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We have a point $P(1,7)$, get the equations of the 2 lines which have a distance of $5$ from point $P$. Both of the lines go through the origin.

So I used the formula $\dfrac{|ax+by-c|}{\sqrt{a^2+b^2}} = 5$

However, I only know $x, y$ and $c$. My teacher however, he said the $b$ must be $-1$. This I don't understand. Why $b = -1$? I know how to solve it once I fill in $-1$ for $b$, I just need some clarification:

Why $b=-1$?

  • I believe it has something to do with the fact that the lines go through the origin.

  • I posted a question about this problem about half a month ago, however, this is a different one and is only about the line in bold, not about the problem itself.

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I have no idea what that formula is. Can you explain it to me? What are the meanings of $a, b, c, x, y$? – Jonathan Christensen Nov 2 '12 at 16:18
If the equation of a line $l$ is $ax+by=c$, then the distance from point $P(x_1,x_2)$ to line $l$ = $\dfrac{|ax_1+bx_2-c|}{\sqrt{a^2+b^2}} = 5$ – JohnPhteven Nov 2 '12 at 16:22
Ok, that's clear. Answer below. – Jonathan Christensen Nov 2 '12 at 16:34
up vote 1 down vote accepted

Let the equation of the line be $A\cdot x+B\cdot y+C=0$

As the lines pass through the origin $(0,0), A\cdot0+B\cdot 0+C=0\implies C=0$

Now, using the same formula you have applied, $$\frac{\mid A\cdot1+B\cdot7\mid}{\sqrt {A^2+B^2}}=5$$

On squaring, $(A+7B)^2=25(A^2+B^2)$

On simplification, $12A^2-7AB-12B^2=0$

Observe that if $A=0,B=0$ and vice versa.Then the equation becomes $0=0$ an identity.

So, $AB\ne0,$ solving for we get, $A=\frac43B $ or $-\frac3 4 B$

If $A=\frac43B,$ the equation becomes $\frac43B\cdot x+B\cdot y=0$ or $4x+3y=0$ as $B\ne 0$

Similarly, if $A=-\frac3 4 B,$ the equation becomes $3x-4y=0$

Observe that we don't have to use any specific value of $B,$ as it ultimately gets cancelled out. So, we can use any specific value $\ne 0$ for $B,$(please test this) and get the corresponding value of $A$ and we are to going to reach at the same two lines as above.

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$b = -1$ is an arbitrary choice. You know that $c = 0$, and you want to solve for $a$ and $b$. But if $(a^*, b^*)$ is a solution, then so is $(2a^*, 2b^*)$, $(3a^*, 3b^*)$, $(\pi a^*, \pi b^*)$, etc., because $a^*x + b^*y = 0$ is the same line as $\pi a^*y + \pi b^*y = 0$. So without putting a restriction on $a$ or $b$, there are infinitely many solutions. As long as neither of the lines is exactly horizontal or vertical, you can get just two solutions by setting either $a$ or $b$ to any non-zero real number and solving for the other one; your teacher probably chose $b = -1$ because $ax + (-1)y = 0 \Leftrightarrow y = ax$, which is a nice standard form for a line.

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But how would you know that the non-zero real number is actually on your line at all? – JohnPhteven Nov 2 '12 at 16:47
Once again, I don't understand what you mean. What does it mean to say that six is or is not on the line $y=ax$? Lines in the plane consist of pairs of real numbers, not single numbers. – Jonathan Christensen Nov 2 '12 at 17:22
You say that $-1$ is an arbitrary choice, which I don't understand and don't believe is true, as I've seen many students make the same choice. However, I believe it is so because we know that the line is $y=ax$ > $ax-y=0$ so therefore $y$ must equal $(-1)$ – JohnPhteven Nov 2 '12 at 19:07
If you set $b = -2$ then you can solve it the same way, but your estimate of $a$ will be twice as large as it will be if you set $b = -1$. When you solve the equation to put it in the form $y = ax$ you'll see that the resulting line is exactly the same. I strongly encourage you to try this if you still don't believe me. – Jonathan Christensen Nov 3 '12 at 0:35
That last comment should say "solution for $a$," not "estimate of $a$." Forgive me for being a statistician... – Jonathan Christensen Nov 3 '12 at 1:56

First of all, it's obvious that any line passing through the origin is of the form $ax+by=0$ or equivalently $(\frac{a}{b})x+y = 0$. Notice that $c$ here is $0$ because the line passes through the origin. Now use that formula for the equation of the line with $x=1$ and $y=7$. Square and simplify, and you'll get a quadratic in $(\frac{a}{b})$. Solve and plug in the solutions, and you'll get the equations of the lines.

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