Indicate in standard form the equation of the line through the given points. P(0, -4), Q(5, 1)
@MMMercado can you help me please
First develop the line equation in any form. then place it in the standard form which is: ax + by = c
do you first put it in point slope form?
The two points given is all you need. Suggest you use the point-slope form to solve for the slope, the slope will become "a" in the standard form.
Yes.
I would go that way. point-slope form.
okay
I will standby for further assistance (if needed).
i put it in point slope form just don't get how to put it in standard once i get it in point
ok what did you get for slope (m) in the point slope form? (-4 - 1) = m(0 -5) -5 = -5m m = ?
1
Correct, now use point slope again using the variables x and y and only one of the given points like this: (y-1) = m (x-5) I used the second point y-1 = 1(x-5) y-1 = x -5 placing in standard form first put the variables on the left side. -x + y -1 = -5 now put the numerical values on the right -x + y = -5 + 1 simplify. and you will be done.
what does the final standard form look like?
So I just simplify it?
Yes combine the numericals on the right side of the equal sign.
So that would equal 6... right?
-x + y = -4 one answer or it could be expressed as: x - y = 4 by multiplying both sides by a -1.
No. -5 + 1 would be a -4. Remember your rules for signs when adding or subtrtacting In this case you are adding a 1 to a -5.
Okay I got it now thank you so much plus my algebra teacher in summer school finally decided to help me
Very good, I like the line equation expressed as x - y = 4 better simply because the coefficient of x is shown as a +1 which is also the value of the slope (+1).
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