write the standard form of the equation of the line throu this pair of points (5, -2) and (-6,10)?

Question

Only interger coefficients

Answer 1

m - slope.
pt. - point.
Let point 1 be pt.(-6,10).
Let point 2 be pt.(5,-2).

Calculate the slope.
m = (y2 - y1) / (x2 - x1)
m = (-2 -10) / (5- (-6))
m = -12/ 11

y = mx + c
y = (-12/11)x + c
Now use one of the points and calculate c.
Using pt.(5,-2).
y = (-12/11)x + c
-2 = (-12/11)(5) + c
-2 = -60/11 + c
c = -2 + 60/11
c = -22/11 + 60/11
c = 38/11

Now fill information into standard equation.
y = mx + c
y = (-12/11)x + c
y = (-12/11)x + 38/11 (x 11)
11y = -12x + 38

Answer 2

First find the slope m, of the line.

m = ∆y/∆x = (-2 -10) / (5 - -6) = -12/11

Now select one of the points and write the equation of the line. Let's choose the point (5,-2).

y - -2 = (-12/11)(x - 5)
y + 2 = (-12/11)(x - 5)

11(y + 2) = -12(x - 5)
12(x - 5) + 11(y + 2) = 0
12x - 60 + 11y + 22 = 0
12x + 11y - 38 = 0

Answer 3

First, find the slope of the line connecting the two point.

m=(10+2)/(-6-5) or -12/11

Then use the point-slope equation to get:

y+2=-12/11*(x-5)

Multiplying by 11 on both sides of the equation:

11*y+22=-12*x+60 or
11*y+12*x-38=0.

Answer 4

m = (10 + 2) / (- 6 - 5)
m = 12 / (-11)
y - 10 = (- 12/11).(x + 6)
11y - 110 = - 12x - 72
12x + 11y - 38 = 0

Answer 5

I am in pre-calculas. I remember slope intercept form. What is standard form again? Find your slope. y2-y1 over x2-x1.