# Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (3, -2).

*print*Print*list*Cite

### 1 Answer

## Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (3, -2).

A point equidistant from the endpoints of a line segment lies on the perpendicular bisector of that segment.

(1) The midpoint of the segment from (2,2) to (3,-2) is found, using the midpoint formula, to be `((2+3)/2,(2-2)/2)=(5/2,0)` .

(2) the slope of the line through (2,2) and (3,-2) is `(2-(-2))/(2-3)=-4` so the slope of the line perpendicular to the segment is `1/4` .

(3) Using the point-slope formula, we find the equation of the line that is the perpendicular bisector of the segment from (2,2) to (3,-2) to be `y-0=1/4(x-5/2)` or `y=1/4 x - 5/8` .

**The y-intercept of this line is `-5/8` which is the value we are seeking.**