Two vertices of a triangle are $$(0, 2)$$ and $$(4, 3)$$. If its orthocenter is at the origin, then its third vertex lies in which quadrant?

Let us denote the vertices of the triangle by $$A(0,\,2),\;B(4,\,3)\; \text{and}\; C(x,\,y),$$ and let the orthocenter be $$H(0,\,0).$$

The orthocenter is the common point of the three altitudes, so every altitude must pass through the origin.

Altitude through $$A$$
We have the two points $$A(0,\,2)\quad\text{and}\quad H(0,\,0).$$ Because their abscissae are equal, the line $$AH$$ is the vertical line $$x=0.$$ An altitude is perpendicular to the opposite side, so the side $$BC$$ must be horizontal, that is, it must have slope $$0$$. Consequently,

$$y_B = y_C \;\Longrightarrow\; y = 3.$$

Altitude through $$B$$
The points of this altitude are $$B(4,\,3)\quad\text{and}\quad H(0,\,0).$$ Its slope is therefore

$$m_{BH} = \frac{0-3}{0-4}= \frac{-3}{-4}= \frac{3}{4}.$$

This altitude is perpendicular to side $$AC$$. If the slope of $$AC$$ is $$m_{AC}$$, the product of the slopes of two perpendicular lines in the plane is $$-1$$. Hence

$$m_{AC}\,(\text{slope of } AC)\times m_{BH}= -1.$$ That is, $$m_{AC}\,\cdot\frac{3}{4}= -1 \;\Longrightarrow\; m_{AC}= -\frac{4}{3}.$$

Now, since $$A(0,\,2)$$ and $$C(x,\,3)$$ lie on side $$AC$$, we can write the slope formula:

$$m_{AC}= \frac{3-2}{x-0}= \frac{1}{x}.$$

Substituting the required slope, we have

$$\frac{1}{x}= -\frac{4}{3} \;\Longrightarrow\; x= -\frac{3}{4}.$$

Coordinates of the third vertex
Combining the results for $$x$$ and $$y$$, we obtain

$$C\!\left(-\frac{3}{4},\,3\right).$$

Quadrant of the point $$C$$
The x-coordinate is negative and the y-coordinate is positive, so the point lies in the second quadrant.

Hence, the correct answer is Option B.