1. Lines and distance Question

The question is;

Find the equations of the lines which pass through the point $(1,2)$ and have perpendicular distance of $1$ from the origin.

The question says "lines" and "equations". The plural makes me think I have gone wrong since I only have one. Here is working....

Let $l$ be a line which passes through $(1,2)$ then $l:~y=(2-a)+ax$ for some $a$. Let the line perpendicular to $l$ passing through the origin be $p$ and thus $p:~y=-\frac{1}{a}\cdot x$ .

Let $l$ and $p$ intersect at $R$ then;

$(2-a)+ax=-\frac{1}{a}\cdot x$

$\implies x=\frac{a(a-2)}{a^2+1}~~,~~y=-\frac{(a-2)}{a^2+1}$

Thus $R:~\left(\frac{a(a-2)}{a^2+1}~,~-\frac{(a-2)}{a^2+1}\right)$

Let the distance between $R$ and the origin be $d$ then,

$d=\sqrt{\left(\frac{a(a-2)}{a^2+1}\right)^2+\left(\frac{(a-2)}{a^2+1}\right)^2}=\frac{(a-2)}{\sqrt{a^2+1}}=1$

$\implies (a-2)=\sqrt{a^2+1} \implies a=\frac{3}{2}$

and therefore the required line is $y=\frac{1}{2}(3x+1)$

The question says lines though. Did I do something wrong? or wrong method?

Edit: Since I defined a as a real number of some sort is it right that a might not be a real number and thus the other line is $x=1$ . Is this right?

2. Originally Posted by Sean12345
The question is;

Find the equations of the lines which pass through the point $(1,2)$ and have perpendicular distance of $1$ from the origin.

The question says "lines" and "equations". The plural makes me think I have gone wrong since I only have one. Here is working....

Let $l$ be a line which passes through $(1,2)$ then $l:~y=(2-a)+ax$ for some $a$. Let the line perpendicular to $l$ passing through the origin be $p$ and thus $p:~y=-\frac{1}{a}\cdot x$ .

Let $l$ and $p$ intersect at $R$ then;

$(2-a)+ax=-\frac{1}{a}\cdot x$

$\implies x=\frac{a(a-2)}{a^2+1}~~,~~y=-\frac{(a-2)}{a^2+1}$

Thus $R:~\left(\frac{a(a-2)}{a^2+1}~,~-\frac{(a-2)}{a^2+1}\right)$

Let the distance between $R$ and the origin be $d$ then,

$d=\sqrt{\left(\frac{a(a-2)}{a^2+1}\right)^2+\left(\frac{(a-2)}{a^2+1}\right)^2}=\frac{(a-2)}{\sqrt{a^2+1}}=1$

$\implies (a-2)=\sqrt{a^2+1} \implies a=\frac{3}{2}$
I get a= 3/4, not 3/2!

and therefore the required line is $y=\frac{1}{2}(3x+1)$

The question says lines though. Did I do something wrong? or wrong method?

Edit: Since I defined a as a real number of some sort is it right that a might not be a real number and thus the other line is $x=1$ . Is this right?
I wondered about that myself. Geometrically, if draw a line from the origin to the point to the origin and look at the line symmetric to your line through that you should get another line satisfying the same properties. And, in fact there is another line!

BUT that other line cannot be written in the form y= ax+ b. What is the one line through (1, 2) that cannot be written y= ax+ b? Do you see that its distance to the origin is also 1, trivially?

I think you stuck in that "edit" while I was typing! Yes, that other line is x= 1.

3. Originally Posted by HallsofIvy
I get a= 3/4, not 3/2!
Silly mistake

Originally Posted by HallsofIvy
I think you stuck in that "edit" while I was typing! Yes, that other line is x= 1
Yes, I was thinking about it through my head when I left.

Thank you.