The slope \( m \) of a line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \( m = \frac{y_2 – y_1}{x_2 – x_1} \).
(i) (4, -8) and (5, -2)
Here, \( (x_1, y_1) = (4, -8) \), \( (x_2, y_2) = (5, -2) \).
Slope = \( \frac{-2 – (-8)}{5 – 4} = \frac{-2 + 8}{1} = \frac{6}{1} = 6 \).
The slope is 6.

(ii) (0, 0) and (\( \sqrt{3}, 3 \))
\( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (\sqrt{3}, 3) \).
Slope = \( \frac{3 – 0}{\sqrt{3} – 0} = \frac{3}{\sqrt{3}} = \sqrt{3} \).
The slope is \( \sqrt{3} \).

(iii) (2a, 3b) and (a, -b)
\( (x_1, y_1) = (2a, 3b) \), \( (x_2, y_2) = (a, -b) \).
Slope = \( \frac{-b – 3b}{a – 2a} = \frac{-4b}{-a} = \frac{4b}{a} \).
The slope is \( \frac{4b}{a} \).

(iv) (a, 0) and (0, b)
\( (x_1, y_1) = (a, 0) \), \( (x_2, y_2) = (0, b) \).
Slope = \( \frac{b – 0}{0 – a} = \frac{b}{-a} = -\frac{b}{a} \).
The slope is \( -\frac{b}{a} \).

(v) A(-1.4, -3.7), B(-2.4, 1.3)
\( (x_1, y_1) = (-1.4, -3.7) \), \( (x_2, y_2) = (-2.4, 1.3) \).
Slope = \( \frac{1.3 – (-3.7)}{-2.4 – (-1.4)} = \frac{1.3 + 3.7}{-2.4 + 1.4} = \frac{5}{-1} = -5 \).
The slope is -5.

(vi) A(3, -2), B(-6, -2)
\( (x_1, y_1) = (3, -2) \), \( (x_2, y_2) = (-6, -2) \).
Slope = \( \frac{-2 – (-2)}{-6 – 3} = \frac{0}{-9} = 0 \).
The slope is 0 (horizontal line).

(vii) A\( \left(-\frac{3}{2}, 3 \right) \), B\( \left(-7, \frac{1}{2} \right) \)
\( (x_1, y_1) = \left(-\frac{3}{2}, 3 \right) \), \( (x_2, y_2) = \left(-7, \frac{1}{2} \right) \).
Slope = \( \frac{\frac{1}{2} – 3}{-7 – \left(-\frac{3}{2}\right)} = \frac{\frac{1}{2} – 3}{-7 + \frac{3}{2}} = \frac{\frac{1 – 6}{2}}{\frac{-14 + 3}{2}} = \frac{\frac{-5}{2}}{\frac{-11}{2}} = \frac{-5}{-11} = \frac{5}{11} \).
The slope is \( \frac{5}{11} \).

(viii) A(0, 4), B(4, 0)
\( (x_1, y_1) = (0, 4) \), \( (x_2, y_2) = (4, 0) \).
Slope = \( \frac{0 – 4}{4 – 0} = \frac{-4}{4} = -1 \).
The slope is -1.

(i) (2, 3), (-1, 0), (2, -4)
Using the area formula: Area = \( \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right| \)
Here, \( (x_1, y_1) = (2, 3) \), \( (x_2, y_2) = (-1, 0) \), \( (x_3, y_3) = (2, -4) \).
Area = \( \frac{1}{2} \left| 2(0 – (-4)) + (-1)(-4 – 3) + 2(3 – 0) \right| = \frac{1}{2} \left| 2(4) + (-1)(-7) + 2(3) \right| = \frac{1}{2} \left| 8 + 7 + 6 \right| = \frac{1}{2} \times 21 = \frac{21}{2} \).
Area = \( \frac{21}{2} \) square units.

(ii) (-5, -1), (3, -5), (5, 2)
\( (x_1, y_1) = (-5, -1) \), \( (x_2, y_2) = (3, -5) \), \( (x_3, y_3) = (5, 2) \).
Area = \( \frac{1}{2} \left| (-5)(-5 – 2) + 3(2 – (-1)) + 5(-1 – (-5)) \right| = \frac{1}{2} \left| (-5)(-7) + 3(3) + 5(4) \right| = \frac{1}{2} \left| 35 + 9 + 20 \right| = \frac{1}{2} \times 64 = 32 \).
Area = 32 square units.

