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)=(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=(|ax+by+c|)/(√(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=(|6x-y+0|)/(√((6)^2+(-1)^2))`


`d=(|6x-y|)/(√(37))`

replacing the point (0,8)


`d=(|6*0-8|)/(√(37))`


`d=(8)/(√(37))`


`d^2=\left((8)/(√(37))\right)^2`


`\boxed{\boxed{d^2=(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=(|C_1-C_2|)/(√(A^2+B^2))`



`A=6;\ B=1\\\\C_1-C_2=8\\\\then:\\\\d=(|8|)/(√(6^2+1^2))=(8)/(√(36+1))=(8)/(√(37))\\\\Answer:d^2=\left((8)/(√(37))\right)^2=(64)/(37)=1(27)/(37)\approx1.73`