question_answer

If \[x+2y+1=0,\] then reflection point of \[(3,2)\] is:

A)  \[(1,4)\]                             

B)  \[(1,-\,4)\]

C)  \[(4,1)\]             

D)         \[\left( -\frac{1}{5},-\frac{22}{5} \right)\]

Correct Answer: D

Solution :

Let the reflection point of (3, 2) is\[P({{x}_{1}},{{y}_{1}})\] Equation of perpendicular line AB i.e., \[x+2y+1=0\]is                  ?.(i) \[2x-y+\lambda =0\] Since, it is passing through the point (3, 2) \[\Rightarrow \]\[6-2+\lambda =0\]\[\Rightarrow \]\[\lambda =-\,4\] \[\therefore \]  \[2x-y-4=0\] It is also passes through the point \[({{x}_{1}},{{y}_{1}})\] \[2{{x}_{1}}-{{y}_{1}}-4=0\]                         ?(ii) Now, mid point of PQ is M \[\left( \frac{{{x}_{1}}+3}{2},\frac{{{y}_{1}}+2}{2} \right)\]it lies on the line AB. \[\Rightarrow \]\[\frac{{{x}_{1}}+3}{2}+2\left( \frac{{{y}_{1}}+2}{2} \right)+1=0\]             [from (i)] \[\Rightarrow \]               \[{{x}_{1}}+2{{y}_{1}}+9=0\]                       ?(iii) On solving Eqs. (ii) and (iii), we get \[{{x}_{1}}=-\frac{1}{5}\]and \[{{y}_{1}}=-\frac{22}{5}\]