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Show that the Straight Lines L1 = (B + C) X + Ay + 1 = 0, L2 = (C + A) X + by + 1 = 0 and L3 = (A + B) X + Cy + 1 = 0 Are Concurrent. - Mathematics

Answer in Brief

Show that the straight lines L1 = (b + c) x + ay + 1 = 0, L2 = (c + a) x + by + 1 = 0 and L3 = (a + b) x + cy + 1 = 0 are concurrent.

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Solution

The given lines can be written as follows:
(b + c) x + ay + 1 = 0           ... (1)
(c + a) x + by + 1 = 0           ... (2)
(a + b) x + cy + 1 = 0           ... (3)
Consider the following determinant.

\[\begin{vmatrix}b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1\end{vmatrix}\]

Applying the transformation  \[C_1 \to C_1 + C_2\] \[\begin{vmatrix}b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1\end{vmatrix} = \begin{vmatrix}a + b + c & a & 1 \\ c + a + b & b & 1 \\ a + b + c & c & 1\end{vmatrix}\]

\[\Rightarrow\] \[\begin{vmatrix}b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1\end{vmatrix}\] =  \[\left( a + b + c \right)\begin{vmatrix}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{vmatrix}\]

\[\Rightarrow\] \[\begin{vmatrix}b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1\end{vmatrix}\] =0

Hence, the given lines are concurrent.

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.11 | Q 5 | Page 83
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