If the lines ax + by + c = 0,bx + cy + a = 0, | vedclass

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(B) For the lines to be concurrent,the determinant of the coefficients must be zero: $\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix

Vedclass · Accepted Answer

$a + b + c = 0$

Vedclass · Answer

(B) For the lines to be concurrent,the determinant of the coefficients must be zero: $\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix} = 0$ Expanding the determinant: $a(cb - a^2) - b(b^2 - ac) + c(ab - c^2) = 0$ $abc - a^3 - b^3 + abc + abc - c^3 = 0$ $3abc - a^3 - b^3 - c^3 = 0$ $a^3 + b^3 + c^3 - 3abc = 0$ Using the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$ This implies $a + b + c = 0$ or $a^2 + b^2 + c^2 - ab - bc - ca = 0$ Since $a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$,the condition is $a + b + c = 0$ (assuming $a, b, c$ are not all equal).

If the lines $ax + by + c = 0$,$bx + cy + a = 0$,and $cx + ay + b = 0$ are concurrent,then:

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