Find the coordinates of foot of perpendicular from the point (2,3) to the line 3x+4y+8=0 - Maths - Straight Lines

Question

Find the coordinates of foot of perpendicular from the point
(2,3) to the line 3x+4y+8=0

Shridhar Kaushik · Accepted Answer

Let the foot of the perpendicular be B (x₁, y₁).

Equation of given line is

$3 x + 4 y + 8 = 0 \Rightarrow 4 y = - 3 x - 8 \Rightarrow y = \frac{- 3 x}{4} - 2$

Slope of the line, m = $\frac{- 3}{4}$ .

As the point B(x₁, y₁) lies on this line, so it satisfies the equation of line.

$∴ 3 x_{1} + 4 y_{1} + 8 = 0 \Rightarrow y_{1} = \frac{- 3 x_{1}}{4} - 2 ....... (1)$

Now, slope of line AB = $\frac{y_{1} - 3}{x_{1} - 2}$

Since AB is normal to the given line.

$∴ (\frac{y_{1} - 3}{x_{1} - 2}) \times (\frac{- 3}{4}) = - 1 \Rightarrow \frac{y_{1} - 3}{x_{1} - 2} = \frac{4}{3} \Rightarrow 3 y_{1} - 9 = 4 x_{1} - 8 \Rightarrow 4 x_{1} - 3 y_{1} = - 1 \Rightarrow 4 x_{1} - 3 (\frac{- 3}{4} x_{1} - 2) = - 1 [Using (1)] \Rightarrow 4 x_{1} + \frac{9}{4} x_{1} + 6 = - 1 \Rightarrow 16 x_{1} + 9 x_{1} + 24 = - 4 \Rightarrow 25 x_{1} = - 28 \Rightarrow x_{1} = \frac{- 28}{25} and ∴ y_{1} = \frac{3}{4} \times \frac{28}{25} - 2 = \frac{- 29}{25} Hence, required foot of the perpendicular is (\frac{- 28}{25}, \frac{- 29}{25}) .$

Find the coordinates of foot of perpendicular from the point (2,3) to the line 3x+4y+8=0.