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

Answer 1

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

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

$x=0$

Substituting the value of $x$ in equation $\left(1\right)$, we have

$0+y-1=0$

$y=1$

Now, the point $(0,1)$ 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 $a$.