Question

Let d be the (shortest) distance between 2 parallel lines with slope 6 whose -intercepts are 8 units apart. Find d^2.

Answer 1

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 $\alpha$

$\alpha=9.46^o$

then we can use the trigonometrical property of sine

$sin(9.46^o)=\frac{C}{8}$

$C\approx1.32$

$\boxed{\boxed{C^2\approx1.73}}$

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Another way to solve this question:

You have to find the line equations

$y=6x\rightarrow6x-y=0$

$y=6x+8\rightarrow6x-y+8=0$

$d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$

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

$d=\frac{|6x-y+0|}{\sqrt{(6)^2+(-1)^2}}$

$d=\frac{|6x-y|}{\sqrt{37}}$

replacing the point (0,8)

$d=\frac{|6*0-8|}{\sqrt{37}}$

$d=\frac{8}{\sqrt{37}}$

$d^2=\left(\frac{8}{\sqrt{37}}\right)^2$

$\boxed{\boxed{d^2=\frac{64}{37}\approx1.73}}$

Answer 2

The distance between 2 parallel lines is constans.

The formula of distance between 2 parallel lines:

$k:Ax+By+C_1=0\ and\ l:Ax+By+C_2=0\\\\d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$


$A=6;\ B=1\\\\C_1-C_2=8\\\\then:\\\\d=\frac{|8|}{\sqrt{6^2+1^2}}=\frac{8}{\sqrt{36+1}}=\frac{8}{\sqrt{37}}\\\\Answer:d^2=\left(\frac{8}{\sqrt{37}}\right)^2=\frac{64}{37}=1\frac{27}{37}\approx1.73$