The orthocentre of the triangle formed by the lines $x y + 3 x + 3 y + 9 = 0$ and $3 x + 4 y - 5 = 0$ is

(0, 0)

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(- 3, - 3)

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(- 3 - 2)

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(9, 12)

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Solution

The correct option is C $(- 3, - 3)$
$\begin{matrix} x y + 3 x + 3 y + 9 = 0 a n d 3 x + 4 y = 0 x (y + 3) + 3 (y + 3) = 0 (x + 3) (y + 3) = 0 \Rightarrow x = - 3; y = - 3 C l e a r l y f r o m t h e g r a p h, o r t h o c e n t r e o f t r i a n g l e i s (- 3, - 3) a s Δ i s r i g h t a n g l e d a t (- 3, - 3) . H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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The orthocentre of the triangle formed by the lines xy+3x+3y+9=0 and 3x+4y−5=0 is

The correct option is C (−3,−3)xy+3x+3y+9=0and3x+4y=0x(y+3)+3(y+3)=0(x+3)(y+3)=0⇒x=−3;y=−3Clearlyfromthegraph,orthocentreoftriangleis(−3,−3)asΔisrightangledat(−3,−3).Hence,optionBiscorrectanswer.

The orthocentre of the triangle formed by the lines $x y + 3 x + 3 y + 9 = 0$ and $3 x + 4 y - 5 = 0$ is