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 $P_A(-2,5) and P_B(6,1)?$

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Answer 1 · 2016-11-03T11:27:12

$-2x+y+1=0$

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 get 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

$-2x+y+1=0$

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