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Q)

Find the equation of straight line which passes through the intersection of the lines $3x-4y+1=0$ and $5x+y-1=0$ and $(2,1)$?

$\begin{array}{1 1}(a)\;(3x-4y+1)-\large\frac{3}{10}\normalsize (5x+y-1)=0\\(b)\;(x-4y+1)-\large\frac{3}{13}\normalsize (5x+y-1)=0\\(c)\;(3x-2y+1)-\large\frac{4}{10}\normalsize (5x+y-1)=0\\(d)\;(x-y+1)-\large\frac{3}{10}\normalsize (5x+y+1)=0\end{array}$ Comment
A)
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• Equation of all the lines passing through the point of intersection of the lines $L_1=0$ and $L_2=0$ is given by $L_1+\lambda L_2=0$
Given: $L_1=3x-4y+1=0$ and $L_2=5x+y-1=0$
Equation of any line passing through the intersection of the given two lines lines is given by
$(3x-4y+1)+\lambda(5x+y-1)=0$..........(i)
Also it is given that (i) passes through $(2,1).$
$\Rightarrow\:(6-4+1)+\lambda(10+1-1)=0$
$3+\lambda (10)=0$
$\lambda =\large\frac{-3}{10}$
Substituting the value of $\lambda$ in (i) we get the required equation as
$(3x-4y+1)-\large\frac{3}{10}$$(5x+y-1)=0$
Hence (a) is the correct answer.