Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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
Solution
$ 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)$
$ \therefore OD = \frac{|4|}{\sqrt{1 + 8}} = \frac{4}{3} \Rightarrow AO = \frac{8}{3} $
So $ AD = \frac{8}{3} + \frac{4}{3} = 4 $
$ \Rightarrow \frac{|\alpha + 2\sqrt{2}\beta - 4|}{3} = 4 \Rightarrow \alpha = \frac{16}{9} \text{ or } -\frac{8}{9} $
$ \because A(\alpha,\beta) $ & $(0,0)$ lies on same side of given line
$ \therefore (\alpha,\beta) = \left( -\frac{8}{9}, -\frac{16\sqrt{2}}{9} \right) $
$ \Rightarrow [\alpha + \sqrt{2}\beta] = \left[ \frac{-8 - 32}{9} \right] = 4 $
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
