How do you write the equation of a line in slope intercept, point slope and standard form given (2,-3) and (4,-1)?

1 Answer
Binayaka C.
Apr 17, 2017

The equation of the line in slope-intercept form is #y= x -5#
in point #(2,-3)# slope form is # y+3 = x-2#
and in standard form is # -x +y = -5#

Explanation:

The slope of the line passing through #(2,-3) and (4,-1)# is #m= (y_2-y_1)/(x_2-x_1)= (-1+3)/(4-2)=2/2=1#

Let the equation of the line in slope-intercept form be #y=mx+c or y=x+c# The point (2,-3) will satisfy the equation . So, # -3= 2+c or c= -5#

Hence the equation of the line in slope-intercept form is #y= x -5#

The equation of the line in point #(2,-3)# slope form is #y-y_1=m(x-x_1) or y - (-3) = 1*(x-2) or y+3 = x-2#

The equation of the line in standard form #(ax+by=c)# is #y= x-5 or -x +y = -5# {Ans]

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