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 the point $(\frac{2}{3},\frac{1}{3}).$

• A. Both the statements are True but Statement 2 is the correct explanation of Statement 1.
• B. Both the statements are True but Statement 2 is not the correct explanation of Statement 1.
• C. Statement 1 is True and Statement 2 is False.
• D. Statement 1 is False and Statement 2 is True.

Answer 1

The correct option is A Both the statements are True but Statement 2 is the correct explanation of Statement 1.

$x+y=1$ ...(i)
$x=2y$ ...(ii)
Solving the above two equations we get the point of intersection as $(\frac{2}{3},\frac{1}{3})$
Hence statement $B$ is true.
The family of above two lines can be written as
$a(x+y-1)+b(x-2y)=0$
$(a+b)x+(a-2b)y-a=0$
$(a+b)x+(a-2b)y=a$
For $a=1$ and $b=1,$ we get
$2x-y=1$
The point $(\frac{2}{3},\frac{1}{3})$ lies on the line $2x-y=1$
Statement 1 is true and follows from statement 2.