Finding a point a certain distance from a line.

**BenBart** · October 14th 2008, 01:34 PM

(I just want to note that I made up this problem. I'm not trying to cheat on a test or anything.)

**Peritus** · October 14th 2008, 01:44 PM

$\sqrt {(x - 1)^2 + (y - 1)^2 } = 2$

**icemanfan** · October 14th 2008, 01:44 PM

The point satisfies $\sqrt{(x-2)^2 + (y-1)^2} = 2$ and $y = 2x - 3$ .

Substitution yields:
$\sqrt{(x-2)^2 + (2x - 4)^2} = 2$

$\sqrt{(x-2)^2 + 4(x-2)^2} = 2$

$5(x-2)^2 = 4$

$(x-2)^2 = \frac{4}{5}$

$x-2 = \sqrt{\frac{4}{5}}$

$x = \sqrt{\frac{4}{5}} + 2$

Now you plug x back into the line equation to find y.

**BenBart** · October 14th 2008, 03:58 PM

I just want to clarify that what I'm looking for is the point where that circle is intersecting the line. I'm not looking to solve for the X coordinate.

So I'm not looking for that answer that's around (4,5) I'm looking for the one that's around (2.9, 2.8)ish – I thought I would need pi.

If this is in fact the correct solution then I guess I need a little help figuring out what the 1's and 2's represent.

**BenBart** · October 14th 2008, 06:49 PM

Problems get bumped fairly quickly in this section. Is there a more appropriate place to post this? Geometry, pre-calc?

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