How do I solve this Algebra 1 Question?

y - y1 = m (x - x1)

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

y + 2 = -3x + 5

3x + y = 3

Step 1: Your equation will ultimately take the form of Ax +By = C

Starting with y - y1 = m(x - x1) where x = 5, y = -2 and m = -3 yields:

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

y + 2 = -3x + 15 which simplifies to y = -3x + 13.

Rewriting by adding -3x to both sides of equation yields:

3x + y = 13 which is now in Standard Form

Hope this helps

y=x+3

Recall that lines have an equation of the form y = mx + c, where m is the slope. Substituting the value of m, we get:

y = -3x + c

To find the value of c, we substitute the coordinates of the given point.

-2 = -3(5) + c

c = 13

So, the equation of the line is y = -3x + 13

Hope this helps.

At point (5,-2),

-2=-3*5+C

therfore, C=15-2=13

Hence general formular for line is

y=-3x+13

(5,-2) Slope -3

y - y1 = m(x - x1)

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

y + 2 = -3x + 15

y = -3x + 15 - 2

y = -3x + 13

The line's equation is: y = n -3x

n is unknown.

-2 = n-3*5

--> n = 13

y = 13 - 3x

y-y1 = m(x-x1)

y+2 = -3(x-5)

y+2= -3x + 15

y+2= -3x + 15 - 2

y= -3x + 13

(y - b) =m(x - a)

)y +2 = -3(x - 5) = -3x + 15

y = -3x + 13

y=5x-3