# How do you find the distance from A (-2,-2) to the line joining B(5,2) and c(-1,4)?

Dec 12, 2016

$d = \frac{19}{\sqrt{10}} = \frac{19 \sqrt{10}}{10}$

#### Explanation:

The equation of the line BC is obtained by the following formula:

$\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}} = \frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}$

where you state B=(x_1;y_1),C=(x_2;y_2)

Then

$\frac{y - 2}{4 - 2} = \frac{x - 5}{- 1 - 5}$

$\frac{y - 2}{2} = \frac{5 - x}{6}$

$3 y - 6 = 5 - x$

$\textcolor{red}{\left(1\right)}$ $x + 3 y - 11 = 0$

in the form

$a x + b y + c = 0$

Then you can find the requested distance by the following formula:

$d = \frac{| a {x}_{0} + b {y}_{0} + c |}{\sqrt{{a}^{2} + {b}^{2}}}$

where A(x_0;y_0)=(-2;-2) is the given point and

a=1;b=3;c=-11 the features of the line BC $\textcolor{red}{\left(1\right)}$

Then the requested distance is:

$d = | 1 \cdot \left(- 2\right) + 3 \cdot \left(- 2\right) - 11 \frac{|}{\sqrt{{1}^{2} + {3}^{2}}}$

$d = | - 2 - 6 - 11 \frac{|}{\sqrt{10}}$

$d = \frac{19}{\sqrt{10}} = \frac{19 \sqrt{10}}{10}$