Question

The distance between the circumcenter and orthocenter of the triangle whose vertices are \((0,0)\), \((6,8)\) and \((-4,3)\) is

• A. \(\frac{125}{8}\) units
• B. \(\frac{\sqrt5}{2}\) units
• C. \(\frac{5\sqrt5}{2}\) units
• D. \(5\sqrt5\) units

Hint:

In a right triangle, the orthocenter is the right-angle vertex and the circumcenter is the midpoint of the hypotenuse.

Answer 1

The Correct Option is C

Concept:
For a right-angled triangle: \[ \text{Orthocenter}=\text{right-angled vertex} \] and \[ \text{Circumcenter}=\text{midpoint of hypotenuse} \] So, first we check whether the given triangle is right-angled or not.

Step 1: Write the given vertices.
Let: \[ A=(0,0) \] \[ B=(6,8) \] \[ C=(-4,3) \]

Step 2: Find the length of \(AB\).
Using the distance formula: \[ AB=\sqrt{(6-0)^2+(8-0)^2} \] \[ AB=\sqrt{6^2+8^2} \] \[ AB=\sqrt{36+64} \] \[ AB=\sqrt{100} \] \[ AB=10 \]

Step 3: Find the length of \(AC\).
\[ AC=\sqrt{(-4-0)^2+(3-0)^2} \] \[ AC=\sqrt{(-4)^2+3^2} \] \[ AC=\sqrt{16+9} \] \[ AC=\sqrt{25} \] \[ AC=5 \]

Step 4: Find the length of \(BC\).
\[ BC=\sqrt{(-4-6)^2+(3-8)^2} \] \[ BC=\sqrt{(-10)^2+(-5)^2} \] \[ BC=\sqrt{100+25} \] \[ BC=\sqrt{125} \] \[ BC=5\sqrt5 \]

Step 5: Check whether the triangle is right-angled.
Now: \[ AB^2=10^2=100 \] \[ AC^2=5^2=25 \] So: \[ AB^2+AC^2=100+25=125 \] Also: \[ BC^2=(5\sqrt5)^2=25\times5=125 \] Therefore: \[ AB^2+AC^2=BC^2 \] Hence, the triangle is right-angled at: \[ A=(0,0) \]

Step 6: Find the orthocenter.
In a right-angled triangle, the orthocenter is the right-angled vertex. Since the right angle is at \(A\), the orthocenter is: \[ H=(0,0) \]

Step 7: Find the circumcenter.
In a right-angled triangle, the circumcenter is the midpoint of the hypotenuse. Here, the hypotenuse is: \[ BC \] So, the circumcenter is the midpoint of \(B(6,8)\) and \(C(-4,3)\). \[ O=\left(\frac{6+(-4)}{2},\frac{8+3}{2}\right) \] \[ O=\left(\frac{2}{2},\frac{11}{2}\right) \] \[ O=\left(1,\frac{11}{2}\right) \]

Step 8: Find the distance between circumcenter and orthocenter.
Now: \[ O=\left(1,\frac{11}{2}\right) \] and \[ H=(0,0) \] Using distance formula: \[ OH=\sqrt{(1-0)^2+\left(\frac{11}{2}-0\right)^2} \] \[ OH=\sqrt{1^2+\left(\frac{11}{2}\right)^2} \] \[ OH=\sqrt{1+\frac{121}{4}} \] \[ OH=\sqrt{\frac{4}{4}+\frac{121}{4}} \] \[ OH=\sqrt{\frac{125}{4}} \] \[ OH=\frac{\sqrt{125}}{2} \] \[ OH=\frac{5\sqrt5}{2} \] Hence, the correct answer is: \[ \boxed{(C)\ \frac{5\sqrt5}{2}\text{ units}} \]