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If two vertices of a triangle are \( \Large \left(-2,\ 3\right)\ and\ \left(5,\ -1\right) \) orthocentre lies at origin and centroid on the line x + y = 7, then the third vertex lies at:

A) \( \Large \left(7,\ 4\right) \)
B) \( \Large \left(8,\ 14\right) \)
C) \( \Large \left(12,\ 21\right) \)
D) none of these

Correct Answer:




D) none of these

Description for Correct answer:

Let \( \Large O \left(0,\ 0\right) \) be the arthocentre \( \Large A \left(h,\ k\right) \) be the third vertex and \( \Large \beta \left(-2,\ 3\right)\ and\ C \left(5,\ -1\right) \) the other two vertices. Then the slope of the line through

A and O is \( \Large \frac{k}{h} \) while the line through B and C has the slope \( \Large \frac{\{ -1-3 \}}{ \left(5+2\right) }=\frac{-4}{7} \). By the property of the orthocentre, these two lines must be Perpendicular, so we have

\( \Large \left(\frac{k}{h}\right) \left(-\frac{4}{7}\right)=-1 => \frac{k}{h}=\frac{7}{4} \) ...)i)

Also, \( \Large \frac{5-2+h}{3}+\frac{-1+3+k}{3}=7 \)

=> h +k = 16 ...(ii)

Which is not satisfied by the points given in the options (a), (b) or (c).

Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products

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