Question

Find the equation of the perpendicular drawn from the point $$\mathrm{P}(2, 3)$$ on the line $$y = 3x + 4$$. Find the co-ordinates of the foot of the perpendicular. 

[Hint: Let $$AB$$ be the line $$y = 3x + 4$$ and let $$PQ \perp AB$$. 
Slope of $$PQ = \frac{-1}{\text {(Slope of AB)}} = \frac{-1}{3}$$. 
Equation of $$PQ$$ is $$\frac{y - 3}{x - 2} = \frac{-1}{3} \Rightarrow x + 3y = 11$$. 
On solving $$y = 3x + 4$$ and $$x + 3y = 11$$, we get co-ordinates of $$Q$$]

Answer 1

Slope of the given line $$y = 3x + 4$$ is $$m_1 = 3$$.

Slope of the perpendicular line $$\mathrm{PQ}$$: 

$$m = -\frac{1}{m_1} = -\frac{1}{3}$$

Equation of $$\mathrm{PQ}$$ passing through $$\mathrm{P}(2, 3)$$:

$$y - 3 = -\frac{1}{3}(x - 2)$$ 

$$3(y - 3) = -(x - 2)$$ 

$$3y - 9 = -x + 2$$

$$x + 3y = 11$$

To find the coordinates of the foot of perpendicular $$\mathrm{Q}$$, solve: 

$$y = 3x + 4 \quad ...(1)$$ 

$$x + 3y = 11 \quad ...(2)$$

Substitute (1) into (2):

$$x + 3(3x + 4) = 11$$

$$x + 9x + 12 = 11$$

$$10x = -1$$

$$\implies x = -\frac{1}{10}$$

Substitute $$x = -\frac{1}{10}$$ into (1): 

$$y = 3\left(-\frac{1}{10}\right) + 4$$

$$= -\frac{3}{10} + 4$$

$$= \frac{37}{10}$$

Thus, the equation of the perpendicular is $$x + 3y = 11$$ and the coordinates of the foot are $$\left(-\frac{1}{10}, \frac{37}{10}\right)$$.