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# Trig

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Find the points where the line goes through the origin with slope -2 intersects the Unit circle. Give exact answers

Guest Apr 9, 2017

#2
+5897
+2

The coordinates of the point that intersect the unit circle are (x,y)

According to the Pythagorean theorem:

x2 + y2 = 12

x2 + y2 = 1

According to the problem statement:

y/x = -2

So,

y = -2x

Substitute:

x2 + (-2x)2 = 1

x2 + 4x2 = 1

5x2 = 1

x2 = 1/5

x = ±√(1/5)

x = ±√(5) / 5

y = -2[ ±√(5) / 5 ]

y = ∓2√(5) / 5

The points are: $$(\frac{\sqrt5}{5},-\frac{2\sqrt5}{5})$$    and    $$(-\frac{\sqrt5}{5},\frac{2\sqrt5}{5})$$

hectictar  Apr 9, 2017
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#1
+80913
+2

The line has the equation  y = -2x      (1)

The circle has the equation  x^2 + y^2  = 1     (2)

Sub (1)  into (2)

x^2 + (-2x)^2   = 1

x^2 + 4x^2  = 1

5x^2  =  1           divide both sides by 5

x^2  =   1/5           take both roots

x  = ± 1/√5

And using (1)   when   x  = 1/√5,   y  = -2/√5

And when  x = -1/√5,  y  =  -2 (-1/ √5)  =   2/√5

So.....the intersection points are   ( 1/√5, -2 /√5 )   and   ( -1/√5,  2/√5)

CPhill  Apr 9, 2017
#2
+5897
+2

The coordinates of the point that intersect the unit circle are (x,y)

According to the Pythagorean theorem:

x2 + y2 = 12

x2 + y2 = 1

According to the problem statement:

y/x = -2

So,

y = -2x

Substitute:

x2 + (-2x)2 = 1

x2 + 4x2 = 1

5x2 = 1

x2 = 1/5

x = ±√(1/5)

x = ±√(5) / 5

y = -2[ ±√(5) / 5 ]

y = ∓2√(5) / 5

The points are: $$(\frac{\sqrt5}{5},-\frac{2\sqrt5}{5})$$    and    $$(-\frac{\sqrt5}{5},\frac{2\sqrt5}{5})$$

hectictar  Apr 9, 2017

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