(i) Slope of a line parallel to X-axis is '0'
i.e., m = 0
Equation of a line passing through (x₁, y₁) and having slope 'm' is
\( y-y_{1}=m\left(x-x_{1}\right) \)
\(\Rightarrow y-3=0(x-4) \)
\(\Rightarrow y-3=0 \)
\(\Rightarrow y=3\)
(ii) Slope of a line perpendicular to X-axis is undefined
\(\text { i. e., } \frac{1}{0}=\mathrm{m}\)
Equation of a straight line is \(y-3=\frac{1}{0}(x-4)\)
\(\Rightarrow x-4=0 \)
\(\Rightarrow x=4 \)

Let the equation of the line be  \(\frac{x}{a}+\frac{y}{b}=1\)
Which meets X - axis at A (a, 0) and Y - axis at B (0, b)
Given that P (- 5, 4) dividing AB in the ratio 1 : 2
\(\text { Using section formula, }\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)
\(\Rightarrow \left(\frac{1(0)+2(a)}{1+2}, \frac{1(b)+2(0)}{1+2}\right)=(-5,4)\)
\(\Rightarrow \left(\frac{2 a}{3}, \frac{b}{3}\right)=(-5,4)\)
\(\frac{2 a}{3}=-5, \frac{b}{3}=4 \)
\(a=-\frac{15}{2}, b=12 \)
\(\therefore \text { The equation of line is } \frac{x}{a}+\frac{y}{b}=1 \text {. }\)
\(\text { i.e., }-\frac{2 x}{15}+\frac{y}{12}=1\)
\(\Rightarrow 8 x-5 y+60=0\)

Assuming F along X-axis and K along Y-axis.
The two points are (32, 273) and (212,373).
The equation of straight line passing through the points (32,273) and (212,373) is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
\(\Rightarrow \frac{K-273}{373-273}=\frac{F-32}{212-32}\)
\(K=\frac{5}{9}(F-32)+273\)
\(\text { Putting } K=0 \text { in }(1), \text { we get }\)
\(0=\frac{5}{9}(F-32)+273\)
\(\Rightarrow F=32-491.4=-459.4\)

\(\text { Slope of } A B=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
\(=\frac{0-4}{-2-2}=\frac{-4}{-4}=1\)
\(\text { Slope of } B C=\frac{4-1}{2-4}=-\frac{3}{2}\)
\(\text { Slope of } \mathrm{CD}=\frac{1+3}{4-0}=\frac{4}{4}=1\)
\(\text { Slope of } A D=\frac{0+3}{-2-0}=-\frac{3}{2}\)
Slope of AB = Slope of CD,
Slope of BC = Slope of AD
Therefore The given points form a parallelogram

\(\text { Slope of } \mathrm{AB}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
\(=\frac{-1-1}{0-2}\)
\(=\frac{-2}{-2}=1\)
\(\text { Slope of } \mathrm{BC}=\frac{1-3}{2+4}\)
\(=-\frac{2}{6}=-\frac{1}{3}\)
\(\text { Slope of } A C=\frac{-1-3}{0+4}\)
\(=-\frac{4}{4}=-1\)
\(\text { (Slope of } A B) \times(\text { Slope of } A C)=1 \times-1=-1\)
\(\left[\therefore \mathrm{m}_{1} \times \mathrm{m}_{2}=-1\right]\)
\(\therefore \text { The given points form a right triangle. }\)

Area of triangle is zero, Since the points are collinear.
\(\text { i.e., } \mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)=0\)
\((p+1)(3-2 p)+(2 p+1)(2 p-1)+(2 p+2)(1-3)=0\)
\(3 p-2 p^{2}+3-2 p+4 p^{2}-1-4 p-4=0\)
\(2 p^{2}-3 p-2=0\)
\((2 p+1)(p-2)=0\)
2P+ 1 = 0, p - 2 = 0
\(p=-\frac{1}{2}, \ p = 2\)

\(\text { Area of triangle }=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+\right.\left.\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right] \text { sq. units }\)
\(=\frac{1}{2}[a(c-c+a)+a(c-a-c-a)-a(c+a-c)]\)
\(=\frac{1}{2}[a(a)+a(-2 a)-a(a)]\)
\(=\frac{1}{2}\left(-2 a^{2}\right)=-a^{2}\)
Area = a² sq. units. [\(\because\) Area cannot be negative]

Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.,
= \(\frac { 1 }{ 2 } \) [2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0
= \(\frac { 1 }{ 2 } \) [-4k] = 0
k = 0 
Area of ΔABC
= \(\frac { 1 }{ 2 } \)[2(0 + 3) + 4(-3 - 3) + 6(3 - 0)] = 0

The area of the triangle formed by the given points is equal to
= \(\frac { 1 }{ 2 } \) [-1.5 (-2 - 4) + 6 (4 - 3) + (-3) (3 + 2)]
= \(\frac { 1 }{ 2 } \) [9 + 6 - 15] = 0
We can have a triangle at area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.

We have Area of the quadrilateral

=\(\frac { 1 }{ 2 } \) [(-12 - 30 - 28 -10) - (+ 10 + 28 + 30 + 12)]
\(\frac { 1 }{ 2 } \) [-80 - (80)]
\(\frac { 1 }{ 2 } \)[-160] = -80 = 80 square units.
(∵ Area is always +ve).

If A(-2, -1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.
We know that the diagonals of a parallelogram bisect each other. Therefore the co-ordinates of the midpoint of AC are same as the co-ordinates of the mid-point of BD. i.e.
\(\left( \frac { -2+4 }{ 2 } ,\frac { -1+b }{ 2 } \right) =\left( \frac { a+1 }{ 2 } ,\frac { 0+2 }{ 2 } \right) \)
⇒ \(\left( 1,\frac { b-1 }{ 2 } \right) =\left( \frac { a+1 }{ 2 } ,1 \right) \)
⇒ \(\frac { a+1 }{ 2 } \) = 1 ⇒ a + 1 = 2 ⇒ a = 1
⇒ \(\frac { b-1 }{ 2 } \) = 1 ⇒ b - 1 = 2 ⇒ b = 3

The area of the triangle formed by the vertices A(1, -1), B(-4, 6) and C(-3, -5), by using the formula above, is given by
= \(\frac { 1 }{ 2 } \)[1(6 + 5) +(-4) (-5 + 1) + (-3)(-1 - 6)]
= \(\frac { 1 }{ 2 } \)[11 + 16 + 21] = 24 square units.

We know that diagonals at a parallelogram bisect each other.
So, the coordinates of the mid-point at AC = coordinates of the mid-point at BD.
i.e., \(\left[ \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 2+3 }{ 2 } \right] \)
\(\left[ \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 5 }{ 2 } \right] \)
\(\frac { 15 }{ 2 } =\frac { 8+P }{ 2 } \)
P = 7

Let P and Q be the points of trisection at AB.
i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1:2. Therefore, the coordinates at P, by applying the section formula, are
\(\left[ \frac { 1(-7)+2(2) }{ 1+2 } ,\frac { 1(7)+2(-2) }{ 1+2 } \right] \) i.e., (-1,10)
Now, Q also divides AB internally in the ratio 2:1, so, the coordinates at Q are
\(\left[ \frac { 2(-7)+1(2) }{ 2+1 } ,\frac { 2(4)+(-2) }{ 2+1 } \right] \) i.e., (-4,2)
Therefore, the coordinates at the points at trisection of the line segment joining A and B are (-1, 0) and (-4, 2).