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 \( A(\alpha, \beta) \) be the foot of the perpendicular from point \( P(2,3) \) to the line \( y = 3x + 4 \).Step 1: Determine the slope of AP The slope of the given line \( y = 3x + 4 \) is \( m_2 = 3 \).The slope of the line segment \( AP \), connecting \( P(2,3) \) and \( A(\alpha, \beta) \), is \( m_1 = \frac{\beta - 3}{\alpha - 2} \).Since \( AP \) is perpendicular to the given line, the product of their slopes is \( -1 \): \( m_1 \times m_2 = -1 \).Substituting the slopes, we get \( \frac{\beta - 3}{\alpha - 2} \times 3 = -1 \).Solving for \( \beta \): \( 3\beta - 9 = -(\alpha - 2) \), which simplifies to \( 3\beta = -\alpha + 11 \).
Step 2: Utilize the equation of the given line As \( A(\alpha, \beta) \) lies on the line \( y = 3x + 4 \), its coordinates satisfy the equation: \( \beta = 3\alpha + 4 \).
Step 3: Solve for \( \alpha \) and \( \beta \) Equating the two expressions for \( \beta \): \( 3\alpha + 4 = -\alpha + 11 \).Rearranging the terms to solve for \( \alpha \): \( 3\alpha + \alpha = 11 - 4 \), which gives \( 4\alpha = 7 \), so \( \alpha = -\frac{1}{10} \).Substitute the value of \( \alpha \) into the equation \( \beta = 3\alpha + 4 \): \( \beta = 3 \times \left( -\frac{1}{10} \right) + 4 = -\frac{3}{10} + \frac{40}{10} = \frac{37}{10} \).Therefore, the coordinates of the foot of the perpendicular are \( \left( \frac{-1}{10}, \frac{37}{10} \right) \).