Straight Line Question 256
Question: Suppose A, B are two points on $ 2x-y+3=0 $ and P (1, 2) is such that $ PA=PB. $ Then the mid-point of AB is
Options:
A) $ ( -\frac{1}{5},\frac{13}{5} ) $
B) $ ( \frac{-7}{5},\frac{9}{5} ) $
C) $ ( \frac{7}{5},\frac{-9}{5} ) $
D) $ ( \frac{-7}{5},\frac{-9}{5} ) $
Correct Answer: A $ \Rightarrow \Delta PAD\cong \Delta PBD $ $ \Rightarrow D $ is foot of perpendicular
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Answer:
Solution:
$ \Rightarrow $ from P to AB $ \frac{\alpha -1}{2}=\frac{\beta -2}{-1}=\frac{-(2\times 1-1\times 2+3)}{4+1} $ $ \frac{\alpha -1}{2}=\frac{\beta -2}{-1}=\frac{-3}{5}\Rightarrow \alpha =\frac{-1}{5},\beta =\frac{13}{5} $
BETA
