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# Determine the standard form of the equation of the line that passes through (-7, 8) and (0, -8)

Another question I am stumped on is determine the standard form of the equation of the line that passes through (9, -4) and (6, 4).

Both of these problems are approached in the same way.  First, you need to find the slope of the line, which is the difference in the y-coordinates divided by the difference in the x-coordinates.  Make sure you use the same point as the first point when subtracting the values, or you will end up with the wrong slope!

Once you have the slope (let's call this "m") you can use the point-slope form of a line as a starting point.  Pick one of the points -- let's say (-7, 8) -- and plug in the values:

y - (-7) = m * (x - 8)

Here, you would insert the value of m that you determined from the slope in the first step.

Now, you need to get this into "standard form".  First distribute, then move the x and y terms to the left-hand side, and the rest to the right-hand side:

y + 7 = mx + 8m
y - mx = 8m - 7

If m is a fraction, you'll want to multiply through by the denominator to end up with integer coefficients.

I hope this helps!

(y+4)=(4+4)/(6-9)*(x-9)

y+4=(-8/3)*(x-9)

3y+12=-8x+72

8x+3y=60