Question

If the family of lines \[x\left( {a + 2b} \right) + y\left( {a + 3b} \right) = a + b\] passes through the point for all values of a and b, then the coordinates of the point is:
A.(2,1)
B.(2,-1)
C.(-2,1)
D.None of these

Answer 1

Hint: In this question, we need to determine the coordinates of the point such that a family of lines equation is given so, by resolving the equation we will find the equation of the lines from the family and then solving those equations we will find the coordinates of the intersection of those points.

Complete step-by-step answer:
Given the equation of the family of the line is \[x\left( {a + 2b} \right) + y\left( {a + 3b} \right) = a + b\]
By solving we can write this equation as
 \[
  xa + 2xb + ya + 3yb = a + b \\
  xa + ya - a + 2xb + 3y - b = 0 \\
  a\left( {x + y -1} \right) + b\left( {2x + 3y -1} \right) = 0
 \]
The family of lines passes through the intersection of the lines \[x + y - 1 - - (i)\] and \[2x + 3y - 1 - - (ii)\] , so we will find the coordinates of the point by solving both the equations.
Now multiply the equation (i) by 3, so we get
 \[3x + 3y - 3 - - (iii)\]
So by solving the equation (ii) and the equation (iii)
 \[
   + 3x + 3y - 3 = 0 \\
   \underline {\mathop + \limits_ - 2x\mathop + \limits_ - 3y\mathop - \limits_ + 1 = 0} \\
   x - 2 = 0 \\
   x = 2
 \]
We get the value of \[x = 2\] , now substitute the value of \[x = 2\] in the equation (i), hence
 \[
  2 + y - 1 = 0 \\
  1 + y = 0 \\
  y = -1
 \]
We get the value of \[y = - 1\]
Therefore, the coordinates of the required point is (2,-1)
So, the correct answer is “Option B”.

Note: A family of lines is a set of lines that have something in common with each other such that all the lines which belong to the same family must have the same slope and the intercept are the same. In other words, a family of lines does not represent a line but it represents a group of lines having similar characteristics.