Answer to Question #146914 in Analytic Geometry for gerome rodriguez

Question

find the general equation of a line parallel to the line 2x+3y-6=0 and passing at a distance 5 square root 13 over 13 from the point (-1,1)

Answer 1

Let us find the general equation of a line parallel to the line $2x+3y-6=0$ and passing at a distance $5\frac{\sqrt{13}}{13}$ from the point $(-1,1)$.

The vector $\overrightarrow{n}=(2,3)$ is perpendicular to all lines parallel to $2x+3y-6=0$ . Thererfore, the euqation of any line parralel to $2x+3y-6=0$ is $2x+3y+c=0$ for some $c\in\mathbb R$.

The distance from the point $(-1,1)$ to the line $2x+3y+c=0$ is $d=\frac{|2(-1)+3\cdot 1+c|}{\sqrt{4+9}}=\frac{|1+c|}{\sqrt{13}}=|1+c|\frac{\sqrt{13}}{13}$. Onthe other hand $d=5\frac{\sqrt{13}}{13}$. Therefore, $|1+c|=5$, and hence $1+c=\pm 5$. So, $c=-6$ or $c=4.$ We conclude that there are two lines parallel to the line $2x+3y-6=0$ and passing at a distance $5\frac{\sqrt{13}}{13}$ from the point $(-1,1)$: $2x+3y-6=0$ and $2x+3y+4=0$. One of them coinside with the initial line $2x+3y-6=0$ .