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

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Answer 1 · 2016-12-12T14:03:17

$d=19/sqrt(10)=(19sqrt(10))/10$

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

$(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)$

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

Then

$(y-2)/(4-2)=(x-5)/(-1-5)$

$(y-2)/2=(5-x)/6$

$3y-6=5-x$

$color(red)((1))$ $x+3y-11=0$

in the form

$ax+by+c=0$

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

$d=(|ax_0+by_0+c|)/sqrt(a^2+b^2)$

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 $color(red)((1))$

Then the requested distance is:

$d=|1*(-2)+3*(-2)-11|/sqrt(1^2+3^2)$

$d=|-2-6-11|/sqrt(10)$

$d=19/sqrt(10)=(19sqrt(10))/10$