Question

Question
B is the midpoint of AC¯¯¯¯¯¯¯¯
and D is the midpoint of CE¯¯¯¯¯¯¯¯
. Solve for BD given that BD = 2y + 8 and AE = 6y +12
Make sure to show all work in order to earn full credit for this answer

Answer 1

To solve for BD, we need to find the values of y, AC, and CE first.

Since B is the midpoint of AC, we can use the midpoint formula:

B = (A + C)/2

Given that B = D, we have:

(A + C)/2 = D

Substituting the values of D and B, we get:

(A + C)/2 = 2y + 8

Multiplying both sides by 2 to eliminate the fraction, we have:

A + C = 4y + 16

Since D is the midpoint of CE, we can use the midpoint formula again:

D = (C + E)/2

Substituting the values of D and B, we get:

(C + E)/2 = 2y + 8

Multiplying both sides by 2, we have:

C + E = 4y + 16

Now we have two equations:

A + C = 4y + 16
C + E = 4y + 16

Adding these two equations together, we get:

(A + C) + (C + E) = 8y + 32

Simplifying, we have:

A + 2C + E = 8y + 32

Since B is the midpoint of AC, we can also use the midpoint formula to find the value of AC:

B = (A + C)/2

Substituting B = D, we have:

(A + C)/2 = (C + E)/2

Multiplying both sides by 2, we get:

A + C = C + E

Subtracting C from both sides, we get:

A = E

Therefore, we can substitute A for E in the equation A + 2C + E = 8y + 32:

A + 2C + A = 8y + 32

Simplifying, we have:

2A + 2C = 8y + 32

Subtracting 2C from both sides, we get:

2A = 8y + 32 - 2C

So:

A = 4y + 16 - C

Now we substitute A = 4y + 16 - C into the equation A + 2C + A = 8y + 32:

2(4y + 16 - C) + 2C = 8y + 32

Simplifying, we have:

8y + 32 - 2C + 2C = 8y + 32

The 2C terms cancel out, so we're left with:

8y + 32 = 8y + 32

This is an identity, which means that the equation is true for all values of y. Therefore, we cannot determine a unique value for BD.

In conclusion, we know that BD = 2y + 8, but we cannot determine a specific value without additional information.