How do you find a standard form equation for the line with (4,-14) (2,-5)?

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Answer 1 · 2018-03-22T03:00:47

#9x+2y=8#

We can use the point-slope form to start with:

#y-y_1=m(x-x_1)# where #(x_1, y_1)# are a point and #m# is the slope.

We can choose either point, so let's pick #(2,-5)#

The slope can be found this way:

#m=(y_2-y_1)/(x_2-x_1)=(-14+5)/(4-2)=-9/2#

So now we have:

#y+5=-9/2(x-2)#

To put this into standard form, we need the equation in the form of:

#Ax+By=C# where #A, B, C in ZZ#:

#y+5=-9/2(x-2)#

Multiply by 2 to get rid of the denominator:

#2(y+5)=2(-9/2(x-2))#

#2y+10=-9(x-2)#

#2y+10=-9x+18#

#9x+2y=8#

The graph of our line looks like this:

graph{(9x+2y-8)((x-4)^2+(y+14)^2-.5)((x-2)^2+(y+5)^2-.5)=0[-20,20,-20,0]}