JL¯¯¯¯¯¯¯ is a perpendicular bisector of KM¯¯¯¯¯¯¯¯¯¯ . Hallie locates point H along line segment JL¯¯¯¯¯¯¯ . If the length of KH¯¯¯¯¯¯¯¯¯ is given by 6y+3 and the length of HM¯¯¯¯¯¯¯¯¯¯ is given by 2y+7 , what is the length in units of segment HM¯¯¯¯¯¯¯¯¯¯ ?

Question

A. 9
B. 8
C. 1
D. 18

Answer 1

Since JL is a perpendicular bisector of KM, it means that point H is the midpoint of segment KM. Thus, we have:

\[ KH = HM \]

You are given:

\[ KH = 6y + 3 \] \[ HM = 2y + 7 \]

Since \(KH = HM\), we can set the two expressions equal to each other:

\[ 6y + 3 = 2y + 7 \]

Now, we will solve for \(y\):

1. Subtract \(2y\) from both sides:

\[ 6y - 2y + 3 = 7 \] \[ 4y + 3 = 7 \]

2. Subtract 3 from both sides:

\[ 4y = 7 - 3 \] \[ 4y = 4 \]

3. Divide by 4:

\[ y = 1 \]

Now, we can substitute this value of \(y\) back into the equation for \(HM\) to find the length of segment \(HM\):

\[ HM = 2y + 7 \] \[ HM = 2(1) + 7 \] \[ HM = 2 + 7 \] \[ HM = 9 \]

Thus, the length of segment \(HM\) is:

\[ \boxed{9} \]