If a $$\triangle ABC$$ has vertices $$A(-1, 7)$$, $$B(-7, 1)$$ and $$C(5, -5)$$, then its orthocentre has coordinates:

First we list the coordinates of the three vertices of the triangle:

$$A(-1,\;7), \qquad B(-7,\;1), \qquad C(5,\;-5).$$

The orthocentre is the common point of the three altitudes. An altitude is a line drawn from one vertex and perpendicular to the opposite side. We shall find two such altitudes and then compute their intersection; that point will automatically be the orthocentre.

Altitude from vertex A

We first need the slope of side $$BC$$. Using the slope formula $$m=\dfrac{y_2-y_1}{x_2-x_1}$$, we have

$$m_{BC}=\dfrac{-5-1}{\,5-(-7)\,}=\dfrac{-6}{12}=-\dfrac12.$$

An altitude from $$A$$ is perpendicular to $$BC$$. For two perpendicular lines the product of their slopes is $$-1$$, that is

$$m_{BC}\;m_{\perp}=-1.$$

Substituting $$m_{BC}=-\dfrac12$$, we get

$$\Bigl(-\dfrac12\Bigr)\,m_{\perp}=-1 \;\Rightarrow\; m_{\perp}=2.$$

The altitude through $$A(-1,7)$$ therefore has slope $$2$$ and its equation (point-slope form $$y-y_1=m(x-x_1)$$) is

$$y-7=2\,(x+1).$$

Simplifying,

$$y-7=2x+2 \;\Longrightarrow\; y=2x+9.$$

Altitude from vertex B

Next we need the slope of side $$AC$$:

$$m_{AC}=\dfrac{-5-7}{\,5-(-1)\,}=\dfrac{-12}{6}=-2.$$

The altitude from $$B$$ is perpendicular to $$AC$$, so using $$m_{AC}\,m_{\perp}=-1$$ we get

$$(-2)\,m_{\perp}=-1 \;\Rightarrow\; m_{\perp}=\dfrac12.$$

This altitude passes through $$B(-7,1)$$ and has slope $$\dfrac12$$, giving

$$y-1=\dfrac12\,(x+7).$$

Simplifying,

$$y-1=\dfrac12x+\dfrac72 \;\Longrightarrow\; y=\dfrac12x+\dfrac72+1=\dfrac12x+\dfrac92.$$

Writing the constant in decimal for convenience, $$\dfrac92=4.5$$, so

$$y=\dfrac12x+4.5.$$

Intersection of the two altitudes

The orthocentre is where the lines

$$y=2x+9 \qquad\text{and}\qquad y=\dfrac12x+4.5$$

meet. Equating their right-hand sides,

$$2x+9=\dfrac12x+4.5.$$

To eliminate the fraction, multiply every term by $$2$$:

$$4x+18 = x+9.$$

Now bring all terms in $$x$$ to one side and constants to the other:

$$4x - x = 9 - 18 \;\Longrightarrow\; 3x = -9.$$

Dividing by $$3$$ gives

$$x = -3.$$

Substituting $$x=-3$$ into either altitude equation, say $$y=2x+9$$, we obtain

$$y = 2(-3)+9 = -6+9 = 3.$$

Thus the coordinates of the orthocentre are

$$(-3,\;3).$$

Comparing with the given options, this matches option A.

Hence, the correct answer is Option A.