Finding length and midpoint help, please

1) What is the length of side AB in the triangle below?
Diagram - http://oi49.tinypic.com/1z2jaet.jpg
a. 19.21
b. 9
c. 3.87
d. 4.12

2) What is the midpoint of TS?
Diagram - http://oi46.tinypic.com/14v1btl.jpg
a. (-1, -3)
b. (-2, -6)
c. (-4, 6)
d. (11, -21)

Thank you.

1) √(1² + 4²) = √17 ≈ 4.123 (d)

2) [(-5+3)/2, (-9+3)/2] = (-1, -3) (a)

According to first diagram, co-ordinate of point A is (-8,-4) and B is (-7,-8)
therefore, distanca AB
=square rt of [{(-8)-(-7)}^2 + {(-4)-(-8)}^2]
=square root of [{-1}^2 + {+4}^2]
=square root of [17]
=4.12


According to second diagram T is (-5,3), and S is (3,-9).
Formula of midpoint of two points is {(x1 + x2)/2, (y1+y2)/2}
Therefore midpoint of TS is,
= (-5+3)/2, (3+(-9))/2
= (-1, -3)
i hope you got it

1) Make a slope triangle of AB. It will be 4 units down and 1 unit across. Then, find the length of AB using pythagorean theorem.

4^2 + 1^2 = x^2
16 + 1 = x^2
17 = x^2

D. 4.12

2) T = (-5, 3)
S = (3, -9)

Midpoint Formula: (x1 + x2/2 , y1 + y2/2)
(-5 + 3/2, 3 + (-9)/2)

A. (-1, -3)

1. By doing the pythagorean theorem (you may also use distance formula),
4^2 + 1^2 = c^2 = sqrt(17) = 4.12, so the answer is D.

2. Since the midpoint formula is (x1+x2)/2 , (y1+y2)/2,
(-5+3)/2 = -1
(3 - 9)/2 = -3

Answer: (-1, -3)