Question

The distance of the point \( (1, 2) \) from the line \( x + y + 5 = 0 \) measured along the line parallel to \( 3x - y = 7 \) is:

• A. \( \frac{4}{\sqrt{10}} \)
• B. 40
• C. \( \sqrt{40} \)
• D. \( \frac{10}{\sqrt{2}} \)

Hint:

To find the distance between a point and a line, find the intersection of the line with the line passing through the point and parallel to the given line.

Answer 1

The Correct Option is C

We are given the point \( (1, 2) \) and the line \( x + y + 5 = 0 \), and we are asked to find the distance of the point from the line measured along the line parallel to \( 3x - y = 7 \). Step 1: Find the slope of the line parallel to \( 3x - y = 7 \) The slope of the line \( 3x - y = 7 \) is given by: \[ \text{slope} = \frac{3}{1} = 3 \] Step 2: Find the equation of the line passing through \( (1, 2) \) with slope 3 The equation of the line passing through \( (1, 2) \) with slope 3 is: \[ y - 2 = 3(x - 1) \quad \Rightarrow \quad y = 3x - 1 \] Step 3: Find the intersection of the two lines To find the distance, we need to find the intersection of the line \( x + y + 5 = 0 \) and the line \( y = 3x - 1 \). Substitute \( y = 3x - 1 \) into \( x + y + 5 = 0 \): \[ x + (3x - 1) + 5 = 0 \quad \Rightarrow \quad 4x + 4 = 0 \quad \Rightarrow \quad x = -1 \] Substitute \( x = -1 \) into \( y = 3x - 1 \): \[ y = 3(-1) - 1 = -3 - 1 = -4 \] Thus, the intersection point is \( (-1, -4) \). Step 4: Calculate the distance The distance between \( (1, 2) \) and \( (-1, -4) \) is given by the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{((-1) - 1)^2 + ((-4) - 2)^2} = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} \] Thus, the correct answer is \( \sqrt{40} \).