Question

If $A(7,9),B(3,-7)$ and $C(-3,3)$ represent the vertices of a triangle, then orthocentre of the triangle is

• A. $(3,-7)$
• B. $(-3,3)$
• C. $(7,9)$
• D. $(\frac{7}{3},\frac{5}{3})$

Answer 1

The correct option is B $(-3,3)$

Given : The vertices of triangle are $A(7,9),B(3,-7)$ and $C(-3,3)$
Now, finding the side lengths,
$AB=\sqrt{4^2+{16}^2}=4\sqrt{17}$ units
$BC=\sqrt{6^2+{10}^2}=2\sqrt{34}$ units
$AC=\sqrt{{10}^2+6^2}=2\sqrt{34}$ units

As $BC=AC$ and $(AB{)}^2=(BC{)}^2+(AC{)}^2,$
it is a right angled isosceles triangle, right angled at $C$.
So, orthocentre $=(-3,3)$