Question

Find the coordinates of the foot of perpendicular from the point \( (–1, 3) \) to the line \(3x – 4y – 16 = 0.\)

Answer 1

Let (a, b) be the coordinates of the foot of the perpendicular from the point (-1, 3) to the line \(3x - 4y - 16 = 0.\)

Slope of the line joining (-1, 3) and (a, b),  \(m_1=\frac{b-3}{a+1}\)

Slope of the line \(3x - 4y - 16 = 0\)  or  \( y=\frac{3}{4}x-4\) , \(m_2 =\frac{ 3}{4}\)

Since these two lines are perpendicular, \(m_1m_2 = -1\)

\(∴\frac{(b-3)}{(a+1)} \times (\frac{3}{4}) = -1\)

\(⇒\frac{\left(3b-9\right)}{\left(4a+4\right)} = -1\)

\(⇒3b–9=-4a – 4\)
\(⇒4a+3b=5 …….(1)\)
Point \((a, b)\) lies on line \(3x - 4y = 16.\)
\(∴3a - 4b = 16 … (2) \)
On solving equations (1) and (2), we obtain

\(a = \frac{68}{25} \) and \(b = \frac{-49}{25}\)

Thus, the required coordinates of the foot of the perpendicular are\( \left(\frac{68}{25},\frac{ -49}{25}\right)\)