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suppose that we are require to write equation of lines,whose go through origin and distance from point $F(-4,3)$ to this line is $1cm$

first of all ,distance from point $F(x_0,y_0)$ to line $A*x+B*y+C=0$ is $d={+,-}(A*x_0+B*y_0+c)/(\sqrt{(A^2+B^2)})$ because we have origin,we have $(x_0,y_0)=(0,0)$ equation of line is $y=k*x$ or $k*x-y=0$ if we put point $F(4,-3)$ ,i have got $K_1=-2/15*(6+\sqrt{6})$ and $k_2=2/15*(\sqrt{6}-6)$ or $y_1=-2/15*(6+\sqrt{6})*x$ and $y_2=2/15*(\sqrt{6}-6)*x$ is that correct or did i make some mistake?

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up vote 2 down vote accepted

Line on the plane through the origin $\,\Longrightarrow mx+y=0\,$ , and we want the distance from the given point to be $\,1\,$: $$\frac{|-4m+3|}{\sqrt{m^2+1}}=1\Longrightarrow 15m^2-24m+8=0\Longrightarrow m_{1,2}=\frac{12\pm 2\sqrt{6}}{15}$$not very nice solutions but nevertheless, solutions.

Since I can't fully understand your solution (please use LaTeX to write mathematics!) I can't be sure whether you got them right...but the signs look fishy there.

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yes equation i have got the same,just you are right my answer seems a little tricky because of LaTeX – dato datuashvili Jun 19 '12 at 11:08
@Dato, yes: I already checked what I think your answer is and it looked fine. You already used the "$" signs, but you didn't use the functions, for example "\frac{}{}" to do fractions... – DonAntonio Jun 19 '12 at 11:11
thanks for reply and trying to help – dato datuashvili Jun 19 '12 at 11:13
+1 for the answer. ;-) – Nancy Rutkowskie Jun 19 '12 at 16:39

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