Question

Statement 1: The lines $(a+b)x+(a-2b)y=a$ are concurrent at the point $(\frac{2}{3},\frac{1}{3})$

Statement 2: The lines $x+y-1=0$ and $x-2y=0$ intersect at point $(\frac{2}{3},\frac{1}{3})$

Answer 1

Given equation is
$(a+b)x+(a-2b)y=a\Rightarrow a(x+y-1)+b(x-2y)=a\Rightarrow a(x+y-1)+b(x-2y)=0\Rightarrow (x+y-1)+\frac{b}{a}(x-2y)=0[\because a\ne 0]$
Which represents a family of line which passes through the point of intersection of the lines $x+y-1=0$ and $x-2y=0$
Solving the above two equations we get,
$x=\frac{2}{3}\text{and}y=\frac{1}{3}$
Therefore, lines are concurrent at $(\frac{2}{3},\frac{1}{3})$
Clearly statement 2 is true and is the correct explanation of statement 1.

Hence option $\left(A\right)$ is the correct option.