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

 Jan 15, 2019
 #1
avatar+101431 
+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)

 

 

cool cool cool

 Jan 15, 2019
edited by CPhill  Jan 15, 2019
 #2
avatar+250 
+1

Thanks!

SydSu22  Jan 15, 2019

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