Question

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

Answer 1

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