Statement-1:The point A(1,0,7) is the mirror image of the point B=(1,6,3) in the line x1=y-12=z-23 Statement-2:The line x1=y-12=z-23 bisects the segment joining A(1,0,7) and B(1,6,3)

D.r's of AB are (0,6,-4),D.r's of givn line are (1,2,3) Now a1a2+b1b2+c1c2=0+12-12=0 ⇒AB ⇀ is perpendicular to the given line ∴Statement-1 is true Mid point of AB is (1,3,5) and 11=3-12=5-23 ⇒(1,3,5) lies on the given line ∴Statement-2 is true Here Statement-2 is not correct explanation of statement-1

Statement-1:The point A(1,0,7) is the mirror image of the point B=(1,6,3) in the line x1=y-12=z-23 Statement-2:The line x1=y-12=z-23 bisects the segment joining A(1,0,7) and B(1,6,3)

D.r's of AB are (0,6,-4),D.r's of givn line are (1,2,3) Now a1a2+b1b2+c1c2=0+12-12=0 ⇒AB ⇀ is perpendicular to the given line ∴Statement-1 is true Mid point of AB is (1,3,5) and 11=3-12=5-23 ⇒(1,3,5) lies on the given line ∴Statement-2 is true Here Statement-2 is not correct explanation of statement-1

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$S t a t e m e n t - 1 : T h e p o i n t A (1, 0, 7) i s t h e m i r r o r i m a g e o f t h e p o i n t B = (1, 6, 3) i n t h e l i n e \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} S t a t e m e n t - 2 : T h e l i n e \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} b i s e c t s t h e s e g m e n t j o i n i n g A (1, 0, 7) a n d B (1, 6, 3)$

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$1) S t a t e m e n t - 1 i s t r u e, S t a t e m e n t - 2 i s f a l s e$

$S t a t e m e n t - 1 i s f a l s e, S t a t e m e n t - 2 i s t r u e$

$S t a t e m e n t - 1 i s t r u e, S t a t e m e n t - 2 i s t r u e; s t a t e m e n t - 2 i s a c o r r e c t e x p l a n a t i o n f o r s t a t e m e n t - 1$

$S t a t e m e n t - 1 i s t r u e, S t a t e m e n t - 2 i s t r u e; s t a t e m e n t - 2 i s n o t c o r r e c t e x p l a n a t i o n f o r s t a t e m e n t - 1$

answer is D.

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Detailed Solution

$D . r' s o f A B a r e (0, 6, - 4), D . r' s o f g i v n l i n e a r e (1, 2, 3) N o w a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 + 12 - 12 = 0 \Rightarrow \overset{⇀}{A B} i s p e r p e n d i c u l a r t o t h e g i v e n l i n e ∴ S t a t e m e n t - 1 i s t r u e M i d p o i n t o f A B i s (1, 3, 5) a n d \frac{1}{1} = \frac{3 - 1}{2} = \frac{5 - 2}{3} \Rightarrow (1, 3, 5) l i e s o n t h e g i v e n l i n e ∴ S t a t e m e n t - 2 i s t r u e H e r e S t a t e m e n t - 2 i s n o t c o r r e c t e x p l a n a t i o n o f s t a t e m e n t - 1$