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