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Question

Find the equation of  the line passing through the point $$(0, -2)$$ and the point of intersection of the lines $$4x+3y=1$$ and $$3x-y+9=0$$.


Solution

Given that,
$$4x+3y=1 \ \longrightarrow(1)$$
$$3x-y+9=0\longrightarrow (2)$$

Solving  the equations simultaneously, we get
$$ 4x+3y=1\times 1\\ 3(3x-y)=-9\times 3\\ \\ 4x+3y=1\\ 9x-3y=-27\\ 13x\quad =-26\\ x=-2\\ y=-3\\ B(-2,-3)$$

$$ \therefore$$ Slope of the passing through the point $$(0,-2) \ and \ (-2,-3)$$
$$\Rightarrow m=\dfrac { -3+2 }{ -2-0 } =\dfrac { -1 }{ -2 }$$ $$\Rightarrow { m }=\dfrac { 1 }{ 2 } $$

So, the equation of line is given as, 
$$y-y_1=m(x-x_1)$$
$$\Rightarrow$$ $$ (y+2)=\cfrac { 1 }{ 2 } (x-0)$$
$$\Rightarrow$$ $$2y+4=x$$
$$\Rightarrow$$ $$ x-2y+4=0$$

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