How would I find the distance?

Question

In a coordinate plane, I want to find the distance from the line y = -x to the point (2,72). I know that this is the perpendicular bisector between them so how do I find the distance of the perpendicular bisector which is the same as finding the distance between the line and the point given?

Answer 1

The line y=-x can be written as
x+y+0=0
equivalent to A=1,B=1,C=0 in the standard equation
Ax+By+C=0
the distance of a point (x,y) from this line is
D=(Ax+By+C)/sqrt(A*A+B*B)
=(1*2+1*72)/(sqrt(1*1+1*1)
=74/sqrt(2)
=52.3259...

Answer 2

The formula for the line says that the value of Y will be positive and equal to the X value which is a negative value. So for X = to -1, -2, -3...-10 Y will be 1, 2, 3...10 etc which makes the line a sloping line rising to the upper left ( I've forgotten the proper terms for the quadrants). If (2,72) represents a point on the perpendicular bisector, then you will need to determine which point on the line will produce a right angle with the point (2,72). I know it can be done mathematically but I have forgotten so you may want to resort to drawing the graph and working it out by sight unless (and it should have been) it was discussed in the text. Once the point is determined, the distance is simply (x2,y2) - (x1,y1).
good luck and I hope I haven't confused you.

Answer 3

Another way you could do it is find a perpendicular thru the point (2.72) that crosses the line (y = -x). Since you want a line perpendicular to that point, it would be a line with slope 1. Solve the system with this line and the given line to find the point of intersection, then use the distance formula to find the distance between (2,72) and the point of intersection.

Answer 4

I agree with the first answer. Another way can be this one:

y = x+70 is the perpendicular through the point (2,72)
y= -x

2y = 70
y = 35

x = -35

The point (-35,35) is the crosspoint. Now find the distance between the both points using the Pythagoras theorem.

Ana