Find the lines which have a certain distance from a certain point

Question

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.

Answer 1

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.

Answer 2

$b = -1$ 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.

Answer 3

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.