Question

What are the coordinates of the foot of perpendicular from the point $(–1,3)$ to the line $3x-4y-16=0$?

Answer 1

Step 1. Obtain the equations required to find coordinates of point.

Let us consider the coordinates of the foot of the perpendicular from $(–1,3)$ to the line $3x-4y-16=0$ be $(a,b)$.

So, the slope of the line join $(-1,3)$ and $(a,b)$ is $m=\frac{b-3}{a+1}$

And the slope of the line $3x-4y-16=0$ or $y=\frac{3}{4}x-4$ is $m\text{'}=\frac{3}{4}$.

Since these two lines are perpendicular,

Thus $m\times m\text{'}=-1$

$$\begin{gathered} \frac{b-3}{a+1}\times \frac{3}{4}=-1 \\ \Rightarrow \frac{3b-9}{4a+4}=-1 \\ \Rightarrow 3b-9=-4a-4 \\ \Rightarrow 4a+3b=5...\left(1\right) \end{gathered}$$

The point $\left(a,b\right)$ lies in the line $3x-4y=16$

So, $3a-4b=16...\left(2\right)$

Step 2. Find the coordinates of the foot of perpendicular.

In order to solve $\left(1\right)$ and $\left(2\right)$ , multiply the equation $\left(1\right)$ with $4$ and equation $\left(2\right)$ with $3$ and add both the equations:

$$\begin{gathered} 16a+12b+9a-12b=20+48 \\ \Rightarrow 25a=68 \\ \Rightarrow a=\frac{68}{25} \end{gathered}$$

Put $a=\frac{68}{25}$ in equation $\left(2\right)$:

$$\begin{gathered} 3\times \frac{68}{25}-4b=16 \\ \Rightarrow \frac{204}{25}-4b=16 \\ \Rightarrow -4b=16-\frac{204}{25} \\ \Rightarrow -4b=\frac{16\times 25-204}{25} \\ \Rightarrow -4b=\frac{400-204}{25} \\ \Rightarrow -4b=\frac{196}{25} \\ \Rightarrow b=\frac{196}{25\times -4} \\ \therefore b=-\frac{49}{25} \end{gathered}$$

Hence $\left(a,b\right)=\left(\frac{68}{25},-\frac{49}{25}\right)$

Hence, the coordinates of the foot of perpendicular are$\left(\frac{68}{25},-\frac{49}{25}\right)$.