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 Distance from a point to a line

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T O P I C    R E V I E W
jfg4707 Posted - 10/17/2007 : 19:18:37
I am having difficulty in finding the denominator of the problem below.
Problem
The distance from the point P(x,y) to the line y = mx + b is given by d = |y - mx - b|/(1 + m). Use the expression for d and show that the distance from the point P(x,y) to the line
Ax + By + C = 0 is d = |Ax + By + C|/(A + B).

Solution
For the numerator, I did the following:

y - mx -b = 0

Also,

Ax + By + C = 0.

So, -m = A/B; -b = C/B

Therefore, -mx + y -b = (A/B)x + y + C/B = 0
Multiplying through by B gives the numerator: |Ax + By + C|.

I have not been successful in showing that 1 + m = A+ B. I hope someone can help me on this.
Thank you.




2   L A T E S T    R E P L I E S    (Newest First)
jfg4707 Posted - 10/18/2007 : 14:04:58
tkhunny, thank you very much. I now see where my error is. We have

d = |y -mx - b|/(1 + m) (Eq. 1)

Ax + By + C = 0, so -m = A/B; -b = C/B. Substitusting these values for -m and -b into Eq. 1, we get

d = |(A/B)x + y + C/B|/(1 + m)
d = |Ax + By + C|/(B(1 + m))
d = |Ax + By + C|/(B(1 + (A/B)) = |Ax + By + C|/(A + B)

tkhunny Posted - 10/18/2007 : 08:20:55
quote:
Multiplying through by B gives the numerator: |Ax + By + C|.
What? Why would you divide by B just to multiply by B? Herein lies the problem.

Multiply the numerator by B, as you did, but you cannot do that by itself. These fractions are NOT the same:

1/2 and 3/2. I just multiplied the numerator by 3. That doesn't work.

If you are going to multiply by something, you must do it in the numerator AND in the denominator. Doing so, in this case, does 2 things:

1) Fixes the numerator, as you hvae demonstrated.
2) Puts and extra B in the denominator. You will need this to finish the problem.

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