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
- \( \frac{13}{2} \)
- 8
- \( \frac{5}{2} \)
- 6
The Correct Option is D
Approach Solution - 1
To find the distance of a point \( P(5, -2) \) from the line \( AB \), we first need to determine the equation of the line \( AB \), where points \( A \) and \( B \) are the intersections of given lines.
- Find the intersection point \( A \) of the lines \( 3x + 2y = 14 \) and \( 5x - y = 6 \).
- Solve the two equations simultaneously. First, multiply the second equation by 2 to eliminate \( y \): \(5x - y = 6 \implies 10x - 2y = 12\)
- Now, add the modified second equation to the first equation: \(3x + 2y = 14\), \(10x - 2y = 12\)
- Adding gives: \(13x = 26 \implies x = 2\)
- Substitute \( x = 2 \) back into the first equation: \(3(2) + 2y = 14 \implies 6 + 2y = 14 \implies 2y = 8 \implies y = 4\)
- Thus, point \( A \) is \( (2, 4) \).
- Find the intersection point \( B \) of the lines \( 4x + 3y = 8 \) and \( 6x + y = 5 \).
- Solve similarly by elimination. Multiply the second equation by 3: \(6x + y = 5 \implies 18x + 3y = 15\)
- Subtract the first equation from this result: \(18x + 3y = 15\), \(4x + 3y = 8\)
- Subtracting gives: \(14x = 7 \implies x = \frac{1}{2}\)
- Substitute \( x = \frac{1}{2} \) back into the first equation: \(4(\frac{1}{2}) + 3y = 8 \implies 2 + 3y = 8 \implies 3y = 6 \implies y = 2\)
- Thus, point \( B \) is \( (\frac{1}{2}, 2) \).
- Determine the equation of line \( AB \) using points \( A(2, 4) \) and \( B(\frac{1}{2}, 2) \).
- The slope \( m \) of \( AB \) is \(m = \frac{2 - 4}{\frac{1}{2} - 2} = \frac{-2}{-\frac{3}{2}} = \frac{4}{3}\)
- Using point-slope form with point \( A(2, 4) \): \(y - 4 = \frac{4}{3}(x - 2)\)
- Simplifying gives the equation of line \( AB \): \(3y = 4x + 4 \implies 4x - 3y + 4 = 0\)
- Find the distance from point \( P(5, -2) \) to the line \( 4x - 3y + 4 = 0 \).
- The distance \( d \) from a point to a line \( Ax + By + C = 0 \) is given by: \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
- 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: \(d = \frac{30}{5} = 6\)
Therefore, the distance of the point \( P(5, -2) \) from the line \( AB \) is 6.
Approach Solution -2
Step 1. Find the coordinates of \( A \) by solving the lines \( L_1: 3x + 2y = 14 \) and \( L_2: 5x - y = 6 \):
Solving these equations gives \( A(2, 4) \).
Step 2. Find the coordinates of \( B \) by solving the lines \( L_3: 4x + 3y = 8 \) and \( L_4: 6x + y = 5 \):
Solving these equations gives \( B\left(\frac{1}{2}, 2\right) \).
Step 3. Determine the equation of line \( AB \) passing through points \( A(2, 4) \) and \( B\left(\frac{1}{2}, 2\right) \):
The equation of \( AB \) is \( 4x - 3y + 4 = 0 \).
Step 4. Calculate the distance from \( P(5, -2) \) to the line \( AB: 4x - 3y + 4 = 0 \):
\(\text{Distance} = \frac{|4(5) - 3(-2) + 4|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 + 6 + 4|}{\sqrt{16 + 9}} = \frac{30}{5} = 6.\)
So, the distance of point \( P \) from the line \( AB \) is 6.
The Correct Answer is: 6
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