Question

What is the standard form of a line that goes through the points 3,-6 and -2.-1 ?

Answer 1

you need the sloe first.
slope is
\[m=\frac{y_2-y_2}{x_2-x_1}=\frac{-1+6}{-2-3}=-1\]

Answer 2

The standard equation of a line is y=mx + b
Where m is the slope of the line and b is the y intercept of the line
To find slope we use (y2-y1) / x2-x1) which gives us a slope of 5/-5 or -1
m = -1
To find b we substitute the m we found into the equation and use one of the points given.
-6 = -1(3) +b
-6 = -3 + b
-3 = b
Therefore the standard equation of a line for this case is
y= -x -3

Answer 3

then use point-slope formula
\[y-y_1=m(x-x_1)\] with
\[m=-1, x_1=-2, y_1=-1\] to get
\[y+1=-(x+2)\] multiply on the right to get
\[y+1=-x+2\]

Answer 4

unfortunately i think "standard form" is
\[ax+by+c=0\] maybe

Answer 5

so write
\[x+y-1=0\] but of course it is all the same

Answer 6

\[y=mx+b\] is "slope- intercept" form

\[ax+by+c=0\] is "standard form" and
\[y-y_1=m(x-x_1)\] is "point-slope" form

Answer 7

although i certainly prefer
\[y=mx+b\] because it is easy to compute points if you just pick a number for x and plug it in