How do you find the equation of the line that goes through #(1,- 3)# and #( 0,8)#?

2 Answers
Oct 1, 2017

#y=-11x+8#

Explanation:

First, we find the slope using #(y^2-y^1)/(x^2-x^1)#

Therefore, #m=(8-(-3))/(0-1)# which is equal to #m=-11#

To get the line, the equation would be #y=mx+b#

#y=11x+b#

We will put in #(1, -3)# as #x=1# and #y=-3#
#-3=(-11)1+b->#
#-3=(-11)*1+b->#
#-3=-11*1+b->#
#-3=-11+b->#
#8=b# (Add #11# to both sides of the equation)

#8=b#

Therefore, the line will be #y=mx+b->#

#y=-11x+8# graph{y=-11x+8 [-10, 10, -5, 5]}

Oct 1, 2017

#11x+1y=8#

Explanation:

For a straight line the slope between any two points is the same for all pairs of points on the line.

By slope we mean the ratio of the difference between the #y# coordinates divided by the difference between the #x# coordinates.

#color(white)("xxxx")"slope"=(delta y)/(delta x)=(y_2-y_1)/(x_2-x_1)#
#color(white)("xxxxxx")#for two points #(x_1,y_1)# and #(x_2,y_2)#

For a general point #(x,y)# and the point #(1,-3)# (this is one of the given points,
we have
#color(white)("xxxx")"slope"=(y-(-3))/(x-1)=(y+3)/(x-1)#
and
for the two given points #(0,8)# and #(1,-3)#
we have
#color(white)("xxxx")"slope" = (8-(-3))/(0-1)=11/(-1)=-11#

Since the slope must be the same for all pairs of points on a straight line
#color(white)("XXX")(y+3)/(x-1)=-11#

While is is a valid equation that answers the given question, it would be normal to convert this into standard #Ax+By=C# form:

#color(white)("XXX")(y+3)=-11(x-1)#

#color(white)("XXX")y+3=-11x+11#

#color(white)("XXX")11x+y+3=11#

#color(white)("XXX")11x+1y=8#

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When doing this kind of problem it is always a good idea to verify that the derived equation satisfies the given points:

For #(x,y)=(1,-3)#
#color(white)("XXX")11 * (1)+1 * (-3)=11-3=8# ... as required
and
for #(x,y)=(0,8)#
#color(white)("XXX")11 * (0) + 1 * (8)= 0+8=8# ...again, as required.