Question

what is the point-slope equation of a line with slope -3 that contains the point (-8,-4)?​

Answer 1

Answer: $(y+4)=-3(x+8)$

Step-by-step explanation:

The equation of the line is point-slope form is:

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

Where m is the slope of the line and ($x_1,y_1$) is a  point of the line.

You know that the slope is -3 and the problem gives you the point (-8,-4), therefore you only need to substitute them into the equation shown above.

Then, you obtain:

$(y-(-4))=-3(x-(-8))$

$(y+4)=-3(x+8)$

Answer 2

Answer:

y+4  = -3(x+8)

Step-by-step explanation:

We have given slope of a line and a point that passes through the line.

slope  = m = -3   and (x₁,y₁)  = (-8,-4)

We have to find the point-slope form of the line.

y-y₁  = m(x-x₁) where m is slope and (x₁,y₁) is a point that passes through the line.

Putting given values in point-slope form ,we have

y-(-4)  = -3(x-(-8))

y+4  = -3(x+8) is point slope form of line having slope equal to -3 and that passes through the point (-8,-4).