To find the nine-point center of the triangle formed by the points \( A(5, -1) \), \( B(-2, 3) \), and \( C(-4, -7) \) with the orthocenter at the origin \( H(0, 0) \), we will follow these steps:
### Step 1: Calculate the Centroid \( G \) of the Triangle
The centroid \( G \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula:
\[
G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
\]
For our triangle:
- \( A(5, -1) \) gives \( x_1 = 5 \), \( y_1 = -1 \)
- \( B(-2, 3) \) gives \( x_2 = -2 \), \( y_2 = 3 \)
- \( C(-4, -7) \) gives \( x_3 = -4 \), \( y_3 = -7 \)
Substituting these values into the centroid formula:
\[
G = \left( \frac{5 + (-2) + (-4)}{3}, \frac{-1 + 3 + (-7)}{3} \right)
\]
Calculating the x-coordinate:
\[
x_G = \frac{5 - 2 - 4}{3} = \frac{-1}{3}
\]
Calculating the y-coordinate:
\[
y_G = \frac{-1 + 3 - 7}{3} = \frac{-5}{3}
\]
Thus, the centroid \( G \) is:
\[
G = \left( -\frac{1}{3}, -\frac{5}{3} \right)
\]
### Step 2: Use the Section Formula to Find the Nine-Point Center \( N \)
The nine-point center \( N \) divides the line segment \( HG \) (from orthocenter \( H(0, 0) \) to centroid \( G(-\frac{1}{3}, -\frac{5}{3}) \)) in the ratio \( 3:1 \).
Using the section formula, the coordinates of point \( N \) are given by:
\[
N = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right)
\]
where \( H(0, 0) \) is \( (x_1, y_1) \) and \( G(-\frac{1}{3}, -\frac{5}{3}) \) is \( (x_2, y_2) \), with \( m = 3 \) and \( n = 1 \).
Substituting the values:
\[
N_x = \frac{3 \left(-\frac{1}{3}\right) + 1(0)}{3 + 1} = \frac{-1 + 0}{4} = -\frac{1}{4}
\]
\[
N_y = \frac{3 \left(-\frac{5}{3}\right) + 1(0)}{3 + 1} = \frac{-5 + 0}{4} = -\frac{5}{4}
\]
Thus, the nine-point center \( N \) is:
\[
N = \left( -\frac{1}{4}, -\frac{5}{4} \right)
\]
### Final Answer
The nine-point center of the triangle is:
\[
\boxed{\left( -\frac{1}{4}, -\frac{5}{4} \right)}
\]
