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Describe and write an equation for the locus of points equidistant form #A(a_x, a_y) and B(b_x,b_y)#? Test what you derived for #P_A(-2,5) and P_B(6,1)? #

Answer 1

Violet Aldrich

#-2x+y+1=0#

Calling #p=(x,y)# we are looking for #p# such that

#norm(p-A)=norm(p-B)# or equivalently

#norm(p-A)^2=norm(p-B)^2# so

#(x-a_x)^2+(y-a_y)^2=(x-b_x)^2+(y-b_y)^2#

Simplifying, we arrive at

#2(a_x-b_x)x+2(a_y-b_y)y-a_x^2-a_y^2+b_x^2+b_y^2=0#

which is the equation of a line equidistant from #A# and #B#

If #A=(-2,5)# and #B=(6,1)# then

#2(-2-6)x+2(5-1)y-4-25+36+1=0# or

#-2x+y+1=0#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.