How far is the point P(3,-5) from the line passing through A(-1,1) and B(-4,-3)?

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Answer 1 · 2018-02-01T06:22:36

# 34/5#.

The eqn. of the line #AB# is given by,

#|(x,y,1),(-1,1,1),(-4,-3,1)|=0, i.e., 4x-3y+7=0#.

Hence, the distance #d# of #P(3,-5)# from the line #AB# is given by,

#d=|4(3)-3(-5)+7|/sqrt(4^2+(-3)^2)=34/5#.

Answer 2 · 2018-02-01T07:04:05

Point #P(3,-5)# is at a distance of #6.8# units from the line joining #A(-1,1)# and #B(-4,-3)#.

Equation of line joining two points #(x_1,y_1)# and #(x_2,y_2)# is

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

therefore equation of line joining #A(-1,1)# and #B(-4,-3)# is

#(y-1)/(-3-1)=(x-(-1))/(-4-(-1))#

or #(y-1)/-4=(x+1)/(-3)#

or #-3y+3=-4x-4# or #4x-3y+7=0#

As distance of a point #(p,q)# from a line #ax+by+c=0# is

#|(ap+bq+c)/sqrt(a^2+b^2)|#

Hence distance of point #P(3,-5)# from line joining #A(-1,1)# and #B(-4,-3)# i.e. #4x-3y+7=0# is

#|(4*3-3*(-5)+7)/sqrt(4^2+(-3)^2)|#

= #(12+15+7)/5|=34/5=6.8# units

graph{((x-3)^2+(y+5)^2-0.03)((x+1)^2+(y-1)^2-0.03)((x+4)^2+(y+3)^2-0.03)(4x-3y+7)=0 [-9.92, 10.08, -7.08, 2.92]}