Question

Find the equation of the perpendicular from the point $$\mathrm{P}(1, -2)$$ on the line $$4x - 3y - 5 = 0$$. Also, find the co-ordinates of the foot of the perpendicular.

Answer 1

The given line is $$4x - 3y - 5 = 0$$: 

$$3y = 4x - 5$$

$$\implies y = \frac{4}{3}x - \frac{5}{3}$$

Slope $$m_1 = \frac{4}{3}$$

Slope of the perpendicular line:

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

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

Equation of the perpendicular line passing through $$\mathrm{P}(1, -2)$$:

$$y - (-2) = -\frac{3}{4}(x - 1)$$

$$4(y + 2) = -3(x - 1)$$

$$4y + 8 = -3x + 3$$

$$3x + 4y + 5 = 0$$

To find the foot of the perpendicular, solve the two linear equations:

$$4x - 3y = 5 \quad ...(1)$$ 

$$3x + 4y = -5 \quad ...(2)$$

Multiply (1) by 4 and (2) by 3:

$$16x - 12y = 20$$

$$9x + 12y = -15$$

Adding them:

$$25x = 5$$ 

$$\implies x = \frac{5}{25} = \frac{1}{5}$$

Substitute $$x = \frac{1}{5}$$ into (1): 

$$4\left(\frac{1}{5}\right) - 3y = 5$$ 

$$\frac{4}{5} - 5 = 3y$$ 

$$-\frac{21}{5} = 3y$$

$$\implies y = -\frac{7}{5}$$

Thus, the equation of the perpendicular is $$3x + 4y + 5 = 0$$ and the coordinates of the foot are $$\left(\frac{1}{5}, -\frac{7}{5}\right)$$.