Find the locus of point $P$ from which distance of two points $A (1, 2)$ and $B (3, 4)$ are always equal.

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Solution

$l e t p o i n t P (h, k) P A = P B t h e n \Rightarrow \sqrt{{(h - 1)}^{2} {(k - 2)}^{2}} = \sqrt{{(3 - h)}^{2} {(4 - k)}^{2}} \Rightarrow \sqrt{h^{2} - 2 h + 1 + k^{2} - 4 k + 4} = \sqrt{9 - 6 h + h^{2} 16 - 8 k + k^{2}} \Rightarrow h^{2} - 2 h + k^{2} - 4 k + 5 = h^{2} + k^{2} - 6 h - 8 k + 25 \Rightarrow 4 h + 4 k = 20 \Rightarrow h + k = 5 l o c u s (h + k) = 5$

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Points A(1,2) and B(3,4) are two ends of a line segment. Find the point which divides AB in the ratio 3:4.

A (1, 2), B (2, - 3), C (- 2, 3)

P A^{2} + P B^{2} = 2 P C^{2}

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(\neq 1)

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