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Prove that the Following Sets of Three Lines Are Concurrent: 15x − 18y + 1 = 0, 12x + 10y − 3 = 0 and 6x + 66y − 11 = 0 - Mathematics

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Question

Prove that the following sets of three lines are concurrent:

 15x − 18y + 1 = 0, 12x + 10y − 3 = 0 and 6x + 66y − 11 = 0

Answer in Brief
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Solution

Given:
15x − 18y + 1 = 0          ... (1)
12x + 10y − 3 = 0          ... (2)
6x + 66y − 11 = 0          ... (3)
Now, consider the following determinant:

\[\begin{vmatrix}15 & - 18 & 1 \\ 12 & 10 & - 3 \\ 6 & 66 & - 11\end{vmatrix} = 15\left( - 110 + 198 \right) + 18\left( - 132 + 18 \right) + 1\left( 792 - 60 \right)\]

\[\Rightarrow \begin{vmatrix}15 & - 18 & 1 \\ 12 & 10 & - 3 \\ 6 & 66 & - 11\end{vmatrix} = 1320 - 2052 + 732 = 0\]

Hence, the given lines are concurrent.

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Chapter 23: The straight lines - Exercise 23.11 [Page 83]

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RD Sharma Mathematics [English] Class 11
Chapter 23 The straight lines
Exercise 23.11 | Q 1.1 | Page 83
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