Question

Find the equation of the perpendicular drawn from the origin to the line 4x – 3y + 5 = 0. Also find the coordinates of the foot of the perpendicular?

Answer 1

First of all find the slope of the line 4x – 3y + 5 = 0 by using the formula $\dfrac{-a}{b}$ where ‘a’ is the coefficient of x and b is the coefficient of y. Now, to find the slope of the perpendicular line, take the reciprocal of $\dfrac{-a}{b}$ and multiply it with -1. Write the equation of the perpendicular line as $y=\left( \dfrac{b}{a} \right)x$. Now, to find the foot of the perpendicular use the relation $\dfrac{\alpha -x}{a}=\dfrac{\beta -y}{b}=\dfrac{a\alpha +b\beta +c}{{{a}^{2}}+{{b}^{2}}}$ where $\left( \alpha ,\beta \right)$ is the point from which perpendicular is drawn, ax + by + c = 0 is the line on which the perpendicular is drawn. Find the values of x and y to get the answer.

Complete step-by-step answer:
Here we have been provided with a line 4x – 3y + 5 = 0 on which the perpendicular is drawn from the origin and we are asked to determine the equation of that perpendicular line and also the coordinates of the foot of perpendicular.
Now, the slope of the line 4x – 3y + 5 = 0 will be given as $\dfrac{-a}{b}$ where ‘a’ is the coefficient of x and b is the coefficient of y. So substituting the values we get,
$\Rightarrow$ Slope of 4x – 3y + 5 = 0 = $\dfrac{-4}{-3}$
$\Rightarrow$ Slope of 4x – 3y + 5 = 0 = $\dfrac{4}{3}$
We know that the product of the slopes of two perpendicular lines is equal to -1, so assuming the slope of the perpendicular line as m we have,
$\begin{aligned} & \Rightarrow m\times \dfrac{4}{3}=-1 \\\ & \Rightarrow m=\dfrac{-3}{4} \\\ \end{aligned}$
Since the given perpendicular line is passing through the origin so its y – intercept will be 0, therefore the equation of the line will be $y=mx$, so substituting the value of m we get,
$\begin{aligned} & \Rightarrow y=\dfrac{-3}{4}x \\\ & \therefore 4y+3x=0 \\\ \end{aligned}$
Therefore, the above equation is the perpendicular line.

Further we need to find the foot of the coordinates of this perpendicular line. We know that the if we draw a perpendicular line from the point $\left( \alpha ,\beta \right)$ on a given line ax + by + c = 0 then coordinates (x, y) of the foot of perpendicular is given as $\dfrac{\alpha -x}{a}=\dfrac{\beta -y}{b}=\dfrac{a\alpha +b\beta +c}{{{a}^{2}}+{{b}^{2}}}$, so substituting the values we get,
$\begin{aligned} & \Rightarrow \dfrac{0-x}{4}=\dfrac{0-y}{-3}=\dfrac{\left( 4 \right)\left( 0 \right)+\left( -3 \right)\left( 0 \right)+5}{{{4}^{2}}+{{\left( -3 \right)}^{2}}} \\\ & \Rightarrow \dfrac{-x}{4}=\dfrac{y}{3}=\dfrac{5}{25} \\\ & \Rightarrow \dfrac{-x}{4}=\dfrac{y}{3}=\dfrac{1}{5} \\\ \end{aligned}$
(i) For the coordinate x we have,
$\begin{aligned} & \Rightarrow \dfrac{-x}{4}=\dfrac{1}{5} \\\ & \Rightarrow x=\dfrac{-4}{5} \\\ \end{aligned}$
(ii) For the coordinate y we have,
$\begin{aligned} & \Rightarrow \dfrac{y}{3}=\dfrac{1}{5} \\\ & \Rightarrow y=\dfrac{3}{5} \\\ \end{aligned}$
Therefore, the coordinates of the foot of perpendicular is given as $\left( \dfrac{-4}{5},\dfrac{3}{5} \right)$.

Note: Note that you can also find the coordinates of the perpendicular line by using the formula $m=\dfrac{\Delta y}{\Delta x}$ where m is the slope of the perpendicular line that we have obtained above. $\Delta y=\left( {{y}_{2}}-{{y}_{1}} \right)$ and similarly $\Delta x=\left( {{x}_{2}}-{{x}_{1}} \right)$. You can assume $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 0,0 \right)$ and find the value of $\left( {{x}_{2}},{{y}_{2}} \right)$ which will be our answer. However, remember the formulas used in the above solution to avoid the mistake in case you forget the actual method.