The point P(–4, 4) that is of the way from A to B on the directed line segment AB.
Solution:
The points of the line segment are A(–8, –2) and B(6, 19).
P is the point that bisect the line segment in .
So, m = 2 and n = 5.
By section formula:
P(x, y) = (–4, 4)
Hence the point P(–4, 4) that is of the way from A to B on the directed line segment AB.
This is a doozy so pay attention. First thing you have to recognize is that is a trig identity, the sum identity for cosine to be exact. That formula is
.
So that means you need to find the cosine of alpha and the sin of beta. We are given the sin of alpha being 4/5 in the first quadrant. If you set up the sin ratio which is side opposite over hypotenuse of a right triangle, you put the 4 on the side opposite the reference angle, alpha, and the hypotenuse of 5 on the terminal ray and then you have to find the missing side of the right triangle you created. Using Pythagorean's theorem, you find that the missing side, which is the adjacent side to alpha, is 3. Now you can find the cosine of alpha as well, since cosine is the side adjacent, 3, over the hypotenuse, 5. So far wwe have
and
for that first angle. Now moving on to the second angle. The cosine of beta is side adjacent, 5, over the hypotenuse, 13, and we are missing the side opposite the reference angle beta. Using Pythagorean's theorem again to find the side opposite, we have that that side measures 12. Now we can find the sine of beta using that opposite side, 12, over the hypotenuse, 13. What we have now is
and
. According to the identity, we have to multiply those ratios now:
.
When you do that you get
.
Of course in order to subtract those 2 fractions you need a common denominator which is 65 so
which gives you a final answer of
which is the first choice given above. There you go!
