Question

The fix point through which the line $x(a+2b)+y(a+3b)=a+b$ always passes for all values of $a$ and $b$, is?

• A. $(2,1)$
• B. $(1,2)$
• C. $(2,-1)$
• D. $(1,-2)$

Answer 1

The correct option is C $(2,-1)$

$x(a+2b)+y(a+3b)=a+b$ value of a&b.

$x(a+2b)+y(a+3b)-a-b=0$

$\therefore xa+2xb+ay+3by-a-b=0$

$\therefore a(x+y-1)+b(2x+3y-1)=0$

hence

$x+y-1=0$ & $2x+3y-1=0$

$x+y=\mathrm{1...}\left(i\right)$

$2x+3y=1...\left(ii\right)$

$\to$solve (i) & (ii)

$2x+2y=2$

$2x+3y=1$

________________

$-y=1$

$y=-1$

$\to$ put value of y in equation (i)

$x-1=1$

$x=2$

$\to$ point (2, -1)