First you have to draw the both lines, like I did
Now you have to imagine that the shortest distance is a perpendicular line (C) like in my drawing.
Using the trigonometrical properties we can find the angle [tex]\alpha[/tex]
[tex]\alpha=9.46^o[/tex]
then we can use the trigonometrical property of sine
[tex]sin(9.46^o)=\frac{C}{8}[/tex]
[tex]C\approx1.32[/tex]
[tex]\boxed{\boxed{C^2\approx1.73}}[/tex]
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Another way to solve this question:
You have to find the line equations
[tex]y=6x\rightarrow6x-y=0[/tex]
[tex]y=6x+8\rightarrow6x-y+8=0[/tex]
[tex]d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}[/tex]
you can chose equation 1 or equation 2, but be careful, the point should be (0,8) if you pick up the equation 1 and the point should be (0,0) if you pick up the equation 2, I prefere to use the first one
[tex]d=\frac{|6x-y+0|}{\sqrt{(6)^2+(-1)^2}}[/tex]
[tex]d=\frac{|6x-y|}{\sqrt{37}}[/tex]
replacing the point (0,8)
[tex]d=\frac{|6*0-8|}{\sqrt{37}}[/tex]
[tex]d=\frac{8}{\sqrt{37}}[/tex]
[tex]d^2=\left(\frac{8}{\sqrt{37}}\right)^2[/tex]
[tex]\boxed{\boxed{d^2=\frac{64}{37}\approx1.73}}[/tex]
