Find the equation of a line whose distance from a point is given The question is:
If the distance of a point $(1, 4)$ from a line  passing through the intersection of the lines $x-2y+3=0$ and $x-y-5=0$ is $4$ units. Find its equation. 
 A: hint.....The point of intersection is $(13,8)$. A line of gradient $m$ through this point is $$y-8=m(x-13)\Rightarrow y-mx+13m-8=0$$
Now you can use the formula for the distance $d$ from a point $(p,q)$ to the line $ax+by+c=0$ which is $$d=\left|\frac{ap+bq+c}{\sqrt{a^2+b^2}}\right|$$
You can now form and solve a quadratic equation to find the possible values of $m$
A: There are $2$ such lines.
The lines with equations $x-2y+3=0$ and $x-y-5=0$ define a pencil of lines:
$$t(x-2y+3)+u(x-y-5)=0\iff (t+u)x-(2t+u)y+3t-5u=0$$
which pass through the intersection point of the given lines.
The distance from the point $(1,4)$ to one of these lines is given by:
$$\frac{\lvert t+u-4(2t+u)+3t-5u\rvert}{\sqrt{(t+u)^2+(2t+u)^2}}=\frac{4\lvert t+2u\rvert}{\sqrt{5t^2+6tu+2u^2}},$$
and it is equal to $4$ if and only if $\;\lvert t+2u\rvert=\sqrt{5t^2+6tu+2u^2}$, i.e. 
$$(t+2u)^2=5t^2+6tu+2u^2\iff 4t^2+2tu-2u^2=0.$$
As this is a homogeneous quadratic equation, we solve it setting $\lambda=\dfrac tu$. We obtain the equation in  $\lambda$: $\;2\lambda^2+\lambda-1=0$. 
It has an integer root: $\lambda=-1$, so the other root, by Vieta's formulae is $\;\lambda=\dfrac12$. Thus  the requested lines have equation:


*

*$y=8$ (for $\lambda=-1$, i.e. $u=-t$),

*$3x-4y-7=0$ (for $\lambda=\dfrac12$, i.e. $u=2t$).

