Straight Line Question 154
Question: If the equation of the locus of a point equidistant from the points ( $ {a _{1,}}b_1 $ ) and ( $ {a _{2,}}b_2 $ ) is ( $ {a _{1,-}}a_2 $ )x+( $ {b _{1,-}}b_2 $ ) $ y+c=0 $ then the value of c is
Options:
A) $ \frac{1}{2}(a_2^{2}+b_2^{2}-a_1^{2}-b_1^{2}) $
B) $ a_1^{2}+a_2^{2}+b_1^{2}-b_2^{2} $
C) $ \frac{1}{2}(a_1^{2}+a_2^{2}-b_1^{2}-b_2^{2}) $
D) $ \sqrt{a_1^{2}+b_2^{2}-a_2^{2}-b_2^{2}} $
Correct Answer: A $ \Rightarrow (a_1-a_2)x+(b_1-b_2)y+\frac{1}{2}(a_2^{2}+b_2^{2}-a_1^{2}-b_1^{2})=0 $ $ \Rightarrow c=\frac{1}{2}(a_2^{2}+b_2^{2}-a_1^{2}-b^2_1) $
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Answer:
Solution:
BETA
