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Question: If the lines \(ax+by+c=0,bx+cy+a=0\) and \(cx+ay+b=0\) be concurrent, then \(A){{a}^{3}}+{{b}^{3}}...

If the lines ax+by+c=0,bx+cy+a=0ax+by+c=0,bx+cy+a=0 and cx+ay+b=0cx+ay+b=0 be concurrent, then
A)a3+b3+c3+3abc=0A){{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3abc=0
B)a3+b3+c3−abc=0B){{a}^{3}}+{{b}^{3}}+{{c}^{3}}-abc=0
C)a3+b3+c3−3abc=0C){{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0
D)D)None of the above

Explanation

Solution

To solve the question we need to have the knowledge of types of lines. The solution will start with knowledge of concurrent lines. Lines are said to be concurrent if more than two lines pass through a common point. To check whether the lines are concurrent or not the determinant should be zero. In this question we will equate the determinant with zero.

Complete step by step answer:

The question asks us to find the relation between a, b, c if the three lines ax+by+c=0,bx+cy+a=0ax+by+c=0,bx+cy+a=0 and cx+ay+b=0cx+ay+b=0 which are given to us are concurrent or not. The steps that should be used to solve this question is that, as we know that for lines to be concurrent the determinant of the concurrent lines is equal to zero. On solving the determinants of the matrix which we get from the coefficient of x,yx,y from the lines given. The matrix formed using the three lines will be:
⇒∣x1y1c1 x2y2c2 x3y3c3 ∣=0\Rightarrow \left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & {{c}_{1}} \\\ {{x}_{2}} & {{y}_{2}} & {{c}_{2}} \\\ {{x}_{3}} & {{y}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|=0
The above is the formula for the coefficient of x,y,cx,y,c present in the line. On substituting the value we get:
⇒∣abc bca cac ∣=0\Rightarrow \left| \begin{matrix} a & b & c \\\ b & c & a \\\ c & a & c \\\ \end{matrix} \right|=0
⇒a(b×c−a×a)−b(b×b−c×a)+c(b×a−a×a)=0\Rightarrow a\left( b\times c-a\times a \right)-b\left( b\times b-c\times a \right)+c\left( b\times a-a\times a \right)=0
⇒a(bc−a2)−b(b2−ca)+c(ba−a2)=0\Rightarrow a\left( bc-{{a}^{2}} \right)-b\left( {{b}^{2}}-ca \right)+c\left( ba-{{a}^{2}} \right)=0
⇒3abc−a3−b3−c3=0\Rightarrow 3abc-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=0
⇒a3+b3+c3−3abc=0\Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0
∴\therefore If the lines ax+by+c=0,bx+cy+a=0ax+by+c=0,bx+cy+a=0 and cx+ay+b=0cx+ay+b=0 be concurrent, thena3+b3+c3−3abc=0{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0

So, the correct answer is “Option C”.

Note: Sometimes it may happen that the equation of the line given to us would have only a single term either of X and Y. If any of the terms will be missing then in that case the coefficient of that term will be taken as zero. For example: if a line a x+c=0x+c=0 is given, it means that the coefficient will be written as zero in the matrix.