(iii) (0, 0), (3, 0), and (0, 2)
\( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (3, 0) \), \( (x_3, y_3) = (0, 2) \).
Area = \( \frac{1}{2} \left| 0(0 – 2) + 3(2 – 0) + 0(0 – 0) \right| = \frac{1}{2} \left| 0 + 3(2) + 0 \right| = \frac{1}{2} \times 6 = 3 \).
Area = 3 square units.

Points are collinear if the area of the triangle formed is zero.
(i) (7, -2), (5, 1), (3, K)
Area = \( \frac{1}{2} \left| 7(1 – K) + 5(K – (-2)) + 3(-2 – 1) \right| = 0 \).
\( \frac{1}{2} \left| 7(1 – K) + 5(K + 2) + 3(-3) \right| = 0 \),
\( 7 – 7K + 5K + 10 – 9 = 0 \),
\( -2K + 8 = 0 \), \( 2K = 8 \), \( K = 4 \).

(ii) (K, K), (2, 3), and (4, -1)
Area = \( \frac{1}{2} \left| K(3 – (-1)) + 2(-1 – K) + 4(K – 3) \right| = 0 \).
\( \frac{1}{2} \left| K(4) + 2(-1 – K) + 4(K – 3) \right| = 0 \),
\( 4K – 2 – 2K + 4K – 12 = 0 \),
\( 6K – 14 = 0 \), \( 6K = 14 \), \( K = \frac{14}{6} = \frac{7}{3} \).

(iii) (8, 1), (K, -4), (2, -5)
Area = \( \frac{1}{2} \left| 8(-4 – (-5)) + K(-5 – 1) + 2(1 – (-4)) \right| = 0 \).
\( \frac{1}{2} \left| 8(1) + K(-6) + 2(5) \right| = 0 \),
\( 8 – 6K + 10 = 0 \),
\( 18 – 6K = 0 \), \( 6K = 18 \), \( K = 3 \).

Step 1: Find the midpoints of the sides.
\( A(0, -1) \), \( B(2, 1) \), \( C(0, 3) \).
Midpoint of \( AB \): \( D = \left( \frac{0 + 2}{2}, \frac{-1 + 1}{2} \right) = (1, 0) \).
Midpoint of \( BC \): \( E = \left( \frac{2 + 0}{2}, \frac{1 + 3}{2} \right) = (1, 2) \).
Midpoint of \( CA \): \( F = \left( \frac{0 + 0}{2}, \frac{3 + (-1)}{2} \right) = (0, 1) \).
Step 2: Area of triangle \( DEF \).
\( D(1, 0) \), \( E(1, 2) \), \( F(0, 1) \).
Area = \( \frac{1}{2} \left| 1(2 – 1) + 1(1 – 0) + 0(0 – 2) \right| = \frac{1}{2} \left| 1(1) + 1(1) + 0 \right| = \frac{1}{2} \times 2 = 1 \).
Area of \( \triangle DEF = 1 \) square unit.
Step 3: Area of triangle \( ABC \).
Area = \( \frac{1}{2} \left| 0(1 – 3) + 2(3 – (-1)) + 0(-1 – 1) \right| = \frac{1}{2} \left| 0 + 2(4) + 0 \right| = \frac{1}{2} \times 8 = 4 \).
Area of \( \triangle ABC = 4 \) square units.
Step 4: Ratio of areas.
Ratio = \( \frac{\text{Area of } \triangle DEF}{\text{Area of } \triangle ABC} = \frac{1}{4} \).
The ratio is 1 : 4.

Divide the quadrilateral into two triangles by drawing a diagonal, say from \( (-4, -2) \) to \( (3, -2) \).
Triangle 1: (-4, -2), (-3, -5), (3, -2)
Area = \( \frac{1}{2} \left| (-4)(-5 – (-2)) + (-3)(-2 – (-2)) + 3(-2 – (-5)) \right| = \frac{1}{2} \left| (-4)(-3) + (-3)(0) + 3(3) \right| = \frac{1}{2} \left| 12 + 0 + 9 \right| = \frac{1}{2} \times 21 = \frac{21}{2} \).
Triangle 2: (-4, -2), (3, -2), (2, 3)
Area = \( \frac{1}{2} \left| (-4)(-2 – 3) + 3(3 – (-2)) + 2(-2 – (-2)) \right| = \frac{1}{2} \left| (-4)(-5) + 3(5) + 2(0) \right| = \frac{1}{2} \left| 20 + 15 + 0 \right| = \frac{1}{2} \times 35 = \frac{35}{2} \).
Total area = \( \frac{21}{2} + \frac{35}{2} = \frac{56}{2} = 28 \).
Area of the quadrilateral = 28 square units.

