Analytic geometry help

[Kanimoto]

The lesson I'm currently studying is on "normal form of equations", and I need help with the following problem.

Find two values of K such that for each the line Kx + y = 5 is distant 4 from the origin.

I tried writing the equation in normal form, but it didn't seem to work out. What am I supposed to do instead? Thanks.

 

[pka]

Using the distance formula for the distance from (0,0) to the line we get:
\(\displaystyle \L \frac{{\left| {K(0) + 1(0) - 5} \right|}}{{\sqrt {K^2 + 1} }} = 4\).

Solve for K.

 

[Kanimoto]

Cool! Thanks.

 

[soroban]

Hello, Kanimoto!

If you don't know (or remember) that point-to-line formula,
. . we can still solve it by elementary methods.

]Find two values of \(\displaystyle K\) such that for each the line \(\displaystyle Kx\,+\,y\:=\:5\)
is distant 4 from the origin.

The given line \(\displaystyle L\) is: \(\displaystyle \:y\:=\:-Kx\,+\,5\) . . . with slope \(\displaystyle \text{-}K\)

We want the line through \(\displaystyle (0,0)\) perpendicular to \(\displaystyle L\).
. . Its slope is \(\displaystyle m = \frac{1}{K}\) and its equation is: \(\displaystyle y\:=\:\frac{1}{K}x\)

Find the intersection \(\displaystyle P\) of the two lines: \(\displaystyle \:\frac{1}{K}x\:=\:-Kx\,+\,5\)

We have: \(\displaystyle \,\frac{1}{K}x\,+\,Kx\:=\:5\;\;\Rightarrow\;\;\left(\frac{1}{K}\,+\,K\right)x\:=\:5\;\;\Rightarrow\;\;\left(\frac{1\,+\,K^2}{K\right)x\:=\:5\)

Hence: \(\displaystyle \:x\,=\,\frac{5K}{1\,+\,K^2}\:\) . . . and: \(\displaystyle \:y\:=\:\frac{5}{1\,+\,k^2}\)

. . We have located point \(\displaystyle P:\;\;\left(\frac{5K}{1+K^2},\,\frac{5}{1+K^2}\right)\)


Since distance \(\displaystyle OP\,=\,4\), we have: \(\displaystyle \:\left(\frac{5K}{1+K^2}\right)^2\,+\,\left(\frac{5}{1+K^2}\right)^2\:=\:4^2\)

. . \(\displaystyle 25K^2\,+\,25 \:=\:16(1\,+\,K^2)^2\;\;\Rightarrow\;\;16k^4\,+\,7K^2\,-\,9\:=\:0\;\;\Rightarrow\;\;(K^2\,+\,1)(16K^2\,-\,9)\:=\:0\)


\(\displaystyle K^2\,+\,1\:=\:0\) has no real roots,

but: \(\displaystyle 16K^2\,-\,9\:=\:0 \;\;\Rightarrow\;\;k^2\,=\,\frac{9}{16}\;\;\Rightarrow\;\;\L\fbox{K\,=\,\pm\frac{3}{4}}\)