Step 1: Understand the given lines.
The given lines are $y=8$ and $y=-2$. Both lines are horizontal and parallel to the x-axis.

Step 2: Find the midpoint between the two lines.
A line equidistant from two parallel horizontal lines will also be horizontal and will lie exactly halfway between them. The y-coordinate of this line is the average of the y-coordinates of the given lines: $$y=\frac{8+(-2)}{2}$$ $$y=\frac{6}{2}=3$$

Step 1: Use the intercept form of a line.
Let the x-intercept be $a$ and the y-intercept be $b$. The equation of the line is: $$\frac{x}{a}+\frac{y}{b}=1$$ Given that the sum of the intercepts is 9, we have $a+b=9$, which gives $b=9-a$.

Step 2: Substitute the known point.
The line passes through the point $(2,2)$. Substituting $x=2$ and $y=2$ into the equation: $$\frac{2}{a}+\frac{2}{9-a}=1$$

Step 3: Solve for $a$.
Multiply through by $a(9-a)$ to clear the denominators: $$2(9-a)+2a=a(9-a)$$ $$18-2a+2a=9a-a^2$$ $$a^2-9a+18=0$$ Factoring the quadratic equation: $$(a-6)(a-3)=0$$ So, $a=6$ or $a=3$.

Step 4: Find the corresponding equations.
Case 1: If $a=6$, then $b=9-6=3$.
Equation: $\frac{x}{6}+\frac{y}{3}=1\Rightarrow x+2y=6$

Case 2: If $a=3$, then $b=9-3=6$.
Equation: $\frac{x}{3}+\frac{y}{6}=1\Rightarrow 2x+y=6$

Step 1: Reduce the equation to intercept form.
The intercept form of a line is $\frac{x}{a}+\frac{y}{b}=1$. To get a 1 on the right-hand side, divide the entire equation by 60: $$\frac{5x}{60}-\frac{12y}{60}=\frac{60}{60}$$ $$\frac{x}{12}+\frac{y}{-5}=1$$ From this, the x-intercept is $a=12$ and the y-intercept is $b=-5$.

Step 2: Find the length of the intercepted portion.
The line intersects the axes at $(12,0)$ and $(0,-5)$. The length of the line segment between these points is calculated using the distance formula or the Pythagorean theorem: $$\text{Length}=\sqrt{a^2+b^2}$$ $$\text{Length}=\sqrt{{12}^2+(-5{)}^2}=\sqrt{144+25}=\sqrt{169}=13$$

Step 1: Determine the angles.
The coordinate axes (x-axis and y-axis) intersect at the origin $(0,0)$ at an angle of ${90}^{\circ }$. The bisectors of the angles between them will pass through the origin and make angles of ${45}^{\circ }$ and ${135}^{\circ }$ with the positive x-axis.

Step 2: Find the slopes.
For the first bisector (quadrants I and III), the slope is $m_1=\tan \left({45}^{\circ }\right)=1$.
For the second bisector (quadrants II and IV), the slope is $m_2=\tan \left({135}^{\circ }\right)=-1$.

Step 3: Write the equations.
Using the point-slope form with the origin $(0,0)$:
$$y-0=1(x-0)\Rightarrow y=x\Rightarrow x-y=0$$ $$y-0=-1(x-0)\Rightarrow y=-x\Rightarrow x+y=0$$

i. Equation of the median through A:
The median through A connects point A to the midpoint M of the opposite side BC.
Midpoint M of B$(2,-3)$ and C$(-1,-2)$: $$M=(\frac{2+(-1)}{2},\frac{-3+(-2)}{2})=(\frac{1}{2},-\frac{5}{2})$$ Slope of the median AM passing through A$(1,4)$ and M$(1/2,-5/2)$: $$m_{AM}=\frac{-5/2-4}{1/2-1}=\frac{-13/2}{-1/2}=13$$ Equation of median AM: $$y-4=13(x-1)\Rightarrow y-4=13x-13\Rightarrow 13x-y-9=0$$

