JEE Mains · Maths · STD 11 - 9. straight line
Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line \(x+2\sqrt{2}y=4\). If the co-ordinates of the vertex A are (\(\alpha, \beta\)), then the greatest integer less than or equal to \(|\alpha+\sqrt{2}\beta|\) is
- A 2
- B 3
- C 5
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
\(\because m _{ BC } \cdot m _{ AD }=-1\) \(\Rightarrow\left(-\frac{1}{2 \sqrt{2}}\right)\left(\frac{\beta}{\alpha}\right)=-1\) \(\Rightarrow \beta=2 \sqrt{2} \alpha\)\(\quad\)....(1) \(\because O D=\left|\frac{-4}{\sqrt{1+8}}\right|=\frac{4}{3} \Rightarrow A O=\frac{8}{3}\) So…
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