Question

find the coordinates of the foot of perpendicular draw from origin upon the lines 3x-5y+2=0 and 4x-3y+5=0 and show that equation of straight line joining the feet is 26 x+53 y=11

Answer 1

$\textbf{Given:}$

$\textsf{Lines are 3x-5y+2=0 and 4x-3y+5=0}$

$\textbf{To find:}$

$\textsf{Coordinates of the foot of the perpendicular drawn from}$

$\textsf{origin to the given lines}$

$\textbf{To prove:}$

$\textsf{Equation of line joining foot of perpendiculars is 26x+53y=11}$

$\textbf{Solution:}$

$\textsf{The line perpendicular to 3x-5y+2=0}$

$\textsf{can be written as}\;\mathsf{-5x-3y=-5x_1-3y_1}$

$\textsf{It passes through (0,0)}$

$\implies\mathsf{5x+3y=0}$

$\textsf{Foot of the perdicular is obtained by solving 3x-5y+2=0 and 5x+3y=0}$

$\mathsf{\dfrac{x}{0-6}=\dfrac{y}{10-0}=\dfrac{1}{9+25}}$

$\mathsf{\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{1}{34}}$

$\mathsf{\dfrac{x}{-3}=\dfrac{y}{5}=\dfrac{1}{17}}$

$\mathsf{Foot\;of\;perpendicular\;\left(\dfrac{-3}{17},\dfrac{5}{17}\right)}$

$\textsf{The line perpendicular to 4x-3y+5=0}$

$\textsf{can be written as}\;\mathsf{-3x-4y=-3x_1-4y_1}$

$\textsf{It passes through (0,0)}$

$\implies\mathsf{3x+4y=0}$

$\textsf{Foot of the perdicular is obtained by solving 4x-3y+5=0 and 3x+4y=0}$

$\mathsf{\dfrac{x}{0-20}=\dfrac{y}{15-0}=\dfrac{1}{16+9}}$

$\mathsf{\dfrac{x}{-20}=\dfrac{y}{15}=\dfrac{1}{25}}$

$\mathsf{\dfrac{x}{-4}=\dfrac{y}{3}=\dfrac{1}{5}}$

$\mathsf{Foot\;of\;perpendicular\;\left(\dfrac{-4}{5},\dfrac{3}{5}\right)}$

$\textsf{Equation of line joining foot of perpendiculars is}$

$\mathsf{\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}}$

$\mathsf{\dfrac{y-\dfrac{5}{17}}{\dfrac{3}{5}-\dfrac{5}{17}}=\dfrac{x+\dfrac{3}{17}}{\dfrac{-4}{5}+\dfrac{3}{17}}}$

$\mathsf{\dfrac{\dfrac{17y-5}{17}}{\dfrac{51-25}{85}}=\dfrac{\dfrac{17x+3}{17}}{\dfrac{-68+15}{85}}}$

$\mathsf{\dfrac{\dfrac{17y-5}{17}}{\dfrac{26}{85}}=\dfrac{\dfrac{17x+3}{17}}{\dfrac{-53}{85}}}$

$\mathsf{\dfrac{17y-5}{26}=\dfrac{17x+3}{-53}}$

$\mathsf{-53{\times}17y+265=26{\times}17y+78}$

$\mathsf{-53{\times}17y+187=26{\times}17x}$

$\mathsf{Divide\;by\;17}$

$\mathsf{-53y+11=26x}$

$\implies\boxed{\mathsf{26x+53y=11}}$