Question

If the line \(L = x \cos \alpha + y \sin \alpha - p = 0\) represents a line perpendicular to the line \(x + y + 1 = 0\) and \(p\) is positive, \(a\) lies in the fourth quadrant and perpendicular distance from \(\left(\sqrt{2}, \sqrt{2}\right)\) to the line \(L = 0\) is 5 units, then find \(p\):

• A. 5
• B. \(\frac{5}{2}\)
• C. 10
• D. \(\frac{15}{2}\)

Hint:

For problems involving perpendicular distance from a point to a line, use the distance formula and apply the correct values for the point and line coefficients.

Answer 1

The Correct Option is A

Step 1: Identify the slope of the given line. The given line equation is: \[ x + y + 1 = 0 \] Rewriting it in slope-intercept form: \[ y = -x - 1 \] Thus, its slope is \( m = -1 \). Step 2: Find the slope of the required line. Since the given line is perpendicular to \( x + y + 1 = 0 \), the slope of the required line is the negative reciprocal of \( -1 \), which is: \[ m' = 1 \] Thus, the equation of the required line is of the form: \[ x \cos\alpha + y \sin\alpha - p = 0 \] Since the slope is 1, comparing with the general form: \[ -\frac{\cos\alpha}{\sin\alpha} = 1 \] \[ \cos\alpha = -\sin\alpha \] Step 3: Use perpendicular distance formula. The given perpendicular distance from \( (\sqrt{2}, \sqrt{2}) \) to the line is 5. The distance formula for a line \( Ax + By + C = 0 \) from a point \( (x_1, y_1) \) is: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] Substituting \( A = \cos\alpha \), \( B = \sin\alpha \), and \( C = -p \): \[ 5 = \frac{|(\cos\alpha)(\sqrt{2}) + (\sin\alpha)(\sqrt{2}) - p|}{\sqrt{\cos^2\alpha + \sin^2\alpha}} \] Since \( \cos^2\alpha + \sin^2\alpha = 1 \), simplifying: \[ 5 = \left| \sqrt{2} (\cos\alpha + \sin\alpha) - p \right| \] From \( \cos\alpha = -\sin\alpha \), we substitute: \[ \cos\alpha + \sin\alpha = 0 \] \[ 5 = \left| -p \right| \] Since \( p \) is positive: \[ p = 5 \] Thus, the correct answer is: \[ \boxed{5} \]