Question

# If $$\begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}=0$$ then all lines represented by $$ax+by+c=0$$ pass through a fixed a point. The fixed point is

A
(1,2)
B
(1,2)
C
(1,2)
D
(1,2)

Solution

## The correct option is A $$(1,-2)$$We have,$$\left| { \begin{array} { *{ 20 }{ c } }{ x+1 } & { x+2 } & { x+a } \\ { x+2 } & { x+3 } & { x+b } \\ { x+3 } & { x+4 } & { x+c } \end{array} } \right| =0$$$${ C_{ 2 } }\to { C_{ 2 } }-{ C_{ 1 } } \\ \left| { \begin{array} { *{ 20 }{ c } }{ x+1 } & 1 & { x+a } \\ { x+2 } & 1 & { x+b } \\ { x+3 } & 1 & { x+c } \end{array} } \right| =0 \\$$$${ R_{ 2 } }\to { R_{ 2 } }-{ R_{ 1 } }, { R_{ 3 } }\to { R_{ 3 } }-{ R_{ 2 } } \\ \left| { \begin{array} { *{ 20 }{ c } }{ x+1 } & 1 & { x+a } \\ 1 & 0 & { b-a } \\ 1 & 0 & { c-b } \end{array} } \right| =0 \\ -\left[ { \left( { c-b } \right) -\left( { b-a } \right) } \right] =0 \\ c+a=2b \\ a-2b+c=0 \\ ax+by+c=0$$$$\\ On\, \, comparing\, \, x=1\, \, \, \, y=-2 \\ fixed\, \, prove\, \, that:\, \, \left( { 1,-2 } \right)$$Option $$A$$ is correct answer.Mathematics

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