What is the equation of the line that passes through the points #(8, -1)# and #(2, -5)# in standard form, given that the point-slope form is #y + 1 = 2/3(x-8)#?

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Answer 1 · 2017-04-06T01:23:42

#2x-3y=19#

We can convert the equation from point slope form to standard form. For us to have standard form, we want the equation in the form of:

#ax+by=c#, where #a# is a positive integer (#a in ZZ^+#), #b# and #c# are integers (#b, c in ZZ#) and #a, b, and c# don't have a common multiple.

Ok, here we go:

#y+1=2/3(x-8)#

Let's first get rid of the fractional slope by multiplying by 3:

#3(y+1)=3(2/3(x-8))#

#3y+3=2(x-8)#

#3y+3=2x-16#

and now let's move #x, y# terms to one side and non #x, y# terms to the other:

#color(red)(-2x)+3y+3color(blue)(-3)=2xcolor(red)(-2x)-16color(blue)(-3)#

#-2x+3y=-19#

and lastly we want the #x# term to be positive, so let's multiply through by #-1#:

#-1(-2x+3y)=-1(-19)#

#2x-3y=19#

Now let's make sure our points work:

#(8,-1)#

#2(8)-3(-1)=19#

#16+3=19#

#19=19 color(white)(00)color(green)sqrt#