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AP EAMCET Engineering 2015 Code C Paper

Section: Mathematics

Question : 11 of 160

Marks: +1, -0

If the lines

x+2ay+a=0,

x+3by+b=0,

x+4cy+c=0

are concurrent, then a, b, c are in

Solution:

Area of the triengle formed by these lines

=0

\Rightarrow \frac{1}{2} \begin{vmatrix} 1 & 2a & a \\ 1 & 36 & b \\ 1 & 4c & c \end{vmatrix}=0

\Rightarrow \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix}=0 .

\Rightarrow 1(3bc-4bc)-2a(c\cdot b)+a(4c-3b)=0 .

\Rightarrow -hc-2ac+2ab+4ac-3ab=0

\Rightarrow bc=-2ac+4ac+2ab-3ab

=2ac-ab

\Rightarrow 2ac=bc+ab

\Rightarrow 2ac=b(a+c)

\Rightarrow b=\frac{2ac}{a+c}=\frac{2}{\frac{1}{a}+\frac{1}{c}}

\therefore b=\frac{2}{\frac{1}{a}+\frac{1}{c}} \Rightarrow \frac{1}{a}+\frac{1}{c}=\frac{2}{b}

\therefore \frac{1}{a}, \frac{1}{b}, \frac{1}{c}

are in

A \cdot P

\therefore a, b, c

are in H.

P

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