The lines (a+2b)x + (a-3b)y = a-b for differe | vedclass

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(C) The given equation of the line is $(a+2b)x + (a-3b)y = a-b$. Rearranging the terms to group coefficients of $a$ and $b$: $ax + 2bx + ay - 3by = a

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

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(C) The given equation of the line is $(a+2b)x + (a-3b)y = a-b$. Rearranging the terms to group coefficients of $a$ and $b$: $ax + 2bx + ay - 3by = a - b$ $a(x + y - 1) + b(2x - 3y + 1) = 0$ For this equation to hold for all values of $a$ and $b$,the coefficients must be zero: $x + y - 1 = 0$  $(i)$ $2x - 3y + 1 = 0$  (ii) From $(i)$,$y = 1 - x$. Substituting into (ii): $2x - 3(1 - x) + 1 = 0$ $2x - 3 + 3x + 1 = 0$ $5x - 2 = 0 \Rightarrow x = \frac{2}{5}$ Substituting $x = \frac{2}{5}$ into $(i)$: $\frac{2}{5} + y - 1 = 0 \Rightarrow y = 1 - \frac{2}{5} = \frac{3}{5}$ Thus,the fixed point is $\left(\frac{2}{5}, \frac{3}{5}\right)$.

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

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