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

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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