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

Solving,

⇒ 4x - 3y - 5 = 0

⇒ 3y = 4x - 5

⇒ y = $\dfrac{4}{3}x - \dfrac{5}{3}$

Comparing above equation with y = mx + c we get, Slope of the line (m₁) = $\dfrac{4}{3}$

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

Then,

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

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $-\dfrac{3}{4}(x - 1)$

⇒ 4(y + 2) = −3(x − 1)

⇒ 4y + 8 = −3x + 3

⇒ 3x + 4y + 5 = 0.

For finding the coordinates of the foot of the perpendicular which is the point of intersection of the lines. Let point of intersection of lines be Q.

4x - 3y - 5 = 0 …….(1)

3x + 4y + 5 = 0 ……..(2)

On multiplying equation (1) by 4, we get :

16x - 12y - 20 = 0 ………..(3)

On multiplying equation (2) by 3, we get :

9x + 12y + 15 = 0 ……….(4)

Adding equations (3) and (4) we get,

⇒ 16x - 12y - 20 + 9x + 12y + 15 = 0

⇒ 25x - 5 = 0

⇒ x = $\dfrac{5}{25}$

⇒ x = $\dfrac{1}{5}$.

Substituting value of x in (1), we get :

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

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

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