How do you write an equation in standard form given point (1,4) and (6,-1)?

1 Answer
Oct 4, 2016

#1x+1y=5#

Explanation:

The slope of the line through #(1,4)# and #(6,-1)# is
#color(white)("XXX")color(green)(m)=(Deltay)/(Deltax)=(4-(-1))/(1-6)=color(green)(-1)#

The slope-point form for a line with slope #color(green)(m)# through a point #(color(red)a,color(blue)b)# is
#color(white)("XXX")y-color(blue)b = color(green)(m)(x-color(red)a)#

Using the previously determined slope #color(green)(m=-1)# and #(color(red)1,color(blue)4)# as our point #(color(red)a,color(blue)b)#
#color(white)("XXX")y-color(blue)4=color(green)(-1)(x-color(red)1)#

Our objective is to convert this into standard form:
#color(white)("XXX")Ax+By=C#

#y-4=-1(x-1)#
#color(white)("XXX")rarr y-4=-x+1#

#color(white)("XXX")rarr x+y=5#

or, if we wish to be explicit in the variable coefficients:
#color(white)("XXX")1x+1y=5#