Distance between two points

Question

A point P is equidistant from R(-2,4) and S(6,-4) and its x-coordinate is twice its y coordinate. Find the coordinates of P

Answer 1

The point to be found must lie of the perpendicular bisector of the line joining the two points.

The line joining the 2 points is: (-4 - 4)/(6+2) = -1 = (y - 4)/(x+2), i.e. y = -x + 2

To find the midpoint of the line average the x-coordinates (2) and average the y co-ordinates (0), i.e.

the midpoint is (2,0)

The slope of the perpendicular bisector is 1 (negative reciprocal of the slope of the original line).

The equation of the perpendicular bisector is 1 = y/(x-2) and the point on this line with co-ordinates (2y,y)

must lie on this line; therefore

1 = y/(2y-2) =>y=2 and x =4.

You should draw a figure using all of these results to see that the results are correct.

Answer 2

Since P is equidistant from R and S, P lies on the perpendicular bisector of line segment RS.

Slope of RS = (-4-4) / [6-(-2)] = -1

Slope of perpendicular bisector = 1

Perpendicular bisector contains the midpoint of RS: midpoint = (2,0)

Equation of perpendicular bisector: y - 0 = 1(x - 2)

y = x - 2

Since x = 2y, y = 2y - 2.

So, y = 2 and x = 4.

P = (4,2)