Question

HELP PLEASE!!!!❕❕❕write in standard form an equation of the line

slope= 2/5;(-1,3)

Answer 1

2x-5y = -17

Explanation:

Using point slope form

y-y1 = m(x-x1)

where m is the slope and (x1,y1) is the point on the line

y-3 = 2/5(x- -1)

y-3 = 2/5(x+1)

Multiply each side by 5 to clear the fraction

5(y-3) = 5* 2/5(x+1)

Distribute

5y-15 = 2x+2

Standard form is Ax + By = C where A is a positive integer and B and C are integers

Subtract 5y from each side

-15 = 2x-5y +2

Subtract 2 from each side

-17 = 2x-5y

2x-5y = -17

Answer 2

Answer: 2x - 5y = -17

Step-by-step explanation: Let's use the point-slope formula.

The point-slope formula is written y - y₁ = m(x - x₁).

In this formula, (x₁, y₁) is our point and m is our slope.

Now let's plug all our information into the formula.

So we have y - 3 = 2/5(x + 1).

Notice that I put x + 1 inside the parenthses and this

is because we have x minus a negative 1 which means x + 1.

Our next step would be to distribute this 2/5through both

terms inside the parenthses to get 2/5x + 2/5 on the right side.

So we have y - 3 = 2/5x + 2/5.

Now in standard form, we cannot have any fractions.

We need to get rid of them by multiplying

both sides of the equation by 5.

When we do this, we get 5y - 15 = 2x + 2.

Now we just move our number to the right by

adding 15 to both sides of the equation.

That gives us 5y = 2x + 17.

Move our 2x to the left side by subtracting

2x from both sides to get -2x + 5y = 17.

In standard form, the coefficient of the x term must be positive.

So our last step is to divide both sides

by -1 so that our answer is 2x - 5y = -17.