ii. Equation of the altitude through A:
The altitude through A is perpendicular to the side BC.
Slope of line BC passing through B$(2,-3)$ and C$(-1,-2)$: $$m_{BC}=\frac{-2-(-3)}{-1-2}=\frac{1}{-3}=-\frac{1}{3}$$ Slope of altitude AD (perpendicular to BC) is $m=3$.
Equation of altitude passing through A$(1,4)$: $$y-4=3(x-1)\Rightarrow y-4=3x-3\Rightarrow 3x-y+1=0$$

iii. Equation of the perpendicular bisector of BC:
This line passes through the midpoint M$(1/2,-5/2)$ and is perpendicular to BC, so its slope is $m=3$.
Equation of the perpendicular bisector: $$y-(-\frac{5}{2})=3(x-\frac{1}{2})$$ $$y+\frac{5}{2}=3x-\frac{3}{2}$$ Multiply by 2: $$2y+5=6x-3\Rightarrow 6x-2y-8=0\Rightarrow 3x-y-4=0$$

Step 1: Identify coordinates.
Let $q$ represent the number of shoes (independent variable on x-axis) and $C$ represent the cost of production (dependent variable on y-axis).
We are given two points: $(q_1,C_1)=(30,1500)$ and $(q_2,C_2)=(50,2000)$.

Step 2: Find the slope (rate of change).
$$m=\frac{C_2-C_1}{q_2-q_1}=\frac{2000-1500}{50-30}=\frac{500}{20}=25$$ This means the marginal cost per pair of shoes is Rs. 25.

Step 3: Write the linear equation.
Using the point-slope form $C-C_1=m(q-q_1)$: $$C-1500=25(q-30)$$ $$C-1500=25q-750$$ $$C=25q-750+1500$$ $$C=25q+750$$

By definition, the inclination of a line is the angle $\theta$ that the line makes with the positive direction of the x-axis, measured in the anticlockwise direction.

A line parallel to the y-axis is vertical and makes an angle of ${90}^{\circ }$ with the positive x-axis. Since the slope $m=\tan \left({90}^{\circ }\right)$, and the tangent of 90 degrees is undefined, the slope is not defined.

Step 1: Calculate the slope of the second line.
Slope $m_2$ for line through $(-1,4)$ and $(0,6)$: $$m_2=\frac{6-4}{0-(-1)}=\frac{2}{1}=2$$

Step 2: Set the slopes equal to each other.
Slope $m_1$ for line through $(3,y)$ and $(2,7)$: $$m_1=\frac{7-y}{2-3}=\frac{7-y}{-1}=y-7$$ Because the lines are parallel, $m_1=m_2$. $$y-7=2\Rightarrow y=9$$

Using the intercept form of the line equation: $$\frac{x}{a}+\frac{y}{b}=1$$ Substitute $a=-2$ and $b=3$: $$\frac{x}{-2}+\frac{y}{3}=1$$ Multiply the entire equation by $-6$ (the least common multiple, keeping standard signs): $$3x-2y=-6$$ Rearranging into general form: $$3x-2y+6=0$$

Step 1: Find the point of intersection.
Solve the system of equations:
$\left(1\right) 2x+3y=5$
$\left(2\right) 3x-2y=1$
Multiply (1) by 2 and (2) by 3 to eliminate y:
$4x+6y=10$
$9x-6y=3$
Add the two equations:
$13x=13\Rightarrow x=1$.
Substitute $x=1$ into equation (1): $2\left(1\right)+3y=5\Rightarrow 3y=3\Rightarrow y=1$.
The point of intersection is $(1,1)$.

Step 2: Find the required slope.
The line must be perpendicular to $5x-3y=7$, which can be written as $3y=5x-7\Rightarrow y=\frac{5}{3}x-\frac{7}{3}$.
The slope of this line is $\frac{5}{3}$. Therefore, the slope of our perpendicular line is $m=-\frac{3}{5}$.

Step 3: Write the equation of the line.
Using point-slope form with $(1,1)$ and $m=-\frac{3}{5}$: $$y-1=-\frac{3}{5}(x-1)$$ $$5y-5=-3x+3$$ $$3x+5y-8=0$$

Step 1: Identify the points and calculate slopes.
Let $A(-2,-1)$, $B(4,0)$, $C(3,3)$, and $D(-3,2)$ be the vertices of the quadrilateral.
We will check if opposite sides are parallel by comparing their slopes using $m=\frac{y_2-y_1}{x_2-x_1}$.

