SOLUTION: write the equation of the line passing through (-6,-3) and (-6,1) i tried -3-1/-6-6= -4/6..i get messed up after this part.

Question 29791: write the equation of the line passing through (-6,-3) and (-6,1)
i tried -3-1/-6-6= -4/6..i get messed up after this part.
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
write the equation of the line passing through (-6,-3) and (-6,1)
The equation of a line joining two given points P(x1,y1) and Q(x2,y2)
is given by (y-y1)/(y1-y2) = (x-x1)/(x1-x2)----(1)
Here P(x1,y1) = (-6,-3) and Q(x2,y2) =(-6,1)meaning x1=-6,y1= -3;x2 = -6,y2 = 1
We observe that both the given points P and Q have their x-coordinate = (-6) which means they are on the line x= -6
Therefore the equation to the line joining the two points is the vertical line
x = -6 which is to the left of the y-axis at a distance 6 units from it
Note:There is no need to apply the standard formula for this special line which is very obvious . But if you still feel that you ought to apply this formula as you would in the case any two general points on the line,since division by zero is not defined and since (x1-x2)=[(-6)-(-6)]= (-6+6) =0
therefore we consider (1) in the equivalent form:
(y-y1)(x1-x2) = (x-x1)(y1-y2)----(1)
[y-(-3)]X(0) = [x-(-6)][(-3)-(1)]
(y+3)X0 = (x+6)X(-4)
0 = -4(x+6)
dividing by (-4)
0 = x+6
That is x+6 =0
x=-6 which is the vertical line to the left of the y-axis at a distance 6 units from it parallel to it.

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