  Question

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

A
All a  B
a=4 only  C
$-1< a<3$  D
a>0 only  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) = 0x = 0$$Substituting the value of$$x$$in equation$$\left( 1 \right)$$, we have$$0 + y - 1 = 0y = 1$$Now, the point$$\left( 0, 1 \right)$$must satisfy$$\left( 2 \right)$$.Therefore,$$a \cdot 0 + 3 \cdot 1 - 3 = 00 = 0$$Hence the given lines will be concurrent for all values of$$a$\$.Maths

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