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  • Using section formula find the foot of perpendicular drawn from the point to the line j

Using section formula find the foot of perpendicular drawn from the point \mathrm{(2,3)} to the line joining the points \mathrm{(2,0)} and \mathrm{\left(\frac{8}{13}, \frac{12}{13}\right)}
 

Option: 1

\left ( \frac{8}{13},\frac{12}{13} \right )


Option: 2

\left ( \frac{8}{13},\frac{-12}{13} \right )


Option: 3

\left ( \frac{1}{4},\frac{12}{13} \right )


Option: 4

\left ( \frac{3}{4},\frac{1}{13} \right )


Answers (1)

best_answer

Let \mathrm{A, B \: and \: C} be the points \mathrm{(2,3),(2,0) \: and \: \left(\frac{8}{13}, \frac{12}{13}\right)}respectively.

Let the foot \mathrm{D} of perpendicular \mathrm{A D \: to\: B \: C \: divides \: B \: C} in the ratio \mathrm{\lambda: 1-\lambda}. Then the coordinates of \mathrm{D} are given by

\mathrm{ \mathrm{x} \text {-coordinate }=\lambda(8 / 13)+(1-\lambda) 2=\frac{-18 \lambda+26}{13} }

\mathrm{ \mathrm{y} \text {-coordinate }=\frac{12 \lambda+(1-\lambda) \times 0}{13}=\frac{12 \lambda}{13} }

\mathrm{ \text { Slope of } A D=\frac{3-\frac{12}{13} \lambda}{2-\left(\frac{-18 \lambda+26}{13}\right)}=\frac{39-12 \lambda}{18 \lambda} }

\mathrm{ \text { Slope of } B C \text { is }=\frac{\frac{12}{13}}{\frac{8}{13}-2}=\frac{12}{-18}=-\frac{2}{3} }

\mathrm{ \text { Since } A D \perp B C }

\mathrm{\therefore \frac{39-12 \lambda}{18 \lambda} \times\left(-\frac{2}{3}\right)=-1 } i.e. \mathrm{\lambda=1 }

\mathrm{\therefore \text { The coordinates of } D \text { are }\left(\frac{8}{13}, \frac{12}{13}\right)}

Hence option 1 is correct.




 

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mansi

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