Point A is the terminal point of angle phi and point B is the terminal point of angle theta. Point A is in the first quadrant and point B is in the second quadrant, while tan phi = 1 and tan theta = -7. Find the slope of AB.
Ok so since A is in the first quadrant and tan phi = 1, thus A must be point (sqrt2 / 2, sqrt2 / 2).
B is in the second quadrant but where?
Thanks
To find the coordinates of point B, we can use the information that tan(theta) = -7. Since tangent is negative in the second quadrant, we know that the y-coordinate of point B will be negative. The x-coordinate can be determined using the fact that tangent is the ratio of the y-coordinate to the x-coordinate.
Let's assume the x-coordinate of point B is x. Then, the y-coordinate of point B will be -7x.
Since point B is in the second quadrant, the x-coordinate will be negative. Therefore, we can say that x < 0.
Now we can use the given information to write an equation. We know that the tangent of angle theta is -7, which can be expressed as:
tan(theta) = y-coordinate of point B / x-coordinate of point B = -7
Replacing y-coordinate with -7x in the equation, we get:
-7 = (-7x) / x
Simplifying the equation:
-7 = -7
This equation holds true for any value of x because the left side is equal to the right side. Therefore, we can say that there are infinitely many possible values for the x-coordinate of point B.
Since we cannot determine the exact value of the x-coordinate of point B using the given information, we cannot determine the exact location of point B. Therefore, we cannot find the slope of AB.
for A, tan phi = 1 = 1/2
so point A is (1,1)
since tanØ = -7 = -7/1 and Ø is in II
then point B is (-7,1)
so slope of AB = (1-1)/(-7-1) = 0
BUT, this is a poorly worded question
What if point A is (5,5)
notice that tan phi = 5/5 = 1
You were also correct in saying A is (√2/2, √2/2)
as a matter of fact , A could be any point on the line
y = x
Same thing is true for B,
B could also be (-14,2) , etc
Really silly question, slope AB could be anything