Answer:
7. C. 6
8. H. √34
9. A. (1, 3.5)
10. J. 10
Step-by-step explanation:
7. AB = 2y, BC = 6y, AC = 48
AB + BC = AC (segment addition theorem)
Substitute the above values into the equation
2y + 6y = 48
Solve for y
8y = 48
Divide both sides by 8
8y/8 = 48/8
y = 6
8. Distance between P(2, 8) and Q(5, 3):
[tex] PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
Let,
[tex] P(2, 8) = (x_1, y_1) [/tex]
[tex] Q(5, 3) = (x_2, y_2) [/tex]
[tex] PQ = \sqrt{(5 - 2)^2 + (3 - 8)^2} [/tex]
[tex] PQ = \sqrt{(3)^2 + (-5)^2} [/tex]
[tex] PQ = \sqrt{9 + 25} [/tex]
[tex] PQ = \sqrt{34} [/tex]
9. Midpoint (M) of segment LB, for L(8, 5) and B(-6, 2) is given as:
[tex] M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) [/tex]
Let [tex] L(8, 5) = (x_1, y_1) [/tex]
[tex] B(-6, 2) = (x_2, y_2) [/tex]
Thus:
[tex] M(\frac{8 + (-6)}{2}, \frac{5 + 2}{2}) [/tex]
[tex] M(\frac{2}{2}, \frac{7}{2}) [/tex]
[tex] M(1, 3.5) [/tex]
10. M = -10, N = -20
Distance between M and N, MN = |-20 - (-10)|
= |-20 + 10| = |-10|
MN = 10
