Question

A line \(L\) passes through the point of intersection of the lines \(3x+y-10=0\) and \(x-y-2=0\). If the perpendicular distance of the line \(L\) from the point \((5,1)\) is exactly \(\dfrac{2}{\sqrt{5}}\) units, which of the following represents the equation of line \(L\)?

• A. \(2x-y-5=0\)
• B. \(x+2y-5=0\)
• C. \(2x+y-7=0\)
• D. \(x-2y+1=0\)

Hint:

When multiple line equations pass through the same point, use the distance condition carefully to identify the correct line.

Answer 1

The Correct Option is B

Concept: To determine the required line, we first find the intersection point of the two given lines. Since the required line passes through this point, each option must satisfy that condition. Then we use the perpendicular distance formula: \[ \text{Distance}= \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} \] for a line \(Ax+By+C=0\) and point \((x_1,y_1)\).

Step 1: Find the intersection point of the given lines. The equations are: \[ 3x+y-10=0 \] \[ x-y-2=0 \] From the second equation, \[ x-y=2 \] \[ y=x-2 \] Substitute into the first equation: \[ 3x+(x-2)-10=0 \] \[ 4x-12=0 \] \[ x=3 \] Then, \[ y=3-2=1 \] Thus, the point of intersection is: \[ (3,1) \]

Step 2: Check which option passes through \((3,1)\). For option (A): \[ 2(3)-1-5=0 \] \[ 6-1-5=0 \] So option (A) passes through the point. For option (B): \[ 3+2(1)-5=0 \] \[ 3+2-5=0 \] Hence option (B) also passes. For option (C): \[ 2(3)+1-7=0 \] \[ 6+1-7=0 \] Hence option (C) also passes. For option (D): \[ 3-2(1)+1=2\neq0 \] So option (D) is rejected.

Step 3: Use the perpendicular distance condition. Distance from point \((5,1)\) to line: \[ x+2y-5=0 \] is \[ \frac{|5+2(1)-5|}{\sqrt{1^2+2^2}} = \frac{2}{\sqrt5} \] which matches the given condition. Hence the required line is: \[ \boxed{x+2y-5=0} \]