Hi, there--
The line through the origin with a slope of 7 has the equation y = 7x.
The line intersects the unit circle at two points exactly i unit from the origin. One point is in
Quadrant I, the other is in Quadrant III.
Quadrant 1
Call the intersection point (n, 7n).
Draw a right triangle with a horizontal leg of length n and a vertical leg of length 7n. The
hypotenuse of the triangle lies between the origin and the point (n, 7n).
Use the Pythagorean Equation to find the value of n when the hypotenuse is 1 unit long.
a^2 + b^2 = c^2
(n)^2 + (7n)^2 = (1)^2
n^2 + 49n^2 = 1
50n^2 = 1
n^2 = 1/50
n = 1/(sqrt(50)) = (sqrt(50))/50 = 5(sqrt(2))/50 = (sqrt(2))/10
The point (n, 7n) in Quadrant I is ((sqrt(2))/10, 7(sqrt(2))/10).
The intersection point in Quadrant III has the same values with opposite signs since x and y in
Quadrant III are both negative. Thus the point is (-(sqrt(2))/10, -7(sqrt(2))/10).
If this is unclear, or you still have questions, you may email me.
Mrs. Figgy
math.in.the.vortex@gmail.com
