triangle ABC has vertices at A(1,-3),B(3,0), and C(5,-3). what is the distance from vertex B to the midpoint of AC

The distance from vertex B to the midpoint of AC is 3.

Step-by-step explanation:

From Linear Algebra we understand that location of the midpoint of AC is determined by the following formula:

(1)

Where:

, - Locations of vertices A and C regarding origin, dimensionless.

- Location of the midpoint regarding origin, dimensionless.

If we know that and , then the midpoint of AC is:

Lastly, the distance from vertex B to the midpoint of AC is calculated from the Pythagorean Theorem:

(2)

If we know that , , and , then the distance is:

The distance from vertex B to the midpoint of AC is 3.

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

42

step-by-step explanation: