The lines (a+2b)x+(a-3b)y=a-b for different values of a and b pass through the fixed point whose coordinates are

Question

The lines (a+2b)x+(a-3b)y=a-b for different values of a and b pass through the fixed point whose coordinates are (A)  ( (2/5),(2/5) )  (B)  ( (3/5),(3/5) )  (C)  ( (1/5),(1/5) )  (D)  ( (3/5),(2/5) )

Tardigrade Admin · Accepted Answer

Given equation is (a+2b) text x+(a-3b) text y=a-b It can be rewritten as a(x+y-1)+b(2x-3y+1)=0 This is the form of intersection of two lines. ∴ x+y-1=0 and 2x-3y+1=0 On solving, we get x=(2/5) and y=(3/5)

Q. The lines $(a + 2 b) x + (a - 3 b) y = a - b$ for different values of $a$ and $b$ pass through the fixed point whose coordinates are

$(\frac{2}{5}, \frac{2}{5})$

$(\frac{3}{5}, \frac{3}{5})$

$(\frac{1}{5}, \frac{1}{5})$

$(\frac{3}{5}, \frac{2}{5})$

$(\frac{2}{5}, \frac{3}{5})$

Given equation is $(a + 2 b) x + (a - 3 b) y = a - b$
It can be rewritten as
$a (x + y - 1) + b (2 x - 3 y + 1) = 0$
This is the form of intersection of two lines.
$∴$ $x + y - 1 = 0$ and $2 x - 3 y + 1 = 0$
On solving, we get
$x = \frac{2}{5}$ and $y = \frac{3}{5}$

Q. The lines (a+2b)x+(a−3b)y=a−b for different values of a and b pass through the fixed point whose coordinates are

(52​,52​)

(53​,53​)

(51​,51​)

(53​,52​)

(52​,53​)

Given equation is (a+2b)x+(a−3b)y=a−b It can be rewritten as a(x+y−1)+b(2x−3y+1)=0 This is the form of intersection of two lines. ∴ x+y−1=0 and 2x−3y+1=0 On solving, we get x=52​ and y=53​