How do you find the equation that joins points #(- 4,6)# and #( 3,- 1)#?

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Answer 1 · 2018-02-24T05:53:47

#y=-x+2#

Given -

#(-4,6); (3,-1)#
#x_1=-4#
#x_2=3#
#y_1=6#
#y_2=-1#

Equation

#(y-y_1)=(y_2-y_1)/(x_2-x_1) (x-x_1)#
#y-6=[(-1)-6]/[3-(-4)] (x-(-4))#
#y-6=[-7]/[7] (x+4)#
#y-6=-1 (x+4))#
#y-6=-x-4#
#y=-x-4+6#

#y=-x+2#

Answer 2 · 2018-02-24T05:59:11

See below:

We can use the point-slope form of a line equation. The general formula is:

#y-y_1=m(x-x_1)#

We can use either point, so let's use #(-4,6)# as our point.

Now we need the slope:

#m=(Deltay)/(Deltax)=(y_2-y_1)/(x_2-x_1)=(-1-6)/(3-(-4))=-7/7=-1#

Substitute:

#y-6=-(x+4)=>y-6=-x-4#

We can put this into other forms:

slope-intercept:

#y=-x+2#

standard:

#x+y=2#

We can graph this to see the points and the line connecting them:

graph{((x+4)^2+(y-6)^2-.5^2)((x-3)^2+(y+1)^2-.5^2)(x+y-2)=0[-15,15,-8,8]}