Step 1: Find the lengths of the sides.
\( A(2, 3) \), \( B(6, 3) \), \( C(2, 6) \).
\( AB = \sqrt{(6 – 2)^2 + (3 – 3)^2} = 4 \),
\( BC = \sqrt{(6 – 2)^2 + (3 – 6)^2} = \sqrt{16 + 9} = 5 \),
\( CA = \sqrt{(2 – 2)^2 + (6 – 3)^2} = 3 \).
Step 2: Apply Heron’s formula.
Semi-perimeter \( s = \frac{4 + 5 + 3}{2} = 6 \).
Area = \( \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{6(6 – 4)(6 – 5)(6 – 3)} = \sqrt{6 \times 2 \times 1 \times 3} = \sqrt{36} = 6 \).
Area = 6 square units.

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: \( \left( \frac{-2}{3}, \frac{5}{3} \right) \).

(i) (2, 3) and (4, 1)
Using the distance formula: \( \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
Here, \( (x_1, y_1) = (2, 3) \), \( (x_2, y_2) = (4, 1) \).
Distance = \( \sqrt{(4 – 2)^2 + (1 – 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).

(ii) (-5, 7) and (-1, 3)
Here, \( (x_1, y_1) = (-5, 7) \), \( (x_2, y_2) = (-1, 3) \).
Distance = \( \sqrt{(-1 – (-5))^2 + (3 – 7)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \).

(iii) (-2, -3) and (3, 2)
Here, \( (x_1, y_1) = (-2, -3) \), \( (x_2, y_2) = (3, 2) \).
Distance = \( \sqrt{(3 – (-2))^2 + (2 – (-3))^2} = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \).

(iv) (a, b) and (-a, -b)
Here, \( (x_1, y_1) = (a, b) \), \( (x_2, y_2) = (-a, -b) \).
Distance = \( \sqrt{(-a – a)^2 + (-b – b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2} \).

Here, \( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (36, 15) \).
Distance = \( \sqrt{(36 – 0)^2 + (15 – 0)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \).

Points are collinear if the area of the triangle formed by them is zero.
Using the formula: Area = \( \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right| \)
Here, \( (x_1, y_1) = (1, 5) \), \( (x_2, y_2) = (2, 3) \), \( (x_3, y_3) = (-2, -1) \).
Area = \( \frac{1}{2} \left| 1(3 – (-1)) + 2(-1 – 5) + (-2)(5 – 3) \right| = \frac{1}{2} \left| 1(4) + 2(-6) + (-2)(2) \right| = \frac{1}{2} \left| 4 – 12 – 4 \right| = \frac{1}{2} \times 12 = 6 \).
Since the area is not zero, the points are not collinear.

Calculate the lengths of the sides using the distance formula.
\( A(5, -2) \), \( B(6, 4) \), \( C(7, -2) \).
\( AB = \sqrt{(6 – 5)^2 + (4 – (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{37} \).
\( BC = \sqrt{(7 – 6)^2 + (-2 – 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{37} \).
\( CA = \sqrt{(7 – 5)^2 + (-2 – (-2))^2} = \sqrt{2^2 + 0^2} = 2 \).
Since \( AB = BC \), the triangle is isosceles.

From the figure: \( A(4, 5) \), \( B(8, 7) \), \( C(8, 5) \), \( D(4, 3) \).
A square has all sides equal and diagonals equal.
Sides: \( AB = \sqrt{(8 – 4)^2 + (7 – 5)^2} = \sqrt{16 + 4} = \sqrt{20} \),
\( BC = \sqrt{(8 – 8)^2 + (7 – 5)^2} = \sqrt{0 + 4} = 2 \),
\( CD = \sqrt{(8 – 4)^2 + (5 – 3)^2} = \sqrt{16 + 4} = \sqrt{20} \),
\( DA = \sqrt{(4 – 4)^2 + (5 – 3)^2} = \sqrt{0 + 4} = 2 \).
Diagonals: \( AC = \sqrt{(8 – 4)^2 + (5 – 5)^2} = 4 \),
\( BD = \sqrt{(8 – 4)^2 + (7 – 3)^2} = \sqrt{16 + 16} = \sqrt{32} \).
Sides \( AB = CD \) and \( BC = DA \), but diagonals are not equal, so ABCD is not a square. Phani is correct.

