[tex] \large \bold{SOLUTION:} [/tex]
Location of each house on the coordinate plane:
- Diana's house is at [tex](-2, 2).[/tex]
- Jolina's house is at [tex](6, 8).[/tex]
- Patricia's house is at [tex](5, 2).[/tex]
Each unit on the grid is equivalent to 2 km.
Required: Distance between Diana's and Jolina's house and the location of your house midway between Jolina’s and Patricia’s house
Distance between Diana's and Jolina's house:
By Distance Formula,
[tex] \quad \:\: \qquad (\overset{x_1}{-2}, \overset{y_1}{2}) \longrightarrow (\overset{x_2}{6}, \overset{y_2}{8}) [/tex]
[tex] \begin{aligned} \qquad d &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &= \sqrt{(6 - (-2))^2 + (8 - 2)^2} \\ d &= \sqrt{8^2 + 6^2} \\ d &=\sqrt{64 + 36} \\ d &= \sqrt{100} \\ d &= 10\:\cancel{\textsf{units}}\left(\frac{2\textsf{ km}}{\cancel{1\textsf{ unit}}}\right) \\ d &= \boxed{20\textsf{ km}}\quad\textit{Answer} \end{aligned} [/tex]
Location of your house midway between Jolina’s house at [tex](\overset{x_1}{6}, \overset{y_1}{8})[/tex] and Patricia’s house at [tex](\overset{x_2}{5}, \overset{y_2}{2})[/tex]:
By Midpoint Formula,
[tex] \begin{array}{l} \qquad M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \\ \\ \qquad M = \left(\dfrac{6 + 5}{2}, \dfrac{8 + 2}{2}\right) \\ \\ \qquad M = \boxed{\left(\frac{11}{2}, 5\right)}\quad\textit{Answer} \end{array} [/tex]
[tex] \blue{\mathfrak{\#CarryOnLearning}} [/tex]
