Question

The coordinates of the foot of perpendicular from the point \( (2, 3) \) on the line \( y = 3x + 4 \) is given by:

• A. \( \left( \frac{37}{10}, \frac{-1}{10} \right) \)
• B. \( \left( \frac{-1}{10}, \frac{37}{10} \right) \)
• C. \( \left( \frac{10}{37}, -10 \right) \)
• D. \( \left( \frac{2}{2}, \frac{-1}{3} \right) \)

Hint:

For finding the foot of the perpendicular from a point to a line, use the slope condition \( m_1 \times m_2 = -1 \) and the fact that the foot of the perpendicular lies on the given line.

Answer 1

The Correct Option is B

Let the coordinates of the foot of the perpendicular from the point \( P(2,3) \) to the given line \( y = 3x + 4 \) be \( A(\alpha, \beta) \). Step 1: Find the slope of AP The slope of line \( y = 3x + 4 \) is: \[ m_2 = 3 \] The slope of line \( AP \) (joining \( P(2,3) \) and \( A(\alpha, \beta) \)) is: \[ m_1 = \frac{\beta - 3}{\alpha - 2} \] Since \( AP \) is perpendicular to the given line, their slopes satisfy: \[ m_1 \times m_2 = -1 \] \[ \frac{\beta - 3}{\alpha - 2} \times 3 = -1 \] Solving for \( \beta \): \[ 3\beta - 9 = -(\alpha - 2) \] \[ 3\beta = -\alpha + 11 \]
Step 2: Use the equation of the given line Since \( A(\alpha, \beta) \) lies on the given line \( y = 3x + 4 \), we substitute \( \alpha \) into the equation: \[ \beta = 3\alpha + 4 \]
Step 3: Solve for \( \alpha \) and \( \beta \) Equating the two expressions for \( \beta \): \[ 3\alpha + 4 = -\alpha + 11 \] \[ 3\alpha + \alpha = 11 - 4 \] \[ 4\alpha = 7 \] \[ \alpha = -\frac{1}{10} \] Substituting \( \alpha = -\frac{1}{10} \) into \( \beta = 3\alpha + 4 \): \[ \beta = 3 \times \left( -\frac{1}{10} \right) + 4 \] \[ = -\frac{3}{10} + \frac{40}{10} \] \[ = \frac{37}{10} \] Thus, the coordinates of the foot of the perpendicular are: \[ \left( \frac{-1}{10}, \frac{37}{10} \right) \] which corresponds to option (b).