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Q.

‘O’ is the orthocenter of the triangle formed by A(1,-3) B(7,2) C(2,5) the distance between the orthocentres of $Δ B O C, Δ A O B$ is $\sqrt{K}$ then $K + 4 =$

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answer is 69.

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Detailed Solution

Orthocentre of $∆ B O C i s A (1, - 3)$

and Orthocentre of $∆ A O B i s C (2, 5)$

Distance between the orthocentres = AC= $\sqrt{{(1 - 2)}^{2} + {(- 3 - 5)}^{2}} = \sqrt{65}$

$∴ K = 65 \Rightarrow K + 4 = 69$

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‘O’ is the orthocenter of the triangle formed by A(1,-3) B(7,2) C(2,5) the distance between the orthocentres of Δ BOC, ΔAOB is Kthen K+4=