MHT CET · Maths · Straight Lines
The orthocenter of the triangle formed by the lines \(x-2 y=10\) and \(6 x^2+x y-y^2=0\) is
- A \((2,-4)\)
- B \((2,4)\)
- C \((-2,-4)\)
- D \((-2,4)\)
Answer & Solution
Correct Answer
(A) \((2,-4)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & 6 x^2+x y-y^2=0 \\ & \Rightarrow(2 x+y)(3 x-y)=0 \\ & \Rightarrow 3 x-y=0 \text { and } 2 x+y=0 \text { and } x-2 y=10 \\ & \because 2 x+y=0 \text { and } x-2 y=10 \text { are perpendicular }\end{aligned}\)
Hence, orthocentre is point of intersection \(2 x+y=0\) and \(x-2 y=10\) i.e. \((2,-4)\)
Hence, orthocentre is point of intersection \(2 x+y=0\) and \(x-2 y=10\) i.e. \((2,-4)\)
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