question_answer

The coordinates of the foot of the  perpendicular drawn from the point \[(3,4)\] on the line \[2x+y-7=0\] is

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 \[k=\frac{17}{5}\]