A) \[\left( \frac{9}{5},\frac{17}{5} \right)\]
B) \[(1,5)\]
C) \[(5,1)\]
D) \[(1,-5)\]
Correct Answer: A
Solution :
We know that foot of the perpendicular (h, k) from \[({{x}_{1}},{{y}_{1}})\]to the line \[ax+by+c=0\]is given by \[\frac{h-{{x}_{1}}}{a}=\frac{k-{{y}_{1}}}{b}=\frac{-(a{{x}_{1}}+b{{y}_{1}}+c)}{{{a}^{2}}+{{b}^{2}}}\] ?.(i) Here, point \[({{x}_{1}},{{y}_{1}})=(3,4)\] and \[ax+by+c=2x++y-7=0\] \[\therefore \] \[a=2,b=1,c=-7\] Then, from Eq. (i) \[\frac{h-3}{2}=\frac{k-4}{1}=\frac{-(2\times 3+1\times 4-7)}{{{2}^{2}}+{{1}^{2}}}\] \[\Rightarrow \] \[\frac{h-3}{2}=\frac{k-4}{1}=\frac{-3}{5}\] \[\Rightarrow \] \[\frac{h-3}{2}=\frac{-3}{5}\] and \[\frac{k-4}{1}=\frac{-3}{5}\] \[\Rightarrow \] \[h=\frac{-6}{5}+3\] and \[k=\frac{-3}{5}+4\] \[\Rightarrow \] \[h=\frac{+9}{5}\] and \[k=\frac{17}{5}\]You need to login to perform this action.
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