anonymous
  • anonymous
Find the point(s) where the line through the origin with slope 7 intersects the unit circle.
Mathematics
katieb
  • katieb
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asnaseer
  • asnaseer
a unit circle has the equation:\[x^2+y^2=1\]a line through the origin with slope 7 would have the equation:\[y=7x\]
asnaseer
  • asnaseer
substitute one into the other and solve for x.
asnaseer
  • asnaseer
do you understand?

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asnaseer
  • asnaseer
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asnaseer
  • asnaseer
do you need more help or can you solve this from here?
anonymous
  • anonymous
I'm still a little confused
anonymous
  • anonymous
i understand what to do, but why do y=7x?
asnaseer
  • asnaseer
ok, we know the equation of the straight line is \(y=7x\). so we can substitute this value of y into the first equation to get:\[x^2+(7x)^2=1\]\[x^2+49x^2=1\]\[50x^2=1\]\[x^2=\frac{1}{50}\]\[x=\pm\frac{1}{\sqrt{50}}\]
asnaseer
  • asnaseer
y=7x is the equation of a straight line through the origin with slope 7.
asnaseer
  • asnaseer
so now you have two values for x, substitute each one into the equation \(y=7x\) to get the corresponding value for y and you have then solved this question.
asnaseer
  • asnaseer
does it make sense?
asnaseer
  • asnaseer
the general equation of a straight line is given by:\[y=mx+c\]where 'm' is the slope and 'c' is the y-intercept. In your case, you are told that the slope is 7 (so m=7) and it passes through the origin (so y-intercept is zero, therefore c=0).
anonymous
  • anonymous
so it would be 7*1/\[\sqrt{50}\] and then 7*-1/\[\sqrt{50}\]
asnaseer
  • asnaseer
correct, so the two values of y will be:\[y=\pm\frac{7}{\sqrt{50}}\]
anonymous
  • anonymous
what would x equal?
anonymous
  • anonymous
1?
asnaseer
  • asnaseer
we found x above
asnaseer
  • asnaseer
\[x=\pm\frac{1}{\sqrt{50}}\]
anonymous
  • anonymous
oh right okay. this is starting to make sense
asnaseer
  • asnaseer
so the two points where the line intersects the circle are:\[(-\frac{1}{\sqrt{50}}, -\frac{7}{\sqrt{50}})\quad\text{and}\quad(\frac{1}{\sqrt{50}}, \frac{7}{\sqrt{50}})\]
anonymous
  • anonymous
thank you so so much! you are a saint
asnaseer
  • asnaseer
you are very welcome - I'm glad I was able to help :-)

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