 Consider the family of lines $5x + 3y - 2 + {\lambda _1}\left( {3x - y - 4} \right) = 0$ and $x - y + 1 + {\lambda _2}\left( {2x - y - 2} \right) = 0$ . Equation of a straight line that belongs to families is: $\left( a \right){\text{ 25x - 62y + 86 = 0}}$  $\left( b \right){\text{ 62x - 25y + 86 = 0}}$  $\left( c \right){\text{ 25x - 62y = 86}}$  $\left( d \right){\text{ 5x - 2y - 7 = 0}}$ Verified
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Hint: For solving it we will first reduce the families of lines to the standard form as $y = ax + b$ and then we will solve it for the value of $x$ and $y$ . So in the end we will substitute the values in a two-point form equation which is $\left( {y - {y_1}} \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$ and we will get the equation of the straight line.
Formula used:
Families of the line to the standard form,
$y = ax + b$
Equation of straight line for the two-point form will be,
$\left( {y - {y_1}} \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$

So we have the families of a line given as $5x + 3y - 2 + {\lambda _1}\left( {3x - y - 4} \right) = 0$ and from this we will get the standard form of the equation.
So by using the formula $y = ax + b$ and equating it, we get
$\Rightarrow 5x + 3y - 2 = 0$ , will be named as equation $1$
$\Rightarrow 3x - y - 4 = 0$ , will be named as equation $2$
Now for solving it, we will multiply the equation $2$ by $3$ , we have the equation as
$\Rightarrow 9x - 3y - 12 = 0$ , will be named as equation $3$
Now solving the equation $1\& 3$ , we get
$\Rightarrow 14x - 14 = 0$
And from here, $x = 1$
Substituting the value $x = 1$ in the equation $1$ , we get
$\Rightarrow y = - 1$
So the point of contact will be ${x_1},{y_1} = \left( {1, - 1} \right)$
Similarly, we have another family of the line which is
$x - y + 1 + {\lambda _2}\left( {2x - y - 2} \right) = 0$
From this, we have the equation as
$\Rightarrow x - y + 1 = 0$ , will be named as equation $4$
And another equation will be $2x - y - 2 = 0$ , will be named as equation $5$
By solving the equation $4\& 5$ , we get
$\Rightarrow x = 3$
And putting it in the equation $4$ , we get
$\Rightarrow y = 4$
Therefore, the contact of point will be ${x_2},{y_2} = 3,4$
Now by using the equation of the straight line for the two-point form which is
$\left( {y - {y_1}} \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$
And on substituting, we get
Equation of straight line for the two-point form will be,
$\Rightarrow \left( {y - \left( { - 1} \right)} \right) = \dfrac{{4 - \left( { - 1} \right)}}{{3 - 1}}\left( {x - 1} \right)$
And on solving it, we get
$\Rightarrow 2\left( {y + 1} \right) = 5\left( {x - 1} \right)$
And on multiplying the above equation, we get
$\Rightarrow 2y + 2 = 5x - 5$
Now on solving it, we get
$\Rightarrow 5x - 2y - 7 = 0$
Hence, the option $\left( d \right)$ is correct.
So, the correct answer is “Option d”.

Note: For solving this type of question, we need to know how we can find the values, and to find it we have to memorize the formula. For solving the equation we can also use the determinant form of the equation is more complex. Also, since, this type of question is lengthy so we should do the process clearly and stepwise so that we don’t get meshed up.