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  • Using section formula find the foot of perpendicular drawn from the point (2,3) to the line joining the points (2,0) and <img alt="\left(\frac{8}{13}, \frac{12}{13}\right)" src="https://entran

Using section formula find the foot of perpendicular drawn from the point (2,3) to the line joining the points (2,0) and \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 A, B and C be the points (2,3),(2,0) and \left(\frac{8}{13}, \frac{12}{13}\right) respectively.
Let the foot D of perpendicular A D to B C divides B C in the ratio\lambda: 1-\lambda. Then the coordinates of D are given by

\begin{aligned} & \mathrm{x} \text {-coordinate }=\lambda(8 / 13)+(1-\lambda) 2=\frac{-18 \lambda+26}{13} \\ & \mathrm{y} \text {-coordinate }=\frac{12 \lambda+(1-\lambda) \times 0}{13}=\frac{12 \lambda}{13} \\ & \text { Slope of AD} =\frac{3-\frac{12}{13} \lambda}{2-\left(\frac{-18 \lambda+26}{13}\right)}=\frac{39-12 \lambda}{18 \lambda} \\ & \text { Slope of BC} \text { is }=\frac{\frac{12}{13}}{13}-\frac{12}{-18}=-\frac{2}{3} \\ & \text { Since AD} \perp \text{BC} \\ & \therefore \quad \frac{39-12 \lambda}{18 \lambda} \times\left(-\frac{2}{3}\right)=-1 \\ & \therefore \quad \text { i.e. } \lambda=1 \\ & \text { The coordinates of D are }\left(\frac{8}{13}, \frac{12}{13}\right) \end{aligned}

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jitender.kumar

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