A) (1, 4)
B) (4,1)
C) (-1,-4)
D) (-4,-1)
Correct Answer: C
Solution :
Let \[Q({{x}_{1}},{{y}_{1}})\] be the image of the point P (3, 8) in the line x + 3 y = 7. Then, PQ is perpendicular to given line. \[\therefore \]\[\frac{{{y}_{1}}-8}{{{x}_{1}}-3}\times \left( -\frac{1}{3} \right)=-1\] \[\Rightarrow \]\[3{{x}_{1}}-{{y}_{1}}=1\] ?(i) Mid point of PQ is\[\left( \frac{{{x}_{1}}+3}{2},\frac{{{y}_{1}}+8}{2} \right)\]which lies on the line x + 3y = 7. \[\therefore \]\[\frac{{{x}_{1}}+3}{2}+3\left( \frac{{{y}_{1}}+8}{2} \right)=7\] \[\Rightarrow \]\[{{x}_{1}}+3+3{{y}_{1}}+24=14\] \[\Rightarrow \] \[{{x}_{1}}+3{{y}_{1}}+13=0\] ?(ii) \[{{x}_{1}}=-1,{{y}_{1}}=-4\] Hence, the coordinates of point Q are (-1, - 4).
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