What is the distance between (-4, 6) and  (3,7) ?

Mar 10, 2018

The distance between the two points is $5 \sqrt{2}$.

Explanation:

Using the distance formula,

$\textcolor{w h i t e}{\implies} d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

we can plug in our points,

$\textcolor{w h i t e}{\implies} {P}_{1} = \left(- 4 , 6\right)$

$\textcolor{w h i t e}{\implies} {P}_{2} = \left(3 , 7\right)$

and solve for the distance:

$\textcolor{w h i t e}{\implies} d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

$\implies d = \sqrt{{\left(\textcolor{red}{- 4 - 3}\right)}^{2} + {\left(\textcolor{b l u e}{6 - 7}\right)}^{2}}$

$\textcolor{w h i t e}{\implies d} = \sqrt{{\left(\textcolor{red}{-} \textcolor{red}{7}\right)}^{2} + {\left(\textcolor{b l u e}{-} \textcolor{b l u e}{1}\right)}^{2}}$

$\textcolor{w h i t e}{\implies d} = \sqrt{\textcolor{red}{49} + \textcolor{b l u e}{1}}$

$\textcolor{w h i t e}{\implies d} = \sqrt{\textcolor{p u r p \le}{50}}$

$\textcolor{w h i t e}{\implies d} = \sqrt{\textcolor{p u r p \le}{25 \cdot 2}}$

$\textcolor{w h i t e}{\implies d} = \sqrt{\textcolor{p u r p \le}{{5}^{2} \cdot 2}}$

$\textcolor{w h i t e}{\implies d} = \sqrt{\textcolor{p u r p \le}{{5}^{2}}} \cdot \sqrt{\textcolor{p u r p \le}{2}}$

$\textcolor{w h i t e}{\implies d} = \textcolor{p u r p \le}{5} \cdot \sqrt{\textcolor{p u r p \le}{2}}$

$\textcolor{w h i t e}{\implies d} = \textcolor{p u r p \le}{5} \sqrt{\textcolor{p u r p \le}{2}}$

Mar 10, 2018

$5 \sqrt{2} \approx 7.07 \text{ to 2 dec. places}$

Explanation:

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

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(-4,6)" and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 7\right)$

$d = \sqrt{{\left(3 - \left(- 4\right)\right)}^{2} + {\left(7 - 6\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{49 + 1} = \sqrt{50} = 5 \sqrt{2} \approx 7.07$