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kap26 [50]
3 years ago
12

Find the midpoint of the segment with the given endpoints.

Mathematics
2 answers:
andrew-mc [135]3 years ago
7 0
To solve for the midpoint of the segment we use the equation that is given as:

(x1 + x2 / 2) , (y1 + y2 / 2)

For the points given,

(x1 + x2 / 2) , (y1 + y2 / 2)
(3+ 2 / 2) , (-5 + 9 / 2)
(5/2 , 2) or (2.5 , 2)

Hope this answers the question. Have a nice day. Feel free to ask more questions.
yanalaym [24]3 years ago
4 0

Answer:

The correct option is C.

Step-by-step explanation:

Given information: End point of a line segment are A(3, –5) and C(2, 9).

We to find the midpoint of line segment AC.

If the end point of a segment are (x_1,y_1) and (x_2,y_2), then the midpoint of that segment is

Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Midpoint=(\frac{3+2}{2},\frac{-5+9}{2})

On simplification, we get

Midpoint=(\frac{5}{2},\frac{4}{2})

Midpoint=(2.5,2)

The midpoint of given segment is (2.5,2). Therefore the correct option is 3.

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Answers: 
33. Angle R is 68 degrees
35. The fraction 21/2 or the decimal 10.5
36. Triangle ACG
37. Segment AB
38. The values are x = 6; y = 2
40. The value of x is x = 29
41. C) 108 degrees
42. The value of x is x = 70
43. The segment WY is 24 units long
------------------------------------------------------
Work Shown:
Problem 33) 
RS = ST, means that the vertex angle is at angle S
Angle S = 44
Angle R = x, angle T = x are the base angles
R+S+T = 180
x+44+x = 180
2x+44 = 180
2x+44-44 = 180-44
2x = 136
2x/2 = 136/2
x = 68
So angle R is 68 degrees
-----------------
Problem 35) 
Angle A = angle H
Angle B = angle I
Angle C = angle J
A = 97
B = 4x+4
C = J = 37
A+B+C = 180
97+4x+4+37 = 180
4x+138 = 180
4x+138-138 = 180-138
4x = 42
4x/4 = 42/4
x = 21/2
x = 10.5
-----------------
Problem 36) 
GD is the median of triangle ACG. It stretches from the vertex G to point D. Point D is the midpoint of segment AC
-----------------
Problem 37)
Segment AB is an altitude of triangle ACG. It is perpendicular to line CG (extend out segment CG) and it goes through vertex A.
-----------------
Problem 38) 
triangle LMN = triangle PQR
LM = PQ
MN = QR
LN = PR
Since LM = PQ, we can say 2x+3 = 5x-15. Let's solve for x
2x+3 = 5x-15
2x-5x = -15-3
-3x = -18
x = -18/(-3)
x = 6
Similarly, MN = QR, so 9 = 3y+3
Solve for y
9 = 3y+3
3y+3 = 9
3y+3-3 = 9-3
3y = 6
3y/3 = 6/3
y = 2
-----------------
Problem 40) 
The remote interior angles (2x and 21) add up to the exterior angle (3x-8)
2x+21 = 3x-8
2x-3x = -8-21
-x = -29
x = 29
-----------------
Problem 41) 
For any quadrilateral, the four angles always add to 360 degrees
J+K+L+M = 360
3x+45+2x+45 = 360
5x+90 = 360
5x+90-90 = 360-90
5x = 270
5x/5 = 270/5
x = 54
Use this to find L
L = 2x
L = 2*54
L = 108
-----------------
Problem 42) 
The adjacent or consecutive angles are supplementary. They add to 180 degrees
K+N = 180
2x+40 = 180
2x+40-40 = 180-40
2x = 140
2x/2 = 140/2
x = 70
-----------------
Problem 43) 
All sides of the rhombus are congruent, so WX = WZ.
Triangle WPZ is a right triangle (right angle at point P).
Use the pythagorean theorem to find PW
a^2+b^2 = c^2
(PW)^2+(PZ)^2 = (WZ)^2
(PW)^2+256 = 400
(PW)^2+256-256 = 400-256
(PW)^2 = 144
PW = sqrt(144)
PW = 12
WY = 2*PW
WY = 2*12
WY = 24
3 0
3 years ago
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