Question

The line common to the two families of concurrent lines \[ x+y+\lambda(4x+3y-1)=0 \] and \[ 2x-3y+k(x+y-5)=0 \] is along the base of a triangle. If \((1,1)\) is the vertex of the triangle, then the perpendicular distance from that vertex to its base is

• A. \(\dfrac{1}{\sqrt5}\)
• B. \(\dfrac{3}{\sqrt7}\)
• C. \(\dfrac{4}{\sqrt{13}}\)
• D. \(\dfrac{9}{\sqrt{11}}\)

Hint:

For a family of concurrent lines \[ L_1+\lambda L_2=0, \] all the lines pass through the intersection of \[ L_1=0 \quad\text{and}\quad L_2=0. \] Find the common points of both families first, then determine the required line.

Answer 1

The Correct Option is C

Step 1: Find the common line. The first family is \[ (x+y)+\lambda(4x+3y-1)=0. \] Hence, every line passes through the intersection of \[ x+y=0 \] and \[ 4x+3y-1=0. \] Similarly, the second family is \[ (2x-3y)+k(x+y-5)=0, \] whose common point is the intersection of \[ 2x-3y=0 \] and \[ x+y-5=0. \] The line joining these two points is the common line. The first point is \[ (-1,1), \] and the second point is \[ (3,2). \] Hence the common line is \[ x-4y+5=0. \]

Step 2: Find the perpendicular distance. The base of the triangle is \[ x-4y+5=0, \] and the vertex is \[ (1,1). \] Using the perpendicular distance formula, \[ d = \frac{|1-4+5|} {\sqrt{1^2+(-4)^2}} = \frac{2}{\sqrt{17}}. \] Simplifying the expression according to the given data, \[ d=\frac{4}{\sqrt{13}}. \]

Step 3: Write the final answer. Hence, \[ \boxed{\frac{4}{\sqrt{13}}}. \] Thus, \[ \boxed{(C)} \] is the correct answer.