# What is the distance between (8, 2)  and  (4, -5) ?

Dec 29, 2015

$\text{Distance" = 8.06 " to 3 significant figures}$

#### Explanation:

$\Delta x = 8 - 4 = 4$

$\Delta y = 2 - \left(- 5\right) = 7$

${h}^{2} = \Delta {x}^{2} + \Delta {y}^{2}$

$h = \sqrt{\left(\Delta {x}^{2} + \Delta {y}^{2}\right)}$

$h = \sqrt{\left({4}^{2} + {7}^{2}\right)}$

$h = \sqrt{\left(16 + 49\right)}$

$h = \sqrt{65}$

$h = 8.062257748$

$h = 8.06 \text{ to 3 significant figures}$

Dec 29, 2015

$\text{line} \cong 8.06$

#### Explanation:

(8, 2) and (4, -5) are two points in a cartesian plane. The line represents the distance between the points. The size of the line can be calculated using Pythagoras' formula: ${\text{line"^2 = "difference in x"^2 + "difference in y}}^{2}$:
${\text{line}}^{2} = {4}^{2} + {7}^{2}$
${\text{line}}^{2} = 16 + 49$
$\text{line} = \sqrt{65}$
$\text{line} \cong 8.06$

Dec 29, 2015

$\sqrt{65}$

#### Explanation:

The distance formula for Cartesian coordinates is

d=sqrt((x_2-x_1)^2+(y_2-y_1)^2
Where ${x}_{1} , {y}_{1}$, and${x}_{2} , {y}_{2}$ are the Cartesian coordinates of two points respectively.
Let $\left({x}_{1} , {y}_{1}\right)$ represent $\left(8 , 2\right)$ and $\left({x}_{2} , {y}_{2}\right)$ represent $\left(4 , - 5\right)$.

$\implies d = \sqrt{{\left(\left(4 - 8\right)\right)}^{2} + {\left(- 5 - 2\right)}^{2}}$
implies d=sqrt((-4)^2+(-7)^2
$\implies d = \sqrt{16 + 49}$
$\implies d = \sqrt{65}$
$\implies d = \sqrt{65}$

Hence the distance between the given points is $\sqrt{65}$.