Question

For what value of a will the lines $3x+4y+1=0,ax+2y-3=0$ and $2x-y-3=0$ be concurrent ?

Answer 1

It is given that lines

$3x+4y+1=0-----\left(1\right)$

$ax+2y-3=0-----\left(2\right)$

$2x-y-3=0------\left(3\right)$

are concurrent.

$\therefore$ We know that,

Three lines are concurrent if they pass through a common point.

then,

Finding point of intersection of lines

$\left(1\right)$ and $\left(3\right)$

$3x+4y+1=0$

$[2x-y-3=0]\times 4$ on solving by direct method.

$3x+4y+1=08x-4y-12=0\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_onadding11x-11=0$

$11x=11$

$x=1$

put in equation $\left(3\right)$

$2x-y-3=0$

$2\times 1-y-3=0$

$-y-1=0$

$y=-1$

substitute value of $x$ and $y$ in equation $\left(2\right)$

$ax+2y-3=0$

$a\times 1+2\times (-1)-3=0$

$a-2-3=0$

$a-5=0$

$a=5$

Hence, this is answer.