Answer:
12.08 units.
Step-by-step explanation:
We have been given that line segments JK and JL in the xy-coordinate plane both have a common endpoint J (-4,11) and midpoints at [tex]M_1 (2,16)[/tex] and [tex]M_2 (-3,5)[/tex].
To find the distance between [tex]M_1[/tex] and [tex]M_2[/tex] we will use distance formula.
[tex]\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}[/tex]
[tex]x_2[/tex] and [tex]x_1[/tex] are x-coordinates of two points.
[tex]y_2[/tex] and [tex]y_1[/tex] are corresponding y-coordinates of two points.
Let [tex]M_2[/tex] our first point, then [tex]x_1=-3[/tex] and [tex]y_1=5[/tex].
Let [tex]M_1[/tex] will be our second point, then [tex]x_2=2[/tex] and [tex]y_2=16[/tex]
Upon substituting coordinates of point [tex]M_1[/tex] and [tex]M_2[/tex] in distance formula we will get,
[tex]\text{Distance}=\sqrt{(2--3)^{2}+(16-5)^{2}}[/tex]
[tex]\text{Distance}=\sqrt{(2+3)^{2}+(11)^{2}}[/tex]
[tex]\text{Distance}=\sqrt{(5)^{2}+(11)^{2}}[/tex]
[tex]\text{Distance}=\sqrt{25+121}[/tex]
[tex]\text{Distance}=\sqrt{146}[/tex]
[tex]\text{Distance}=12.0830459735945721\approx 12.08[/tex]
Therefore, distance between [tex]M_1[/tex] and [tex]M_2[/tex] is 12.08 units.
