Question

What are the coordinates of the orthocentre of a triangle formed using the points A(0, 0), B(1, 0) and C(0, 1)?

• A. (0, 0)
• B. (1, 0)
• C. (0, 1)
• D. (1, 1)

Answer 1

The correct option is A (0, 0)

First, let’s find the type of triangle formed using the points A(0, 0), B(1, 0) and C(0, 1). Using the distance formula, we get,

$AB=\sqrt{(0-1{)}^2+(0-0{)}^2}=\sqrt{1}=1unit$
$BC=\sqrt{(1-0{)}^2+(0-1{)}^2}=\sqrt{1+1}=\sqrt{2}units$
$AB=\sqrt{(0-0{)}^2+(0-1{)}^2}=\sqrt{1}=1unit$
Now,
$A{B}^2+A{C}^2=1^2+1^2=2$
$B{C}^2=(\sqrt{2}{)}^2=2$
$\therefore A{B}^2+A{C}^2=B{C}^2$
So, ABC is a right $\Delta$

We know that, orthocentre is the point of intersection of three altitudes.
Also, in a right triangle the right vertex is said to be the orthocentre as this is the point where the altitudes will intersect each other.


So, the orthocentre is A(0, 0).
Thus, option (a) is correct.