All sides of an equilateral triangle are equal.
\( AB = \sqrt{(-a – a)^2 + (0 – 0)^2} = \sqrt{(-2a)^2} = 2a \),
\( BC = \sqrt{(0 – (-a))^2 + (a\sqrt{3} – 0)^2} = \sqrt{a^2 + (a\sqrt{3})^2} = \sqrt{a^2 + 3a^2} = \sqrt{4a^2} = 2a \),
\( CA = \sqrt{(0 – a)^2 + (a\sqrt{3} – 0)^2} = \sqrt{a^2 + 3a^2} = 2a \).
Since \( AB = BC = CA \), the triangle is equilateral.

A parallelogram has opposite sides equal and parallel.
\( A(-7, -3) \), \( B(5, 10) \), \( C(15, 8) \), \( D(3, -5) \).
\( AB = \sqrt{(5 – (-7))^2 + (10 – (-3))^2} = \sqrt{12^2 + 13^2} = \sqrt{313} \),
\( CD = \sqrt{(15 – 3)^2 + (8 – (-5))^2} = \sqrt{12^2 + 13^2} = \sqrt{313} \),
\( BC = \sqrt{(15 – 5)^2 + (8 – 10)^2} = \sqrt{10^2 + (-2)^2} = \sqrt{104} \),
\( DA = \sqrt{(3 – (-7))^2 + (-5 – (-3))^2} = \sqrt{10^2 + (-2)^2} = \sqrt{104} \).
Since \( AB = CD \) and \( BC = DA \), the points form a parallelogram.

A rhombus has all sides equal. Area = \( \frac{1}{2} \times \text{product of diagonals} \).
\( A(-4, -7) \), \( B(-1, 2) \), \( C(8, 5) \), \( D(5, -4) \).
Sides: \( AB = \sqrt{(-1 – (-4))^2 + (2 – (-7))^2} = \sqrt{3^2 + 9^2} = \sqrt{90} \),
\( BC = \sqrt{(8 – (-1))^2 + (5 – 2)^2} = \sqrt{9^2 + 3^2} = \sqrt{90} \),
\( CD = \sqrt{(8 – 5)^2 + (5 – (-4))^2} = \sqrt{3^2 + 9^2} = \sqrt{90} \),
\( DA = \sqrt{(5 – (-4))^2 + (-4 – (-7))^2} = \sqrt{9^2 + 3^2} = \sqrt{90} \).
All sides are equal, so it’s a rhombus.
Diagonals: \( AC = \sqrt{(8 – (-4))^2 + (5 – (-7))^2} = \sqrt{12^2 + 12^2} = 12\sqrt{2} \),
\( BD = \sqrt{(5 – (-1))^2 + (-4 – 2)^2} = \sqrt{6^2 + (-6)^2} = 6\sqrt{2} \).
Area = \( \frac{1}{2} \times (12\sqrt{2}) \times (6\sqrt{2}) = \frac{1}{2} \times 72 \times 2 = 72 \) square units.

(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
\( A(-1, -2) \), \( B(1, 0) \), \( C(-1, 2) \), \( D(-3, 0) \).
Sides: \( AB = \sqrt{(1 – (-1))^2 + (0 – (-2))^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( BC = \sqrt{(-1 – 1)^2 + (2 – 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( CD = \sqrt{(-1 – (-3))^2 + (2 – 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( DA = \sqrt{(-3 – (-1))^2 + (0 – (-2))^2} = \sqrt{4 + 4} = 2\sqrt{2} \).
All sides are equal, so it’s a rhombus.
Diagonals: \( AC = \sqrt{(-1 – (-1))^2 + (2 – (-2))^2} = 4 \), \( BD = \sqrt{(-3 – 1)^2 + (0 – 0)^2} = 4 \).
Diagonals are equal, so it’s a square.

(ii) (-3, 5), (3, 1), (1, -3), (-5, 1)
\( A(-3, 5) \), \( B(3, 1) \), \( C(1, -3) \), \( D(-5, 1) \).
Sides: \( AB = \sqrt{(3 – (-3))^2 + (1 – 5)^2} = \sqrt{36 + 16} = \sqrt{52} \),
\( BC = \sqrt{(1 – 3)^2 + (-3 – 1)^2} = \sqrt{4 + 16} = \sqrt{20} \),
\( CD = \sqrt{(-5 – 1)^2 + (1 – (-3))^2} = \sqrt{36 + 16} = \sqrt{52} \),
\( DA = \sqrt{(-5 – (-3))^2 + (1 – 5)^2} = \sqrt{4 + 16} = \sqrt{20} \).
Opposite sides are equal (\( AB = CD \), \( BC = DA \)), so it’s a parallelogram.

