If a and b are two arbitrary constants,then t | vedclass

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(A) The given equation is $(a - 2b)x + (a + 3b)y + 3a + 4b = 0$.Rearranging the terms with respect to $a$ and $b$,we get:$a(x + y + 3) + b(-2x + 3y +

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$(-1, -2)$

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(A) The given equation is $(a - 2b)x + (a + 3b)y + 3a + 4b = 0$.Rearranging the terms with respect to $a$ and $b$,we get:$a(x + y + 3) + b(-2x + 3y + 4) = 0$.This represents a family of straight lines passing through the intersection of the lines $x + y + 3 = 0$ and $-2x + 3y + 4 = 0$.Solving the system of equations:$x + y = -3$ $(i)$$-2x + 3y = -4$ (ii)Multiplying $(i)$ by $2$: $2x + 2y = -6$.Adding this to (ii): $(2x - 2x) + (2y + 3y) = -6 - 4$,which gives $5y = -10$,so $y = -2$.Substituting $y = -2$ into $(i)$: $x - 2 = -3$,so $x = -1$.Thus,the point of intersection is $(-1, -2)$.

If $a$ and $b$ are two arbitrary constants,then the straight line $(a - 2b)x + (a + 3b)y + 3a + 4b = 0$ will pass through

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