Question

Find the ortho-centre of the $△$ABC formed by the vertices A(2,2), B(6,3) and C(4,11).

Answer 1

The given vertices are A(2,2), B(6,3) and C(4,11).

Length of AB = $\sqrt{(6-2{)}^2+(3-2{)}^2}=\sqrt{17}units$

Length of 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 angled at B.

Hence, ortho-centre is the vertex containing right angle, i.e., $B(6,3)$.