Find a point in Quadrant 4 that is equally distant from (4,1) and the Y-axis.

SydSu22 Jan 15, 2019

#1**+2 **

Let the coordinates of this point be ( 4, y)

The point on the y axis closest to this point is (0, y)

So... we have this equation

Distance from (4, y) to (4, 1) = Distance from (4, y) to (0, y)

( 4 - 4)^2 + (1 - y)^2 = (4 - 0)^2 + ( y - y)^2 smplify

y^2 - 2y + 1 = 16

y^2 - 2y - 15 = 0 factor

(y - 5) ( y + 3) = 0

Since y must be negative......y = -3

So.....the point is ( 4, -3)

CPhill Jan 15, 2019