Using the section formula: If a point divides the line segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio \( m : n \), the coordinates are \( \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) \).
Here, \( (x_1, y_1) = (-1, 7) \), \( (x_2, y_2) = (4, -3) \), \( m = 2 \), \( n = 3 \).
x-coordinate = \( \frac{2 \cdot 4 + 3 \cdot (-1)}{2 + 3} = \frac{8 – 3}{5} = 1 \).
y-coordinate = \( \frac{2 \cdot (-3) + 3 \cdot 7}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3 \).
The coordinates are \( (1, 3) \).

Points of trisection divide the segment into three equal parts, i.e., in the ratios 1:2 and 2:1.
\( A(4, -1) \), \( B(-2, -3) \).
First point (1:2):
x-coordinate = \( \frac{1 \cdot (-2) + 2 \cdot 4}{1 + 2} = \frac{-2 + 8}{3} = 2 \).
y-coordinate = \( \frac{1 \cdot (-3) + 2 \cdot (-1)}{1 + 2} = \frac{-3 – 2}{3} = \frac{-5}{3} \).
First point: \( \left(2, \frac{-5}{3}\right) \).
Second point (2:1):
x-coordinate = \( \frac{2 \cdot (-2) + 1 \cdot 4}{2 + 1} = \frac{-4 + 4}{3} = 0 \).
y-coordinate = \( \frac{2 \cdot (-3) + 1 \cdot (-1)}{2 + 1} = \frac{-6 – 1}{3} = \frac{-7}{3} \).
Second point: \( \left(0, \frac{-7}{3}\right) \).

Let the point \( (-1, 6) \) divide the segment joining \( A(-3, 10) \) and \( B(6, -8) \) in the ratio \( k : 1 \).
Using the section formula: x-coordinate = \( \frac{k \cdot 6 + 1 \cdot (-3)}{k + 1} = -1 \).
\( \frac{6k – 3}{k + 1} = -1 \), \( 6k – 3 = -k – 1 \), \( 7k = 2 \), \( k = \frac{2}{7} \).
The ratio is \( \frac{2}{7} : 1 \), or 2 : 7.

In a parallelogram, the midpoints of the diagonals coincide.
Diagonal \( AC \): Midpoint of \( (1, 2) \) and \( (x, 6) \): \( \left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left( \frac{1 + x}{2}, 4 \right) \).
Diagonal \( BD \): Midpoint of \( (4, y) \) and \( (3, 5) \): \( \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right) = \left( \frac{7}{2}, \frac{y + 5}{2} \right) \).
Equate the midpoints: \( \frac{1 + x}{2} = \frac{7}{2} \), \( 1 + x = 7 \), \( x = 6 \).
\( 4 = \frac{y + 5}{2} \), \( 8 = y + 5 \), \( y = 3 \).
Thus, \( x = 6 \), \( y = 3 \).

The center is the midpoint of the diameter \( AB \).
Center \( (2, -3) \), \( B(1, 4) \), let \( A(x, y) \).
Midpoint: \( \left( \frac{x + 1}{2}, \frac{y + 4}{2} \right) = (2, -3) \).
\( \frac{x + 1}{2} = 2 \), \( x + 1 = 4 \), \( x = 3 \).
\( \frac{y + 4}{2} = -3 \), \( y + 4 = -6 \), \( y = -10 \).
The coordinates of \( A \) are \( (3, -10) \).

\( AP = \frac{3}{7} AB \), so \( \frac{AP}{AB} = \frac{3}{7} \), meaning \( P \) divides \( AB \) in the ratio 3 : 4.
\( A(-2, -2) \), \( B(2, -4) \).
x-coordinate = \( \frac{3 \cdot 2 + 4 \cdot (-2)}{3 + 4} = \frac{6 – 8}{7} = \frac{-2}{7} \).
y-coordinate = \( \frac{3 \cdot (-4) + 4 \cdot (-2)}{3 + 4} = \frac{-12 – 8}{7} = \frac{-20}{7} \).
The coordinates of \( P \) are \( \left( \frac{-2}{7}, \frac{-20}{7} \right) \).

