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

Question

1447 views around the world

You can reuse this answer
Creative Commons License

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#