KCET · Maths · Three Dimensional Geometry
The coordinates of the foot of the perpendicular drawn from the point \((3,4)\) on the line \(2 x+y-7=0\) is
- A \(\left(\frac{9}{5}, \frac{17}{5}\right)\)
- B \((1,5)\)
- C \((-5,1)\)
- D \((1,-5)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{9}{5}, \frac{17}{5}\right)\)
Step-by-step Solution
Detailed explanation
We know that foot of the perpendicular \((h, k)\) from \(\left(x_{1}, y_{1}\right)\) to the line \(a x+b y+c=0\) is given by
\(\frac{h-x_{1}}{a}=\frac{k-y_{1}}{b}\)
\(=\frac{-\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\)
Here, point \(\left(x_{1}, y_{1}\right)=(3,4)\)
and \(a x+b y+c=2 x+y-7=0\)
\(\therefore \quad a=2, b=1, c=-7\)
then, from Eq. (i)
\(\frac{h-3}{2}=\frac{k-4}{1}=\frac{-(2 \times 3+1 \times 4-7)}{2^{2}+1^{2}}\)
\(\Rightarrow \quad \frac{h-3}{2}=\frac{k-4}{1}=\frac{-3}{5}\)
\(\Rightarrow \quad h=\frac{+9}{5} \quad\) and \(\quad k=\frac{17}{5}\)
\(\frac{h-x_{1}}{a}=\frac{k-y_{1}}{b}\)
\(=\frac{-\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\)
Here, point \(\left(x_{1}, y_{1}\right)=(3,4)\)
and \(a x+b y+c=2 x+y-7=0\)
\(\therefore \quad a=2, b=1, c=-7\)
then, from Eq. (i)
\(\frac{h-3}{2}=\frac{k-4}{1}=\frac{-(2 \times 3+1 \times 4-7)}{2^{2}+1^{2}}\)
\(\Rightarrow \quad \frac{h-3}{2}=\frac{k-4}{1}=\frac{-3}{5}\)
\(\Rightarrow \quad h=\frac{+9}{5} \quad\) and \(\quad k=\frac{17}{5}\)
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