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

12.08 units.

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}=√(25+121)`

`\text{Distance}=√(146)`

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

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

Answer 2

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 <=