(iii) (4, 5), (7, 6), (4, 3), (1, 2)
\( A(4, 5) \), \( B(7, 6) \), \( C(4, 3) \), \( D(1, 2) \).
Sides: \( AB = \sqrt{(7 – 4)^2 + (6 – 5)^2} = \sqrt{9 + 1} = \sqrt{10} \),
\( BC = \sqrt{(4 – 7)^2 + (3 – 6)^2} = \sqrt{9 + 9} = 3\sqrt{2} \),
\( CD = \sqrt{(4 – 1)^2 + (3 – 2)^2} = \sqrt{9 + 1} = \sqrt{10} \),
\( DA = \sqrt{(1 – 4)^2 + (2 – 5)^2} = \sqrt{9 + 9} = 3\sqrt{2} \).
Opposite sides are equal, so it’s a parallelogram.

Let the point on the X-axis be \( (x, 0) \).
Distance from \( (x, 0) \) to \( (2, -5) \): \( \sqrt{(x – 2)^2 + (0 – (-5))^2} = \sqrt{(x – 2)^2 + 25} \).
Distance from \( (x, 0) \) to \( (-2, 9) \): \( \sqrt{(x – (-2))^2 + (0 – 9)^2} = \sqrt{(x + 2)^2 + 81} \).
Since the distances are equal: \( \sqrt{(x – 2)^2 + 25} = \sqrt{(x + 2)^2 + 81} \).
Square both sides: \( (x – 2)^2 + 25 = (x + 2)^2 + 81 \),
\( x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81 \),
\( -4x + 29 = 4x + 85 \),
\( -8x = 56 \), \( x = -7 \).
The point is \( (-7, 0) \).

Distance = \( \sqrt{(x – 1)^2 + (7 – 15)^2} = 10 \).
\( \sqrt{(x – 1)^2 + (-8)^2} = 10 \),
\( (x – 1)^2 + 64 = 100 \),
\( (x – 1)^2 = 36 \),
\( x – 1 = \pm 6 \),
\( x = 7 \) or \( x = -5 \).

Distance = \( \sqrt{(10 – 2)^2 + (y – (-3))^2} = 10 \).
\( \sqrt{8^2 + (y + 3)^2} = 10 \),
\( 64 + (y + 3)^2 = 100 \),
\( (y + 3)^2 = 36 \),
\( y + 3 = \pm 6 \),
\( y = 3 \) or \( y = -9 \).

Radius = distance between the center \( (3, 2) \) and the point \( (-5, 6) \).
\( \sqrt{(-5 – 3)^2 + (6 – 2)^2} = \sqrt{(-8)^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \).
The radius is \( 4\sqrt{5} \).

Check if the points are collinear using the area formula.
\( (x_1, y_1) = (1, 5) \), \( (x_2, y_2) = (5, 8) \), \( (x_3, y_3) = (13, 14) \).
Area = \( \frac{1}{2} \left| 1(8 – 14) + 5(14 – 5) + 13(5 – 8) \right| = \frac{1}{2} \left| 1(-6) + 5(9) + 13(-3) \right| = \frac{1}{2} \left| -6 + 45 – 39 \right| = 0 \).
Since the area is zero, the points are collinear, and a triangle cannot be formed.

Distance from \( (x, y) \) to \( (-2, 8) \): \( \sqrt{(x – (-2))^2 + (y – 8)^2} \).
Distance from \( (x, y) \) to \( (-3, -5) \): \( \sqrt{(x – (-3))^2 + (y – (-5))^2} \).
\( \sqrt{(x + 2)^2 + (y – 8)^2} = \sqrt{(x + 3)^2 + (y + 5)^2} \).
Square both sides: \( (x + 2)^2 + (y – 8)^2 = (x + 3)^2 + (y + 5)^2 \),
\( x^2 + 4x + 4 + y^2 – 16y + 64 = x^2 + 6x + 9 + y^2 + 10y + 25 \),
\( 4x – 16y + 68 = 6x + 10y + 34 \),
\( -2x – 26y + 34 = 0 \),
\( x + 13y – 17 = 0 \).