Question

Consider the family of lines \[(x + y - 1) + \lambda (2x + 3y - 5) = 0\] and \[(3x + 2y - 4) + \mu (x + 2y - 6) = 0\]. Find the equation of the straight line that belongs to both the families.
A. \[x - 2y - 8 = 0\]
B. \[x - 2y + 8 = 0\]
C. \[2x + y - 8 = 0\]
D. \[2x + y - 8 = 0\]

Answer 1

Hint: For any family of lines \[{L_1} + \lambda {L_2} = 0\]all the lines belong to the family that passes through the points \[{L_1} = 0\] and \[{L_2} = 0\]. For this question, equate the \[{L_1}\]and \[{L_2}\]for both the family of lines to 0. You will get two passing points from the given two equation then calculate the equation of straight line by using \[\dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]

Complete step by step solution:
It is given that Family of lines is:
\[(x + y - 1) + \lambda (2x + 3y - 5) = 0\] …… (i)
\[(3x + 2y - 4) + \mu (x + 2y - 6) = 0\] ………… (ii)
A line passes belongs to both these families
For any family of lines,\[{L_1} + \lambda {L_2} = 0\]
We equate, \[{L_1} = 0\]and \[{L_2} = 0\]
For family of lines (i),
\[{L_1} = x + y - 1\] and \[{L_2} = 2x + 3y - 5\]
Equating to \[{L_1} = 0\]
\[x + y - 1 = 0\] …………… (iii)
Equating to \[{L_2} = 0\]
\[2x + 3y - 5 = 0\] ………… (iv)
So, we will calculate the value of y in terms of x so we get from equation (iii) ,
\[ \Rightarrow y = - x + 1\]
We will substitute the value of y in equation (iv)
\[ \Rightarrow 2x + 3\left( { - x + 1} \right) - 5 = 0\]
\[ \Rightarrow 2x - 3x + 3 - 5 = 0\]
On simplifying we get,
\[ \Rightarrow - x - 2 = 0\]
Therefore, \[x = - 2\]
Hence, on substituting the value of x we get \[y = 3\]
Therefore, we can say that the first family of lines will pass through the points \[\left( { - 2,3} \right)\]
For family of lines (ii),
\[{L_1} = 3x + 2y - 4\] and \[{L_2} = x + 2y - 6\]
Equating to \[{L_1} = 0\]
\[3x + 2y - 4 = 0\] …………… (v)
Equating to \[{L_2} = 0\]
\[x + 2y - 6 = 0\] ………… (vi)
So, we will calculate the value of x in terms of y so we get from equation (vi) ,
\[ \Rightarrow x = - 2y + 6\]
We will substitute the value of x in equation (v)
\[ \Rightarrow 3\left( { - 2y + 6} \right) + 2y - 4 = 0\]
\[ \Rightarrow - 6y + 18 + 2y - 4 = 0\]
On simplifying we get,
\[ \Rightarrow - 4y + 14 = 0\]
Therefore, \[y = \dfrac{7}{2}\]
Hence, on substituting the value of y we get \[x = - 1\]
Therefore, we can say that the second family of lines will pass through the points \[\left( { - 1,\dfrac{{ - 7}}{2}} \right)\]
Hence, the equation of line passing through the points \[\left( { - 2,3} \right) = \left( {{x_1},{y_1}} \right)\] and \[\left( { - 1,\dfrac{{ - 7}}{2}} \right) = \left( {{x_2},{y_2}} \right)\] is given by
\[\dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
On substituting the values we get,
\[ \Rightarrow \dfrac{{y - 3}}{{x + 2}} = \dfrac{{\dfrac{{ - 7}}{2} - 3}}{{ - 1 + 2}}\]
After solving right hand side we get,

\[ \Rightarrow \dfrac{{y - 3}}{{x + 2}} = \dfrac{1}{2}\]
On cross multiplying we get,
\[ \Rightarrow 2y - 6 = x + 2\]
Hence, the equation of straight line that belongs to both the families is \[x - 2y + 8 = 0\]

So, the correct answer is “Option B”.

Note: Remember that for every family of lines \[{L_1} + \lambda {L_2} = 0\], all the lines passing through \[{L_1} = 0\] and \[{L_2} = 0\]belong to the family. Also, all lines that satisfy \[{L_1} = 0\] and \[{L_2} = 0\] must belong to the family of lines. You must be able to identify \[{L_1}\] and \[{L_2}\] in a given family of lines. To find the equation of lines, you must remember the general equation of a straight line that is \[y - {y_1} = m\left( {x - {x_1}} \right)\] .