Find-locus-of-a-point-P-which-moves-such-that-its-distance-from-the-line-y-3-x-7-is-same-as-its-distance-from-2-3-1-

Question Number 47438 by rahul 19 last updated on 10/Nov/18

F i n d l o c u s o f a p o i n t P w h i c h m o v e s

s u c h t h a t i t s d i s t a n c e f r o m t h e l i n e

y = \sqrt{3} x - 7 i s s a m e a s i t s d i s t a n c e f r o m

(2 \sqrt{3}, - 1) ?

Answered by ajfour last updated on 10/Nov/18

\frac{{(y - x \sqrt{3} + 7)}^{2}}{4} = {(x - 2 \sqrt{3})}^{2} + {(y + 1)}^{2}

y^{2} + 3 x^{2} + 49 - 2 \sqrt{3} x y - 14 \sqrt{3} x

+ 14 y = 4 x^{2} + 4 y^{2} - 16 \sqrt{3} x + 8 y + 52

\Rightarrow x^{2} + 3 y^{2} + 2 \sqrt{3} x y - 2 \sqrt{3} x - 6 y + 3 = 0

\Rightarrow {(x + \sqrt{3} y)}^{2} - 2 \sqrt{3} (x + \sqrt{3} y) + 3 = 0

\Rightarrow {(x + \sqrt{3} y - \sqrt{3})}^{2} = 0

\Rightarrow \sqrt{3} y = \sqrt{3} - x .

Commented by rahul 19 last updated on 10/Nov/18

thanks sir��

Answered by rahul 19 last updated on 10/Nov/18

s i n c e t h e p o i n t s a t i s i f e s t h e l i n e .

l o c u s i s s t . l i n e p e r p e n d i c u l a r t o g i v e n

l i n e p a s s i n g t h r o u g h g i v e n p o i n t .

\Rightarrow l o c u s = (y + 1) = - \frac{1}{\sqrt{3}} (x - 2 \sqrt{3}) .

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