Find the orthocentre of the $△ A B C$ formed by the vertices $A (2, 2), B (6, 3)$ and $C (4, 11)$

Open in App

Solution

The given vertices of $△ A B C$ are A(2, 2), B(6, 3) and C(4, 11)
$A B = \sqrt{{(6 - 2)}^{2} + {(3 - 2)}^{2}}$
$= \sqrt{17}$ units
$B C = \sqrt{{(6 - 4)}^{2} + {(3 - 11)}^{2}} = \sqrt{68}$ units
Length of $A C = \sqrt{{(4 - 2)}^{2} + {(11 - 2)}^{2}}$
$= \sqrt{85}$ units
Clearly $A C^{2} = A B^{2} + B C^{2} . A B C$ is a right triangle right angle at $B$
Hence orthocentre is the vertex containing right angle i.e $B (6, 3)$

Suggest Corrections

Similar questions

△

Δ A B C

△ A B C

A (2, 1), B (6, - 1)

C (4, 11)

A

△ A B C

A (3, 4) B (5 sin t, 5 c o s t)

C (5 cos t, - 5 sin t)

△ A B C

Join BYJU'S Learning Program