Question

Orthocentre of the triangle with vertices $(0,0),(3,4)$ and $(4,0)$ is

• A. $(3,\frac{5}{4})$
• B. $(3,12)$
• C. $(3,\frac{3}{4})$
• D. $(3,9)$

Answer 1

Answer: (C)

$(3,\frac{3}{4})$

Let $O(0,0),A(3,4)$ and $B(4,0)$ then equation of $OB$ is $y=0$ Equations of the altitudes from $A$ and $0$ are respectively $x=3$ and $y=\left(1/4\right)$ $x$. So the orthocentre is $(3,3/4)$, the point of intersection of these altitudes.