Given vertices A(1, - 4) , B(2, - 3) and C(4, - 7)
Slope of the line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
Slope of AB = \(\frac{-4+3}{1-2}=\frac{-1}{-1}=1 \)
Slope of BC = \(\frac{-3+7}{2-4}=\frac{4}{-2}=-2 \)
Slope of AC = \(\frac{-4+7}{1-4}=\frac{3}{-3}=-1 \)
(Slope of AB) x (Slope of AC) = - 1
AB is perpendicular to AC.
Hence, the given vertices form a right angled triangle
Distance between the points (x₁, y₁) and (x₂, y₂) is \(\sqrt{\left.\left(x_{1}-x_{2}\right)^{2}+y_{1}-y_{2}\right)^{2}}\) units
\(A B =\sqrt{(1-2)^{2}+(-4+3)^{2}}=\sqrt{1+1}=\sqrt{2} \)
\(A B^{2} =(\sqrt{2})^{2}=2 \)
\(B C =\sqrt{(2-4)^{2}+(-3+7)^{2}}=\sqrt{4+16}=\sqrt{20} \)
BC² = 20
\(A C=\sqrt{(1-4)^{2}+(-4+7)^{2}}=\sqrt{9+9}=\sqrt{18}\)
AC² = 18
Now, AB² + AC² = BC²
Hence, the Pythagoras theorem is satisfied.

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 (- 2, a) and (9, 3) and Slope = - \(\frac { 1 }{ 2 } \).
Slope of the line = - \(\frac { 1 }{ 2 } \)
\(\frac{y_{1}-y_{2}}{x_{1}-x_{2}} =-\frac{1}{2} \)
\(\frac{a-3}{-2-9} =-\frac{1}{2} \)
\(\frac{a-3}{-11} =-\frac{1}{2} \)
2(a - 3) = 11
2a = 11 + 6
a = \(\frac{17}{2}\)

Given points (3, - 1), (a, 3) and (1, - 3)
Let the points be A (3, - 1), B (a, 3) and C (1,- 3)
Slope of AB = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-1-3}{3-a}=\frac{-4}{3-a} \)
Slope of BC = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{3+3}{a-1}=\frac{6}{a-1} \)
Since, the points A, B and C are collinear.
Slope of AB = Slope of BC
\(\frac{-4}{3-a}=\frac{6}{a-1}\)
-2 (a - 1) = 3(3 - a)
-2a + 2 = 9 - 3a
3a - 2a = 9 - 2
a = 7

Given points (- 3, - 4), (7, 2) and (12, 5)
Let the points be A (- 3, - 4),8 (2, 2) and C (12, 5)
Slope of a line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
Slope of AB = \(\frac{-4-2}{-3-7}=\frac{-6}{-10}=\frac{3}{5} \)
Slope of BC = \(\frac{2-5}{7-12}=\frac{-3}{-5}=\frac{3}{5} \)
Slope of AB = Slope of BC
The points A, B and C are collinear

Mid point of line segment joining (4, 2) and (-6, 4)
\(\text { Mid point } =\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \)
\(=P\left(\frac{4-6}{2}, \frac{2+4}{2}\right) \)
\(=P\left(-\frac{2}{2}, \frac{6}{2}\right)=\mathrm{P}(-1,3) \)
Now, slope of a line joining A (5, 1) and P (- 1, 3)
\(\mathrm{m}=\frac{y_{1}-y_{2}}{x_{1}-x}=\frac{1-3}{5+1}=\frac{-2}{6}=-\frac{1}{3}\)
Slope of a perpendicular to the line joining A and P
\(=-\frac{1}{m}=-\frac{1}{\left(-\frac{1}{3}\right)}=3\)

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.

The slope of line p is m₁ = \(\frac { 4+2 }{ 12-3 } =\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
The slope of line p is m₂ = \(\frac { 2+2 }{ 12-6 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Thus, slope of line p = slope of line q.
Therefore, the line p is parallel to the line q.

Th e slope of line r is m₁  \(=\frac { 8-2 }{ 5+2 } =\frac { 6 }{ 7 } \)
The slope of line θ is m₂  \(=\frac { 0-7 }{ -2+8 } =\frac { -7 }{ 6 } \)
The product of slopes \(=\frac { 6 }{ 7 } \times \frac { -7 }{ 6 } =-1\)
That is, m₁m₂ = -1

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