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svet-max [94.6K]
3 years ago
11

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

Mathematics
2 answers:
lara [203]3 years ago
8 0
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}}

____________________________________________________

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}}

rewona [7]3 years ago
3 0
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
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