1. Given vertices

$A(-1,7),\quad B(-7,1),\quad C(5,-5)$

We need the orthocentre, i.e. the point of intersection of the altitudes.

2. Find slope of side $BC$

Points $B(-7,1)$ and $C(5,-5)$ give

$m_{BC}=\frac{-5-1}{5-(-7)}=\frac{-6}{12}=-\frac12$

So the altitude from $A$ is perpendicular to $BC$, hence its slope is

$m_{A\text{-alt}}=2$

Equation of altitude through $A(-1,7)$:

$y-7=2(x+1)$ $y=2x+9$

3. Find slope of side $AC$

Points $A(-1,7)$ and $C(5,-5)$ give

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

So the altitude from $B$ is perpendicular to $AC$, hence its slope is

$m_{B\text{-alt}}=\frac12$

Equation of altitude through $B(-7,1)$:

$y-1=\frac12(x+7)$

$2y-2=x+7$ $x-2y+9=0$

or

$y=\frac12x+\frac92$

4. Find intersection of the two altitudes

Solve

$y=2x+9$ $y=\frac12x+\frac92$

Equating,

$2x+9=\frac12x+\frac92$

Multiply by $2$:

$4x+18=x+9$ $3x=-9$ $x=-3$

Then

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

So the orthocentre is

$(-3,3)$

5. Check with options

Option A: $(-3,3)$ ✅

Hence the correct answer is A.

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They match.