The locus of a point equidistant from (a+b , b-a ) and (a-b , a+b) is ?

2 Answers
Mar 28, 2017

#y = b/a(x-a)+b#

Explanation:

The slope of the line connecting the two points is

#m_1 = ((a+b)-(b-a))/((a-b)-(a+b)#

#m_1 = (a+b-b+a)/(a-b-a-b)#

#m_1 = (2a)/(-2b)#

#m_1 = -a/b#

The x coordinate goes from #a+b# to #a-b#; this means that the x coordinate of the midpoint is #a#

The y coordinate goes from #b-a# to #b+a#; this means that the y coordinate of the midpoint is #b#

The midpoint is #(a, b)#

The slope of the perpendicular bisector is:

#m_2 = -1/m_1#

#m_2 = -1/(-a/b)#

#m_2 = b/a#

Use the point-slope form of the equation of a line to write the equation of the perpendicular bisector:

#y = b/a(x-a)+b#

Mar 29, 2017

#bx=ay#

Explanation:

Let the point be #(x,y)#. As it is equidistant from #(a+b,b-a)# and #(a-b,a+b)#, we should have

#(x-a-b)^2+(y-b+a)^2=(x-a+b)^2+(y-a-b)^2#

or #x^2+a^2+b^2-2ax-2bx+2ab+y^2+b^2+a^2-2by+2ay-2ab=x^2+a^2+b^2-2ax+2bx-2ab+y^2+a^2+b^2-2ay-2by+2ab#

or #-2bx+2ay=+2bx-2ay#

or #4bx=4ay# i.e. #bx=ay#