## anonymous 5 years ago Find the point(s) where the line through the origin with slope 7 intersects the unit circle.

1. asnaseer

a unit circle has the equation:$x^2+y^2=1$a line through the origin with slope 7 would have the equation:$y=7x$

2. asnaseer

substitute one into the other and solve for x.

3. asnaseer

do you understand?

4. asnaseer

|dw:1326759365741:dw|

5. asnaseer

do you need more help or can you solve this from here?

6. anonymous

I'm still a little confused

7. anonymous

i understand what to do, but why do y=7x?

8. 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}}$

9. asnaseer

y=7x is the equation of a straight line through the origin with slope 7.

10. 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.

11. asnaseer

does it make sense?

12. 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).

13. anonymous

so it would be 7*1/$\sqrt{50}$ and then 7*-1/$\sqrt{50}$

14. asnaseer

correct, so the two values of y will be:$y=\pm\frac{7}{\sqrt{50}}$

15. anonymous

what would x equal?

16. anonymous

1?

17. asnaseer

we found x above

18. asnaseer

$x=\pm\frac{1}{\sqrt{50}}$

19. anonymous

oh right okay. this is starting to make sense

20. asnaseer

so the two points where the line intersects the circle are:$(-\frac{1}{\sqrt{50}}, -\frac{7}{\sqrt{50}})\quad\text{and}\quad(\frac{1}{\sqrt{50}}, \frac{7}{\sqrt{50}})$

21. anonymous

thank you so so much! you are a saint

22. asnaseer

you are very welcome - I'm glad I was able to help :-)