JL¯¯¯¯¯¯¯

Question

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¯¯¯¯¯¯¯¯¯¯
? (1 point)
Responses

8
8

1
1

18
18

9

Answer 1

Since \( JL \) is the perpendicular bisector of \( KM \), we have:

\[ KH = HM \]

Given the expressions for the lengths:

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

Setting these two expressions equal to each other since \( KH = HM \):

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

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

1. Subtract \( 2y \) from both sides: \[ 4y + 3 = 7 \]

2. Subtract 3 from both sides: \[ 4y = 4 \]

3. Divide both sides by 4: \[ y = 1 \]

Now substitute \( y = 1 \) back into the expression for \( HM \):

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

Thus, the length of segment \( HM \) is \( \boxed{9} \) units.