The vertices are A(-1, 2), B(k, -2) and C(7, 4)
Area of triangle ABC is 22 sq. units
\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) } = 22
\(\frac{1}{2}\) { (2 + 4k + 14) - (2k - 14 - 4) } = 22
2k + 34 = 44 gives 2k = 10 so k = 5.

slope of the line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
Slope of the line joining the points (- 2,6) and (4, 8)
\(m_{1}=\frac{6-8}{-2-4}=\frac{-2}{-6}=\frac{1}{3}\)
Slope of the line joining the points (8, 12) and (x, 24)
\(m_{2}=\frac{12-24}{8-x}=-\frac{12}{8-x}\)
Given that the lines are Perpendicular
m₁ x m₂ = -1
\(\frac{1}{3} \times \frac{-12}{8-x}=-1\)
4 = 8 - x
x = 8 - 4
X = 4.

Given points \(\left( 5,\sqrt { 5 } \right) \) and (0, 0)
Slope of a line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}(\text { or }) \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
\(=\frac{\sqrt{5}-0}{5-0}=\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)

Given slope 'm' = 0
tan θ = 0 = tan 0⁰
θ = 0⁰

Given angle of inclination θ = 90⁰
Slope of a line = tan θ
= tan90⁰ = ∝ (undefined)

Th e vertices are A(-2, 5) , B(6, -1) and C(2, 2).
Slope of AB = \(\frac { -1-5 }{ 6+2 } =\frac { -6 }{ 8 } =\frac { -3 }{ 4 } \)
Slope of BC = \(\frac { 2+1 }{ 2-6 } =\frac { 3 }{ -4 } =\frac { -3 }{ 4 } \)
We get, Slope of AB = Slope of BC
Therefore, the points A, B, C all lie in a same straight line.
Hence the points A, B and C are collinear.

Given points are (2, 3), (4, a) and (6, - 3)
Since the points are colinear, Area of triangle is zero
\(\text { i.e., } \frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
2(a + 3) + 4(- 3 -3) + 6(3 - a) = 0
2a + 6 - 24 + 18 - 6a = 0
-4a + 0 = 0
-4a = 0
a = 0

(i) Given vertices are (0, 0), (P, 8) and (6,2)
Area of triangle = 20 sq. units.
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+\right. \left.x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)
\(\frac{1}{2}\) [(8 - 2) + p (2 - 0) + 6 (0 - 8)] = 20
2p - 48 = 40
2P = 40 + 48
2P = 88
\(p=\frac{88}{2}=44\)
(ii) Given vertices are (p, p), (5,6) and (5, - 2)
Area of triangle = 32 sq. units
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+\right. \left.x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)
\(\frac{1}{2}\) [p( 6 + 2) + 5(- 2 -p) + 5(P - 6) = 32
8p -10 - 5P + 5P - 30 = 64
8p - 40 = 64
8P = 64 + 40 = 104
\(p=\frac{104}{8}=13\)

Given points are \((-\frac12 ,3)\), (- 5, 6) and (-8, 8)
Let us use area of triangle formula
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+\right. \left.\quad x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)
\(=\frac{1}{2}\left[-\frac{1}{2}(6-8)-5(8-3)-8(3-6)\right] \)
\(=\frac{1}{2}\left[-\frac{1}{2}(-2)-5(5)-8(-3)\right] \)
\(=\frac{1}{2}[1-25+24]=\frac{1}{2}(0)=0 \)
Since, the area of triangle is zero, the given points are collinear.

(1,–1), (–4, 6) and (–3, –5)

 A(-4, 6), B(-3, -5), C(1, -1)
Area of triangle ABC \( =\frac{1}{2}\left[\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)\right. \left.-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right] \)
\(=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)\right. \left.+x_{3}\left(y_{1}-y_{2}\right)\right] \text { sq. units } \)
\(=\frac{1}{2}[-4(-5+1)-3(-1-6)+1(6+5)] \)
\(=\frac{1}{2}[-4 \times(-4)-3 \times(-7)+1 \times(11)] \)
\(=\frac{1}{2}[16+21+11] \)
\(=\frac{1}{2}(48)=24 \text { sq. units. } \)