The locus of a point equidistant from two giv | vedclass

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Question

(A) Let $P(r)$ be a point equidistant from points $A(a)$ and $B(b)$.Then,the distance $PA = PB$,which implies $PA^2 = PB^2$.In vector form,this is $|r

Vedclass · Accepted Answer

$[r - \frac{1}{2}(a + b)] \cdot (a - b) = 0$

Vedclass · Answer

(A) Let $P(r)$ be a point equidistant from points $A(a)$ and $B(b)$.Then,the distance $PA = PB$,which implies $PA^2 = PB^2$.In vector form,this is $|r - a|^2 = |r - b|^2$.Expanding this,we get $(r - a) \cdot (r - a) = (r - b) \cdot (r - b)$.$|r|^2 - 2(r \cdot a) + |a|^2 = |r|^2 - 2(r \cdot b) + |b|^2$.$-2(r \cdot a) + 2(r \cdot b) = |b|^2 - |a|^2$.$2r \cdot (b - a) = |b|^2 - |a|^2$.Alternatively,the locus is the perpendicular bisector of the segment $AB$.Let $M$ be the midpoint of $AB$. The position vector of $M$ is $\frac{1}{2}(a + b)$.Since $PM$ is perpendicular to $AB$,the vector $PM$ is perpendicular to the vector $AB$.Thus,$(r - \frac{1}{2}(a + b)) \cdot (b - a) = 0$.Multiplying by $-1$,we get $[r - \frac{1}{2}(a + b)] \cdot (a - b) = 0$.

The locus of a point equidistant from two given points $a$ and $b$ is given by

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