Find the equation of the perpendicular from the point (1, -2) on the line 4x - 3y - 5 = 0. Also find the coordinates of the foot of perpendicular.

Converting 4x - 3y - 5 = 0 in the form of y = mx + c.

⇒ 4x - 3y - 5 = 0

⇒ 3y = 4x - 5

⇒ y = $\dfrac{4}{3}x - \dfrac{5}{3}$

Slope of the line (m₁) = $\dfrac{4}{3}$.

Let the slope of the line perpendicular to 4x - 3y - 5 = 0 be m₂.

Then, m₁ × m₂ = -1.

$\Rightarrow \dfrac{4}{3} \times m$2 = -1 \\[1em] \Rightarrow m2 = -\dfrac{3}{4}.

The equation of the line having slope m₂ and passing through the point (1, -2) can be given by point-slope form i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - (-2) = -\dfrac{3}{4}(x - 1) \\[1em] \Rightarrow 4(y + 2) = -3(x - 1) \\[1em] \Rightarrow 4y + 8 = -3x + 3 \\[1em] \Rightarrow 3x + 4y + 5 = 0

For finding the coordinates of the foot of the perpendicular which is the point of intersection of the lines

4x - 3y - 5 = 0 ….(i)
3x + 4y + 5 = 0 ….(ii)

On multiplying (i) by 4 and (ii) by 3 we get,

16x - 12y - 20 = 0 ….(iii)
9x + 12y + 15 = 0 ….(iv)

Adding (iii) and (iv) we get,

⇒ 16x - 12y - 20 + 9x + 12y + 15 = 0

⇒ 25x - 5 = 0

⇒ x = $\dfrac{5}{25}$

⇒ x = $\dfrac{1}{5}$.

Putting value of x in (i), we have

$\Rightarrow 4 \times \dfrac{1}{5} - 3y - 5 = 0 \\[1em] \Rightarrow \dfrac{4}{5} - 3y - 5 = 0 \\[1em] \Rightarrow 3y = \dfrac{4}{5} - 5 \\[1em] \Rightarrow 3y = \dfrac{4 - 25}{5} \\[1em] \Rightarrow 3y = -\dfrac{21}{5} \\[1em] \Rightarrow y = -\dfrac{7}{5}.$

∴ Coordinates = $\Big(\dfrac{1}{5}, -\dfrac{7}{5}\Big)$.

Hence, the equation of the new line is 3x + 4y + 5 = 0 and coordinates of the foot of perpendicular (i.e., its intersection with 4x - 3y - 5 = 0) are $\Big(\dfrac{1}{5}, -\dfrac{7}{5}\Big)$.