Question

Line segments JK and JL in the xy-coordinate plane both have a common endpoint J (-4,11) and midpoints at M1 (2,16) and M2 (-3,5). What's the distance between M1 and M2?

Answer 1

Distance formula : sqrt ((x2 - x1)^2 + (y2 - y1)^2)
(2,16)...x1 = 2 and y1 = 16
(-3,5)...x2 = -3 and y2 = 5
time to sub and solve
d = sqrt ((-3 - 2)^2 + (5 - 16)^2)
d = sqrt ((-5^2) + (-11^2))
d = sqrt (25 + 121)
d = sqrt 146
d = 12.08 <=

Answer 2

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 $M_1 (2,16)$ and $M_2 (-3,5)$.

To find the distance between $M_1$ and $M_2$ we will use distance formula.

$\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

$x_2$ and $x_1$ are x-coordinates of two points.

$y_2$ and $y_1$ are corresponding y-coordinates of two points.    

Let $M_2$ our first point, then  $x_1=-3$ and  $y_1=5$.

Let $M_1$ will be our second point, then $x_2=2$ and $y_2=16$  

Upon substituting coordinates of point $M_1$ and $M_2$ in distance formula we will get,    

$\text{Distance}=\sqrt{(2--3)^{2}+(16-5)^{2}}$

$\text{Distance}=\sqrt{(2+3)^{2}+(11)^{2}}$

$\text{Distance}=\sqrt{(5)^{2}+(11)^{2}}$  

$\text{Distance}=\sqrt{25+121}$

$\text{Distance}=\sqrt{146}$

$\text{Distance}=12.0830459735945721\approx 12.08$  

Therefore, distance between $M_1$ and $M_2$ is 12.08 units.