Coordinate Geometry Question 122
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) $ a_1^{2}-a_2^{2}+b_1^{2}-b_2^{2} $
B) $ \sqrt{a_1^{2}+b_1^{2}-a_2^{2}-b_2^{2}} $
C) $ \frac{1}{2}(a_1^{2}+a_2^{2}+b_1^{2}+b_2^{2}) $
D) $ \frac{1}{2}(a_2^{2}+b_2^{2}-a_1^{2}-b_1^{2}) $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ (h,k) $ be the point on the locus, then by the given conditions $ {{(h-a_1)}^{2}}+{{(k-b_1)}^{2}}={{(h-a_2)}^{2}}+{{(k-b_2)}^{2}} $
$ \Rightarrow 2h(a_1-a_2)+2k(b_1-b_2)+a_2^{2}+b_2^{2}-a_1^{2}-b_1^{2}=0 $
$ \Rightarrow h(a_1-a_2)+k(b_1-b_2)+\frac{1}{2}(a_2^{2}+b_2^{2}-a_1^{2}-b_1^{2})=0 $ ….(i) Also, since (h, k) lies on the given locus, therefore $ (a_1-a_2)h+(b_1-b_2)k+c=0 $
…..(ii) Comparing (i) and (ii), we get $ c=\frac{1}{2}(a_2^{2}+b_2^{2}-a_1^{2}-b_1^{2}) $ .
BETA
