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

1
1

18
18

8
8

9

Answer 1

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

\[ KH = HM \]

Given that \( KH = 6y + 3 \) and \( HM = 2y + 7 \), we can set up the equation:

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

To solve for \( y \), we first subtract \( 2y \) from both sides:

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

This simplifies to:

\[ 4y + 3 = 7 \]

Next, we subtract 3 from both sides:

\[ 4y = 4 \]

Dividing both sides by 4 gives us:

\[ y = 1 \]

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

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

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

\[ \boxed{9} \]