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

2 Answers
Feb 1, 2018

34/5.

Explanation:

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.

Feb 1, 2018

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

Explanation:

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]}