The slope intercept form of equation of a line is y=mx+cy=mx+c, where mm is the slope of the line and cc is its intercept on yy-axis.
Point slope form of equation of a line is y-y_1=m(x-x_1)y−y1=m(x−x1), where mm is the slope of the line and (x_1,y_1)(x1,y1) are the coordinates of the point through which the line passes.
The general form of equation of a line is ax+by+c=0ax+by+c=0.
For the equation of a line passing through two points (x_1,y_1)(x1,y1) and (x_2,y_2)(x2,y2), the formula is (y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)y−y1y2−y1=x−x1x2−x1. However, we first find the slope of line (x_1,y_1)(x1,y1) and (x_2,y_2)(x2,y2), which is (y_2-y_1)/(x_2-x_1)y2−y1x2−x1. For points (1,-2)(1,−2) and (3,-8)(3,−8) is (-8-(-2))/(3-1)=-6/2=-3−8−(−2)3−1=−62=−3.
Hence selecting the point slope form of equation of desired line (choosing point (1,-2)(1,−2)) is y-(-2)=(-3)(x-1)y−(−2)=(−3)(x−1) or y+2=-3x+3y+2=−3x+3,
which in general form is 3x+y-1=03x+y−1=0
Note that had we chosen (3,-8)(3,−8), the point slope form of equation could also have been y-(-8)=(-3)(x-3)y−(−8)=(−3)(x−3) or y+8=-3x+9y+8=−3x+9 which again gives us 3x+y-1=03x+y−1=0.
We can also write the equation as y=-3x+1y=−3x+1 in slope intercept form which shows slope as -3−3 and yy-intercept as 11.
graph{-3x+1 [-10, 10, -5, 5]}
