2,-1) into point slope formula:

`y-y_1=m\left(x-x_1\right)`

`y-\left(-1\right)=-(3)/(5)\left(x-2\right)`

Now we simplify it to get equation of the standard form Ax+By=C

`y+1=-(3)/(5)\left(x-2\right)`

`5\left(y+1\right)=-3\left(x-2\right)`

`5y+5=-3x+6`

`3x+5y+5=+6`

3x+5y=+6-5

3x+5y=1

Hence final answer is 3x+5y=1.