How do you write the equation of a line in slope intercept, point slope and standard form given (1,-2), and (3,-8)?

1 Answer
Jan 17, 2018

Please see below.

Explanation:

The slope intercept form of equation of a line is #y=mx+c#, where #m# is the slope of the line and #c# is its intercept on #y#-axis.

Point slope form of equation of a line is #y-y_1=m(x-x_1)#, where #m# is the slope of the line and #(x_1,y_1)# are the coordinates of the point through which the line passes.

The general form of equation of a line is #ax+by+c=0#.

For the equation of a line passing through two points #(x_1,y_1)# and #(x_2,y_2)#, the formula is #(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)#. However, we first find the slope of line #(x_1,y_1)# and #(x_2,y_2)#, which is #(y_2-y_1)/(x_2-x_1)#. For points #(1,-2)# and #(3,-8)# is #(-8-(-2))/(3-1)=-6/2=-3#.

Hence selecting the point slope form of equation of desired line (choosing point #(1,-2)#) is #y-(-2)=(-3)(x-1)# or #y+2=-3x+3#,

which in general form is #3x+y-1=0#

Note that had we chosen #(3,-8)#, the point slope form of equation could also have been #y-(-8)=(-3)(x-3)# or #y+8=-3x+9# which again gives us #3x+y-1=0#.

We can also write the equation as #y=-3x+1# in slope intercept form which shows slope as #-3# and #y#-intercept as #1#.

graph{-3x+1 [-10, 10, -5, 5]}