How would you write an equation of a line with a slope of -3 passing through the point (3,-4) in standard form?

When you know the slope m of a line and the coordinates (a, b) of a point on the line, you can write the equation in point-slope form: y - be = m(x - a) y + 4 = (-3)(x - 3) 3x + y = 5.

First write it in point slope form y-y1=m(x-x1). Y1 and x1 are just the coordinate points. M is the slope.

Y+4= -3(x-3) Distribute y+4= -3x+9 subtract 4 to get to slope intercept form. Y= -3x+5 add 3x Answer: 3x+y=5.

Put into y=mx+b form. M=-3 and b=-4 y=-3x -4 Standard form is Ax+By=C (the A must always be positive) move the mx (-3x) to the other side The y value in the cordnants is C 3x+y=-4 Done. This **** is really hard tho, so id be too lazy to do it too lol.

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