Answer:
[tex]\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}[/tex]
Step-by-step explanation:
The equation of a line:
[tex]y=mx+b[/tex]
We have
[tex]m=\dfrac{2}{3}[/tex]
substitute:
[tex]y=\dfrac{2}{3}x+b[/tex]
The formula of a distance between a point and a line:
General form of a line:
[tex]Ax+By+C=0[/tex]
Point:
[tex](x_0,\ y_0)[/tex]
Distance:
[tex]d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}[/tex]
Convert the equation:
[tex]y=\dfrac{2}{3}x+b[/tex] |subtract [tex]y[/tex] from both sides
[tex]\dfrac{2}{3}x-y+b=0[/tex] |multiply both sides by 3
[tex]2x-3y+3b=0\to A=2,\ B=-3,\ C=3b[/tex]
Coordinates of the point:
[tex](0,\ 0)\to x_0=0,\ y_0=0[/tex]
substitute:
[tex]d=4[/tex]
[tex]4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}[/tex]
[tex]4=\dfrac{|3b|}{\sqrt{13}}\qquad|[/tex] |multiply both sides by [tex]\sqrt{13}[/tex]
[tex]4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}[/tex] |divide both sides by 3
[tex]b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}[/tex]
Finally:
[tex]y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}[/tex]
