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प्रश्न
If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in
पर्याय
H.P.
G.P.
A.P.
none of these
उत्तर
A.P.
The given lines are
ax + 12y + 1 = 0 ... (1)
bx + 13y + 1 = 0 ... (2)
cx + 14y + 1 = 0 ... (3)
It is given that (1), (2) and (3) are concurrent.
\[\begin{vmatrix}a & 12 & 1 \\ b & 13 & 1 \\ c & 14 & 1\end{vmatrix} = 0\]
\[ \Rightarrow a\left( 13 - 14 \right) - 12\left( b - c \right) + 14b - 13c = 0\]
\[ \Rightarrow - a - 12b + 12c + 14b - 13c = 0\]
\[ \Rightarrow - a + 2b - c = 0\]
\[ \Rightarrow 2b = a + c\]
Hence, a, b and c are in AP.
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