SATHEE: Straight Line Question 41

Straight Line Question 41

Question: The lines $ (p-q)x+(q-r)y+(r-p)=0 $ $ (q-r)x+(r-p)y+(p-q)=0 $ $ (r-p)x+(p-q)y+(q-r)=0 $ are

Options:

A)Parallel

B)Perpendicular

C)Concurrent

D)None of these

Show Answer

Answer:

Correct Answer: C

Solution:

$ | \begin{matrix} p-q & q-r & r-p \\ q-r & r-p & p-q \\ r-p & p-q & q-r \\ \end{matrix} |=| \begin{matrix} 0 & q-r & r-p \\ 0 & r-p & p-q \\ 0 & p-q & q-r \\ \end{matrix} |=0 $ Hence the lines are concurrent.Aliter: Since sum of the coefficient of x, y and the constant term is zero, hence the lines are concurrent.

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Straight Line Question 41

Question: The lines $ (p-q)x+(q-r)y+(r-p)=0 $ $ (q-r)x+(r-p)y+(p-q)=0 $ $ (r-p)x+(p-q)y+(q-r)=0 $ are

Options:

Answer:

Solution:

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