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

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 $y - {y_1} = m\left( {x - {x_1}} \right)$ .