The lines $p x + q y + r = 0, q x + r y + p = 0 a n d r x + p y + q = 0$ are concurrent then

p + q + r = 0

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p^{3} + q^{3} + r^{3} = 3 p q r

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p^{2} + q^{2} + r^{2} - p q - q r - r p = 0

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p^{2} + q^{2} + r^{2} = 2 (p q + q r + r p)

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Solution

The correct options are
A $p + q + r = 0$
B $p^{3} + q^{3} + r^{3} = 3 p q r$
C $p^{2} + q^{2} + r^{2} - p q - q r - r p = 0$
$∣ ∣ ∣ ∣ \begin{matrix} p & q & r q & r & p r & p & q \end{matrix} ∣ ∣ ∣ ∣ = 0$

$p (r q - p^{2}) - q (q^{2} - r p) + r (q p - r^{2}) = 0$

$3 p q r - p^{3} - q^{3} - r^{3} = 0$

$p^{3} + q^{3} + r^{3} = 3 p q r$

$p^{3} + q^{3} + r^{3} - 3 p q r = 0$

$(p + q + r) (p^{2} + q^{2} + r^{2} - p a - q r - p r) = 0$

$\Rightarrow p + q + r = 0$ or

$p^{2} + q^{2} + r^{2} - p q - q r - p r = 0$

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Similar questions

Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent, if

p³+ q³ + r³ – 3pqr = ?

(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

p = - 2, q = 1

r = 3

p^{2} + q^{2} - r^{2}

p - q - r

p^{3} + q^{3} + r^{3} + 3 p q r

p x + q y + r = 0

q x + r y + p = 0

r x + p y + q = 0

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