Answer:
√65 or 8.06 units
Step-by-step explanation:
Use the distance formula to find the length of both these congruent segments. You first get the coordinate points of the starting and ending point of the line and then plug it into the distance formula to get the length.
[tex]\fbox{% \parbox{\textwidth}{% \textbf{The distance formula:} \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Where: \begin{itemize} \item \( d \) = distance between the two points \item \( (x_1, y_1) \) = coordinates of the first point \item \( (x_2, y_2) \) = coordinates of the second point \end{itemize} }%}[/tex]
Solving:
[tex]\section*{Coordinates:}\begin{itemize} \item \( A = (-1, 5) \) \item \( B = (7, 4) \) \item \( J = (-2, 1) \) \item \( K = (6, 2) \)\end{itemize}[/tex]
[tex]\section*{\( AB \):}Using the distance formula:\[AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substitute the coordinates of \( A \) and \( B \):\[AB = \sqrt{(7 - (-1))^2 + (4 - 5)^2}\]\[AB = \sqrt{(7 + 1)^2 + (-1)^2}\]\[AB = \sqrt{8^2 + (-1)^2}\]\[AB = \sqrt{64 + 1}\]\[AB = \sqrt{65}\]\[\boxed{AB \approx 8.06}\][/tex]
Just to verify we can solve for JK as well (both are congruent).
[tex]\section*{\( JK \):}Using the distance formula:\[JK = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substitute the coordinates of \( J \) and \( K \):\[JK = \sqrt{(6 - (-2))^2 + (2 - 1)^2}\]\[JK = \sqrt{(6 + 2)^2 + 1^2}\]\[JK = \sqrt{8^2 + 1^2}\]\[JK = \sqrt{64 + 1}\]\[JK = \sqrt{65}\]\[\boxed{JK \approx 8.06}\][/tex]
If the answer is asking for exact use √65 for approximate use 8.06.
