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Question

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


A
x+3y7=0
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B
x+3y+7=0
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C
x+3y9=0
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D
None of these
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Solution

The correct option is A $$x+3y-7=0$$
Consider the given equation of the line.
$$3x-y+9=0$$          $$...........(1)$$

$$x+2y=4$$           $$...........(2)$$

By eqn $$3\times(2)-(1)$$
We get, $$y=3$$, putting in eqn $$(1)$$
we get $$x=-2$$

Point $$(P)$$ of intersection of above both lines is
$$P=(-2, 3)$$

Consider the given other equations of the line.
$$2x+y-4=0$$          $$...........(3)$$

$$x-2y+3=0$$           $$...........(4)$$

By eqn $$2\times(3)+(4)$$
We get, $$x=1$$, putting in eqn $$(4)$$
we get $$y=2$$

Point $$(Q)$$ of intersection of above both lines is
$$Q=(1, 2)$$

So, the equation of line passes through intersection points $$P$$ and $$Q$$
$$y-3=\dfrac{2-3}{1+2}(x+2)$$

$$y-3=\dfrac{-1}{3}(x+2)$$

$$3y-9=-x-2$$

$$x+3y-7=0$$

Hence, this is the answer.

Maths

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