Statement (A): If 3a - 2b + 5c = 0,then the l | vedclass

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(A) Given the line equation: $ax + by + c = 0$ $(i)$Given the condition: $3a - 2b + 5c = 0$Dividing by $5$,we get: $\frac{3}{5}a - \frac{2}{5}b + c =

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Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

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(A) Given the line equation: $ax + by + c = 0$ $(i)$Given the condition: $3a - 2b + 5c = 0$Dividing by $5$,we get: $\frac{3}{5}a - \frac{2}{5}b + c = 0$ $(ii)$Comparing $(i)$ and $(ii)$,we can write the equation as:$a(x - \frac{3}{5}) + b(y + \frac{2}{5}) = 0$This represents a family of lines passing through the intersection of $x - \frac{3}{5} = 0$ and $y + \frac{2}{5} = 0$.Thus,the lines are concurrent at the point $(\frac{3}{5}, -\frac{2}{5})$.Since the family of lines $L_1 + \lambda L_2 = 0$ is concurrent at the intersection of $L_1$ and $L_2$,$(R)$ is the correct explanation for $(A)$.

Statement $(A)$: If $3a - 2b + 5c = 0$,then the line $ax + by + c = 0$ is always concurrent at a point.
Reason $(R)$: If $L_1 = 0$ and $L_2 = 0$ are two lines,then the family of lines $L_1 + \lambda L_2 = 0$ is concurrent at the intersection of $L_1$ and $L_2$.

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Statement $(A)$: If $3a - 2b + 5c = 0$,then the line $ax + by + c = 0$ is always concurrent at a point.Reason $(R)$: If $L_1 = 0$ and $L_2 = 0$ are two lines,then the family of lines $L_1 + \lambda L_2 = 0$ is concurrent at the intersection of $L_1$ and $L_2$.