Respuesta :

W0lf93
B is the correct answer. To find out how long AC is, you must first find the length of AB. To do that, form a triangle by finding the point where perpendicular lines from points A and B meet. According to Pythagoras’ theorem, a^2 + b^2 = c^2. Thus 16^2 + 8^2 = c^2. 256 + 64 = c^2. The square root of this is 17.9. AB is therefore 17.9 units long. 3/4 of 17.9 is 13.4 AC is 13.4 units long. The co ordinate that fits this is (2,0).

The correct answer is:


B) (2,0)


Explanation:


First we find the distance of AB using the distance formula:

[tex] d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2} [/tex]


Using the coordinates of A and B, we have:

[tex] d=\sqrt{(2--6)^2+(6--10)^2}=\sqrt{8^2+16^2}
\\
\\=\sqrt{64+256}=\sqrt{320} [/tex]


To simplify this radical, we find the prime factorization of 320:

320 = 10*32

10 = 5*2

32 = 16*2

16 = 2*8

8 = 2*4

4 = 2*2

320 = 2*2*2*2*2*2*5


For a square root, we want pairs of factors. There are 3 pairs of 2's, so we take 3 2's out of the radical and leave the 5 in:

[tex] 2*2*2\sqrt{5}=8\sqrt{5} [/tex]


We want the length of AC to be 3/4 of the length of AB:

[tex] \frac{3}{4}\times 8\sqrt{5} = \frac{24}{4}\sqrt{5}=6\sqrt{5} [/tex]


If the coordinates of B are (0, 2), using these and the coordinates of A in the distance formula gives us:

[tex] d=\sqrt{(0--6)^2+(2--10)^2}=\sqrt{6^2+12^2}=\sqrt{36+144}
\\
\\=\sqrt{180} [/tex]


To simplify this, we find the prime factorization of 180:

180 = 10*18

10 = 5*2

18 = 2*9

9 = 3*3

180 = 3*3*2*2*5


We want pairs. We have a pair of 3's and a pair of 2's; this means a 3 and a 2 come out and the 5 stays in:

[tex] 3*2\sqrt{5} = 6\sqrt{5} [/tex]


This is the correct length.