If a + b + c = 0 and p neq 0,the lines ax + ( | vedclass

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(A) Given the condition $a + b + c = 0$,we can substitute $(b + c) = -a$,$(c + a) = -b$,and $(a + b) = -c$.The equations of the lines become:$ax - ay

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Do not intersect

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(A) Given the condition $a + b + c = 0$,we can substitute $(b + c) = -a$,$(c + a) = -b$,and $(a + b) = -c$.The equations of the lines become:$ax - ay = p \implies x - y = \frac{p}{a}$$bx - by = p \implies x - y = \frac{p}{b}$$cx - cy = p \implies x - y = \frac{p}{c}$Since $p 
eq 0$ and $a, b, c$ are distinct (assuming non-zero values for the lines to be distinct),the slopes of all three lines are $1$. Because the lines have the same slope but different $y$-intercepts (since $\frac{p}{a} 
eq \frac{p}{b} 
eq \frac{p}{c}$),they are parallel to each other.Therefore,the lines do not intersect in the finite plane.

If $a + b + c = 0$ and $p \neq 0$,the lines $ax + (b + c)y = p$,$bx + (c + a)y = p$ and $cx + (a + b)y = p$ are:

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