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Question

$$77/90$$ The lines $$2x+y-1=0,ax+3y-3=0$$ and $$3x+2y-2=0$$ are concurrent for 


A
All a
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B
a=4 only
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C
$-1< a<3$
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D
a>0 only
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Solution

The correct option is A All $$a$$
$$2x + y - 1 = 0 ..... \left( 1 \right)$$
$$ax + 3y - 3 = 0 ..... \left( 2 \right)$$
$$3x + 2y - 2 = 0 ..... \left( 3 \right)$$
These lines are concurrent, i.e., point of intersection of any two lines lies on the third line.
Intersection point of $$\left( 1 \right)$$ and $$\left( 3 \right)$$-
Multiplying eqiuation $$\left( 1 \right)$$ by $42$$, we get
$$4x + 2y - 2 = 0 ..... \left( 4 \right)$$
Subtracting equaton $$\left( 3 \right)$$ from $$\left( 4 \right)$$, we have
$$\left( 4x + 2y - 2 \right) - \left( 3x + 2y - 2 \right) = 0$$
$$x = 0$$
Substituting the value of $$x$$ in equation $$\left( 1 \right)$$, we have
$$0 + y - 1 = 0$$
$$y = 1$$
Now, the point $$\left( 0, 1 \right)$$ must satisfy $$\left( 2 \right)$$.
Therefore,
$$a \cdot 0 + 3 \cdot 1 - 3 = 0$$
$$0 = 0$$
Hence the given lines will be concurrent for all values of $$a$$.

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