Divide into four equal parts, so the ratios are 1:3, 2:2, and 3:1.
First point (1:3):
x-coordinate = \( \frac{1 \cdot 0 + 3 \cdot (-4)}{1 + 3} = \frac{-12}{4} = -3 \).
y-coordinate = \( \frac{1 \cdot 6 + 3 \cdot 0}{1 + 3} = \frac{6}{4} = \frac{3}{2} \).
First point: \( \left(-3, \frac{3}{2}\right) \).
Second point (2:2):
x-coordinate = \( \frac{2 \cdot 0 + 2 \cdot (-4)}{2 + 2} = \frac{-8}{4} = -2 \).
y-coordinate = \( \frac{2 \cdot 6 + 2 \cdot 0}{2 + 2} = \frac{12}{4} = 3 \).
Second point: \( (-2, 3) \).
Third point (3:1):
x-coordinate = \( \frac{3 \cdot 0 + 1 \cdot (-4)}{3 + 1} = \frac{-4}{4} = -1 \).
y-coordinate = \( \frac{3 \cdot 6 + 1 \cdot 0}{3 + 1} = \frac{18}{4} = \frac{9}{2} \).
Third point: \( \left(-1, \frac{9}{2}\right) \).

Ratios are 1:3, 2:2, and 3:1.
First point (1:3):
x-coordinate = \( \frac{1 \cdot 2 + 3 \cdot (-2)}{1 + 3} = \frac{2 – 6}{4} = -1 \).
y-coordinate = \( \frac{1 \cdot 8 + 3 \cdot 2}{1 + 3} = \frac{8 + 6}{4} = \frac{14}{4} = \frac{7}{2} \).
First point: \( \left(-1, \frac{7}{2}\right) \).
Second point (2:2):
x-coordinate = \( \frac{2 \cdot 2 + 2 \cdot (-2)}{2 + 2} = \frac{4 – 4}{4} = 0 \).
y-coordinate = \( \frac{2 \cdot 8 + 2 \cdot 2}{2 + 2} = \frac{16 + 4}{4} = 5 \).
Second point: \( (0, 5) \).
Third point (3:1):
x-coordinate = \( \frac{3 \cdot 2 + 1 \cdot (-2)}{3 + 1} = \frac{6 – 2}{4} = 1 \).
y-coordinate = \( \frac{3 \cdot 8 + 1 \cdot 2}{3 + 1} = \frac{24 + 2}{4} = \frac{26}{4} = \frac{13}{2} \).
Third point: \( \left(1, \frac{13}{2}\right) \).

\( A(a + b, a – b) \), \( B(a – b, a + b) \), ratio 3:2.
x-coordinate = \( \frac{3 \cdot (a – b) + 2 \cdot (a + b)}{3 + 2} = \frac{3a – 3b + 2a + 2b}{5} = \frac{5a – b}{5} \).
y-coordinate = \( \frac{3 \cdot (a + b) + 2 \cdot (a – b)}{3 + 2} = \frac{3a + 3b + 2a – 2b}{5} = \frac{5a + b}{5} \).
The coordinates are \( \left( \frac{5a – b}{5}, \frac{5a + b}{5} \right) \).

Centroid formula: \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \).
(i) (-1, 3), (6, -3), and (3, 6)
x-coordinate = \( \frac{-1 + 6 + 3}{3} = \frac{8}{3} \).
y-coordinate = \( \frac{3 + (-3) + 6}{3} = \frac{6}{3} = 2 \).
Centroid: \( \left( \frac{8}{3}, 2 \right) \).
(ii) (6, 2), (0, 0), and (4, -7)
x-coordinate = \( \frac{6 + 0 + 4}{3} = \frac{10}{3} \).
y-coordinate = \( \frac{2 + 0 + (-7)}{3} = \frac{-5}{3} \).
Centroid: \( \left( \frac{10}{3}, \frac{-5}{3} \right) \).
(iii) (1, -1), (0, 6), and (-3, 0)
x-coordinate = \( \frac{1 + 0 + (-3)}{3} = \frac{-2}{3} \).
y-coordinate = \( \frac{-1 + 6 + 0}{3} = \frac{5}{3} \).
Centroid: \( \left( \frac{-2}{3}, \frac{5}{3} \right) \).