Okay so, I have to write an equation in standard form, of a line that passes through the given point and has the given slope, ( M) or that passes through the given points.
Heres what I think:
1)( 4,-1) M= 3
So, it's in y=3x+B
and then, I would plug in
-1 for y, and 4 for x.
So, -1=3(4)+B
-1=12+B
Subtract 12 from -1
and get 13.
B=13
Full equation:
y=3x+13
But, how would I write this in standard form?
Could a tutor please explain to me?
Thank you.
-Allyson
<<Full equation:
y = 3x+13
But, how would I write this in standard form? >>
It is already in the standard form, but the correct answer is
y = 3x - 13
You made an error after the line
-1 = 12 + b
I sure you will see it right away when you check that step.
no, that form is slope intercept form.
Isn't it?
Some people consider that the standard form. If you want it in Ax + By = C form,
or Ax + By + C = 0, it is easily rearranged.
But first correct the error.
3x - y = 13
To write the equation in standard form, we need to rearrange the equation so that the x and y terms are on the same side, and the constant term is on the other side.
1) Start with the equation y = 3x + 13.
2) To remove the fractions, we can multiply the entire equation by 3 (the coefficient of x).
3(y) = 3(3x + 13)
Simplifying, we get 3y = 9x + 39.
3) Next, we want to rearrange the equation to get it in the form Ax + By = C, where A, B, and C are integers.
-9x + 3y = 39.
4) Multiply the entire equation by -1 to make the coefficient of x positive.
9x - 3y = -39.
5) Since the standard form requires that the coefficient of x is positive, we can divide the entire equation by -3 to make it positive.
-3x + y = -13.
Therefore, the equation of the line in standard form that passes through the point (4, -1) and has a slope of 3 is -3x + y = -13.