Find the equation of the locus of points equidistant from the point A(4, −1) and the line x−y+2=0.


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Question Number 175176 by nadovic last updated on 21/Aug/22

$Find the equation of the locus of$ $points equidistant from the point$ $A (4, - 1) and the line x - y + 2 = 0 .$
	Answered by som(math1967) last updated on 22/Aug/22

	$l e t m o v a l e p o i n t (h, k)$ $f r o m t h e c o n d i t i o n$ $\sqrt{{(h - 4)}^{2} + {(k + 1)}^{2}} = \frac{∣ h - k + 2 ∣}{\sqrt{{(1)}^{2} + {(- 1)}^{2}}}$ $2 {(h - 4)}^{2} + 2 {(k + 1)}^{2} = h^{2} + k^{2} + 4 - 2 h k$ $+ 4 h - 4 k$ $[s q u a r i n g b o t h s i d e]$ $h^{2} + k^{2} + 12 h - 8 k - 2 h k - 30 = 0$ $∴ e q u a t i o n o f t h e l o c u s$ $x^{2} + y^{2} + 12 x - 8 y - 2 x y - 30 = 0$ $l o c u s p a t h i s p a r a b o l a$
	Commented by nadovic last updated on 22/Aug/22

	$T h a n k y o u S i r$
	Answered by a.lgnaoui last updated on 22/Aug/22

	$l e n s e m b l e d e s p o i n t s s i t u e s a l a m e m e d i s t a n c e d e A a l a d r o i t e (D)$ $y = x + 2 a u p o n t M (4, - 1) s o n t l e s p o i n t s d e l a d r o i t e p a r a l l e l e a (D) n o t e (D^{'}) q u i s$ $q u i p a s s e p a r M y = x + m$ $x = 4 y = - 1$ $4 + m = - 1 m = - 5$ $y = x - 5$
	Commented by som(math1967) last updated on 22/Aug/22

	$(4, - 1) n o t e q u i d i s t a c e t o y = x + 2$ $a n d y = x - 5$
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