Question

JKL has vertices at J(2, 4), K(2, -3), and L(-6, -3). What is the approximate length of line segment JL?

Answer 1

`sqrt(113) " units "~~10.63 " units"`

Explanation:

To find the length of a line segment from two points, we can form a vector and find the length of the vector.

The vector from two points `A (x_1,y_1)` and `B(x_2,y_2)`, is
`vec(AB)=B-A`

`=>vec(AB)=((x_2-x_1),(y_2-y_1))`

So to find `vec(JL)` from points `J(2,4)` and `L(-6,-3)` we would do the following steps:

`vec(JL)=((-6-2),(-3-4))`

`=>vec(JL)=((-8),(-7))`

We have found the vector `vec(JL)`. Now we need to find the length of the vector. To do this, use the following:

If `vec(AB)=((x),(y))`

Then length of `vec(AB)=|vec(AB)|=sqrt(x^2+y^2)`

Hence for JL:
`|vec(JL)|=sqrt((-8)^2+(-7)^2)`

`|vec(JL)|=sqrt(64+49)`

`|vec(JL)|=sqrt(113) " units "~~10.63 " units"`

Answer 2

`JL~~10.63" to 2 decimal places"`

Explanation:

`"to calculate the length use the "color(blue)"distance formula"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(2/2)|)))`
where `(x_1,y_1),(x_2,y_2)" are 2 points"`

`"the 2 points are " J(2,4),L(-6,-3)`

`"let " (x_1,y_1)=(2,4),(x_2,y_2)=(-6,-3)`

`d=sqrt((-6-2)^2+(-3-4)^2)`

`color(white)(d)=sqrt(64+49)`

`color(white)(d)=sqrt113larrcolor(red)" exact value"`

`color(white)(d)~~10.63" to 2 decimal places"`