Given the 3 points A(2,16), B(32,-64), C(8,0).?

Question

Find the distance between the midpoints of line AC and line CB. Show work

Answer 1

ah goodness me , am last one again .

■ Step 1
Midpoint of AC =( x A + x C ) / 2 = 2 + 8 / 2 = 10 / 2 = + 5
Midpoint of AC = ( y A + y C ) / 2 = 16 + 0 / 2 = 16 / 2 = +8
Let X1 = + 5
Let Y1 = + 8
( 5 , 8 )

■ Step 2
Midpoint of CB = ( x C + x B ) / 2 = 8 + 32 / 2 = 40 / 2 = +20
Midpoint of CB = ( y C + y B ) / 2 = 0 - 64 / 2 = -64 / 2 = - 32
Let X2 = + 20
Let Y2 = - 32
( 20 , -32 )

■ Step 3
Distance Formula = √ [ ( x2 - x1 ) ² + ( y2 - y1 )² ]
D = √ [ ( 20 -5) ² + ( -32 - 8 )² ]
D = √ [ ( 15 ) ² + ( -40 )² ]
D = √ [ 225 + 1600 ]
D = √ [1825 ]
D = + 42.72

Good Luck dear ♣

Answer 2

First find the midpoints. The coordinates for the midpoints of a line segment would be (x1+x2)/2 , (y1 + y2)/2. For |AC, that would be (5,8). For |CB that would be (20, -32).

Next use the distance formula to determine the linear distance between the two midpoints. First, check for the easy cases: same x values (a vertical line), or same y values (a horizontal line). If you got lucky the distance is just |y1-y2| for vertical lines, or |x1-x2| for horizontals. Notice that a distance an absolute or "scalar" value, so we take the absolute value of the difference.

Didn't get lucky? No sweat. If you don't remember the distance formula, you can derive it by envisioning the line segment for which you want the distance as the hypoteneuse of a right triangle. Make the right triangle by droping a vertical line down from the higher point, and a horizontal line across from the remaining point. Recall that the length of the hypotenuese is easily calculated using the Pythagorean theorem: A^2 + B^2 = C^2, where C is the hypotenuese. Solving for C: C= (A^2 + B^2)^0.5. You can arbitrarily assign A and B -- they will work either way. Find A as the absolute value of the difference of the x's: in this case |5 - 20| = 15. Find B as the absolute value of the difference of the y's: |8 - (-32)| = 40.

Substituting these values for the Pythagorean solution, you get:

(15^2 + 40^2)^0.5 ~~ 42.72

Answer 3

Mpt of AC = [(2+8)/2, (16+0)/2]=[5, 8]
Mpt of CB = [(32+8)/2, (-64+0)/2]=[20,-32]
D^2=(20-5)^2+(-32-8)^2=225+1600=1825; D^2=42.72^2; D=42.72
what's remarkable! pt C belongs to line AB, that is all 3 points are on the same line!

Answer 4

use midpoint formula... 2 point of x / 2
2 points of y/ 2

n then distance formula

for for AC find midpoint 2+ 8/ 2 = 5
16+0 / 2= 8
midpoint of AC= 5,8

CB midpoint
32+8 /2 = 20
-64+ 0 /2= -32
midpoint of CB= 20, -32

distane formula= y(2) minus y(1)/ x(2) minus X(1)
so -32 minus 8 / 20 minus 5
= -40/15
= -2.67

Answer 5

The midpoint of AC is (5,8)

The midpoint of CB is (20, -32)

The distance between them is the square root of
(20-5)^2 + (-32-8)^2
which is the square root of 225 + 1600
which is the square root of 1825 which is 42.7