Question

$\huge\rm\underline\blue{Question:-}$

The lines (a+2b)x + (a-3b)y = a - b ,For different values of a and b passes through the fixed point

1] $\large\rm{(\frac{2}{5},\frac{2}{5})}$

2] $\large\rm{(\frac{2}{5},\frac{3}{5})}$

3] $\large\rm{\frac{3}{5},\frac{3}{5})}$

4] $\large\rm{\frac{3}{5},\frac{2}{5})}$

(Solve this question using family of lines through the intersection of lines concept . Please don't spam if u do so ur answers will be reported)


​

Answer 1

Answer:

Equation of line is

(a+2b)x+(a−3b)y=a−b

⇒(a+2b)x+(a−3b)y−a+b=0

⇒ax+2bx+ay−3by−a+b=0

⇒(x+y−1)a+(2x−3y+1)b=0

Hence,

x+y−1=0......(1)

2x−3y+1=0......(2)

On solving ( 1) and ( 2) to and we get,

The answer of the solution which is option(b)- 2/5,3/5..

Hence, this is the answer.

Answer 2

Answer:

$\longrightarrow\sf(a+2b)x + (a-3b)y = a-b} \\ \\ \longrightarrow\sf{(a+2b)x + (a-3b)y - a+b = 0} \\ \\ \longrightarrow\sf{ax+2bx+ay-3by-a+b = 0} \\ \\ \longrightarrow\sf{(x+y-1)a+(2x-3y+1)b=0}$

2] $\large\rm{(\frac{2}{5},\frac{3}{5})}$ is correct one.

$\sf\underline\red{Thanks}$