Question

Let \( A \) be the point of intersection of the lines \( 3x + 2y = 14 \), \( 5x - y = 6 \) and \( B \) be the point of intersection of the lines \( 4x + 3y = 8 \), \( 6x + y = 5 \). The distance of the point \( P(5, -2) \) from the line \( AB \) is

• A. \( \frac{13}{2} \)
• B. 8
• C. \( \frac{5}{2} \)
• D. 6

Answer 1

The Correct Option is D

The objective is to calculate the distance between point \( P(5, -2) \) and line \( AB \). This requires first establishing the equation of line \( AB \), where points \( A \) and \( B \) are the intersection points of specified pairs of lines.

1. Determine intersection point \( A \) of lines \( 3x + 2y = 14 \) and \( 5x - y = 6 \).
   • Solve the system of equations. Multiply the second equation by 2: \( 5x - y = 6 \implies 10x - 2y = 12 \).
   • Add this modified equation to the first equation: \( 3x + 2y = 14 \) and \( 10x - 2y = 12 \).
   • Summing yields: \( 13x = 26 \implies x = 2 \).
   • Substitute \( x = 2 \) into the first equation: \( 3(2) + 2y = 14 \implies 6 + 2y = 14 \implies 2y = 8 \implies y = 4 \).
   • Therefore, point \( A \) is \( (2, 4) \).
2. Determine intersection point \( B \) of lines \( 4x + 3y = 8 \) and \( 6x + y = 5 \).
   • Solve by elimination. Multiply the second equation by 3: \( 6x + y = 5 \implies 18x + 3y = 15 \).
   • Subtract the first equation from the result: \( 18x + 3y = 15 \) and \( 4x + 3y = 8 \).
   • Subtracting yields: \( 14x = 7 \implies x = \frac{1}{2} \).
   • Substitute \( x = \frac{1}{2} \) into the first equation: \( 4(\frac{1}{2}) + 3y = 8 \implies 2 + 3y = 8 \implies 3y = 6 \implies y = 2 \).
   • Therefore, point \( B \) is \( (\frac{1}{2}, 2) \).
3. Derive the equation of line \( AB \) using points \( A(2, 4) \) and \( B(\frac{1}{2}, 2) \).
   • Calculate the slope \( m \) of \( AB \): \( m = \frac{2 - 4}{\frac{1}{2} - 2} = \frac{-2}{-\frac{3}{2}} = \frac{4}{3} \).
   • Apply the point-slope form using point \( A(2, 4) \): \( y - 4 = \frac{4}{3}(x - 2) \).
   • Simplify to obtain the equation of line \( AB \): \( 3y = 4x + 4 \implies 4x - 3y + 4 = 0 \).
4. Calculate the distance from point \( P(5, -2) \) to the line \( 4x - 3y + 4 = 0 \).
   • The distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \).
   • Substitute \( A = 4 \), \( B = -3 \), \( C = 4 \), \( x_1 = 5 \), \( y_1 = -2 \): \( d = \frac{|4(5) + (-3)(-2) + 4|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 + 6 + 4|}{\sqrt{16 + 9}} \).
   • Simplify: \( d = \frac{30}{5} = 6 \).

The distance of point \( P(5, -2) \) from line \( AB \) is 6.