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

Jun 19, 2017

$\sqrt{113} \text{ units "~~10.63 " units}$

#### 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 $A \left({x}_{1} , {y}_{1}\right)$ and $B \left({x}_{2} , {y}_{2}\right)$, is
$\vec{A B} = B - A$

$\implies \vec{A B} = \left(\begin{matrix}{x}_{2} - {x}_{1} \\ {y}_{2} - {y}_{1}\end{matrix}\right)$

So to find $\vec{J L}$ from points $J \left(2 , 4\right)$ and $L \left(- 6 , - 3\right)$ we would do the following steps:

$\vec{J L} = \left(\begin{matrix}- 6 - 2 \\ - 3 - 4\end{matrix}\right)$

$\implies \vec{J L} = \left(\begin{matrix}- 8 \\ - 7\end{matrix}\right)$

We have found the vector $\vec{J L}$. Now we need to find the length of the vector. To do this, use the following:

If $\vec{A B} = \left(\begin{matrix}x \\ y\end{matrix}\right)$

Then length of $\vec{A B} = | \vec{A B} | = \sqrt{{x}^{2} + {y}^{2}}$

Hence for JL:
$| \vec{J L} | = \sqrt{{\left(- 8\right)}^{2} + {\left(- 7\right)}^{2}}$

$| \vec{J L} | = \sqrt{64 + 49}$

$| \vec{J L} | = \sqrt{113} \text{ units "~~10.63 " units}$

Jun 19, 2017

$J L \approx 10.63 \text{ to 2 decimal places}$

#### Explanation:

$\text{to calculate the length use the "color(blue)"distance formula}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ are 2 points}$

$\text{the 2 points are } J \left(2 , 4\right) , L \left(- 6 , - 3\right)$

$\text{let } \left({x}_{1} , {y}_{1}\right) = \left(2 , 4\right) , \left({x}_{2} , {y}_{2}\right) = \left(- 6 , - 3\right)$

$d = \sqrt{{\left(- 6 - 2\right)}^{2} + {\left(- 3 - 4\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{64 + 49}$

$\textcolor{w h i t e}{d} = \sqrt{113} \leftarrow \textcolor{red}{\text{ exact value}}$

$\textcolor{w h i t e}{d} \approx 10.63 \text{ to 2 decimal places}$