A point which is equidistant from the points A (5, 4) and B (1, 6) is (3,5)
If true then enter $1$ and if false then enter $0$

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Solution

$L e t u s t a k e t h e p o i n t a s P (x, y) w h i c h s h o u l d b e e q u i d i s t a n t f r o m t h e g i v e n p o i n t s A (x_{1}, y_{1}) = (5, 4) & B (x_{2}, y_{2}) = (1, 6) . U s i n g t h e d i s t a n c e f o r m u l a d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}, w e g e t A P = \sqrt{{(x - x_{1})}^{2} + {(y - y_{1})}^{2}} = \sqrt{{(x - 5)}^{2} + {(y - 4)}^{2}} a n d B P = \sqrt{{(x - x_{2})}^{2} + {(y - y_{2})}^{2}} = \sqrt{{(x - 1)}^{2} + {(y - 6)}^{2}} . S i n c e P i s e q u i d i s t a n t f r o m t h e p o i n t s A & B, A P = B P . ⟹ \sqrt{{(x - 5)}^{2} + {(y - 4)}^{2}} = \sqrt{{(x - 1)}^{2} + {(y - 6)}^{2}} ⟹ 2 x - y - 1 = 0........ (i) T h i s i s t h e l o c u s (L) o f P . I t w i l l p a s s t h r o u g h t h e m i d p o i n t b e t w e e n A & B t h e c o - o r d i n a t e s o f w h i c h a r e (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) = (\frac{5 + 1}{2}, \frac{4 + 6}{2}) = (3, 5) . P u t t i n g (x, y) = (3, 5) i n (i) w e h a v e L . H . S . = 2 \times 3 - 5 - 1 = 0 = R . H . S . S o L p a s s e s t h r o u g h t h e m i d p o i n t o f A B . ∴ L i s t h e p e r p e n d i c u l a r b i s e c t o r o f A B . T h e r e a r e i n f i n i t e n u m b e r o f p o i n t s o n L o n e o f w h i c h i s (3, 5) . A n s - T h e r e a r e i n f i n i t e n u m b e r o f p o i n t s o n L o n e o f w h i c h i s (3, 5) .$

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