• Slope of AB: $m_{AB}=\frac{0-(-1)}{4-(-2)}=\frac{1}{6}$
• Slope of DC: $m_{DC}=\frac{3-2}{3-(-3)}=\frac{1}{6}$

Since $m_{AB}=m_{DC}$, side AB is parallel to side DC.

• Slope of BC: $m_{BC}=\frac{3-0}{3-4}=\frac{3}{-1}=-3$
• Slope of AD: $m_{AD}=\frac{2-(-1)}{-3-(-2)}=\frac{3}{-1}=-3$

Since $m_{BC}=m_{AD}$, side BC is parallel to side AD.

Step 1: Set up the coordinates.
Let the line intersect the x-axis at $A(a,0)$ and the y-axis at $B(0,b)$. The line segment is between these two points. Let the point $P(h,k)$ divide the segment AB in the ratio 1:2. (Generally, the ratio is read from the x-axis to the y-axis).

Step 2: Apply the section formula.
$$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$$ Here, $m=1$, $n=2$, $(x_1,y_1)=(a,0)$, and $(x_2,y_2)=(0,b)$. $$h=\frac{1\left(0\right)+2\left(a\right)}{1+2}\Rightarrow h=\frac{2a}{3}\Rightarrow a=\frac{3h}{2}$$ $$k=\frac{1\left(b\right)+2\left(0\right)}{1+2}\Rightarrow k=\frac{b}{3}\Rightarrow b=3k$$

Step 3: Write the intercept form of the line.
Using $\frac{x}{a}+\frac{y}{b}=1$: $$\frac{x}{3h/2}+\frac{y}{3k}=1$$ $$\frac{2x}{3h}+\frac{y}{3k}=1$$ Multiply through by $3hk$: $$2kx+hy=3hk$$

Step 1: Find the slope of the given line.
The line equation is $x-2y+3=0$. Rewriting in slope-intercept form ($y=mx+c$): $$2y=x+3\Rightarrow y=\frac{1}{2}x+\frac{3}{2}$$ The slope of this line is $m_1=\frac{1}{2}$.

Step 2: Find the slope of the perpendicular line.
For perpendicular lines, $m_1\times m_2=-1$. $$m_2=\frac{-1}{1/2}=-2$$

Step 3: Write the equation of the line.
Use the point-slope form with point $(1,-2)$ and slope $m=-2$: $$y-(-2)=-2(x-1)$$ $$y+2=-2x+2$$ $$2x+y=0$$

Step 1: Find the midpoint of BC.
The median through A connects vertex A to the midpoint M of the opposite side BC. Let's calculate the midpoint M of B$(4,6)$ and C$(6,4)$: $$M=(\frac{4+6}{2},\frac{6+4}{2})=(\frac{10}{2},\frac{10}{2})=(5,5)$$

Step 2: Find the equation of the line passing through A and M.
We need the equation of the line passing through A$(1,2)$ and M$(5,5)$.
First, find the slope $m$: $$m=\frac{5-2}{5-1}=\frac{3}{4}$$ Using point-slope form with A$(1,2)$: $$y-2=\frac{3}{4}(x-1)$$ $$4(y-2)=3(x-1)$$ $$4y-8=3x-3$$ $$3x-4y+5=0$$

Step 1: Set up intercepts.
Let the line intersect the x-axis at $A(a,0)$ and the y-axis at $B(0,b)$. The portion of the line between the axes is the segment AB.

Step 2: Use the midpoint formula.
The midpoint of AB is given as $(2,2)$. $$(\frac{a+0}{2},\frac{0+b}{2})=(2,2)$$ Equating the coordinates: $$\frac{a}{2}=2\Rightarrow a=4$$ $$\frac{b}{2}=2\Rightarrow b=4$$

Step 3: Write the equation of the line.
Using the intercept form $\frac{x}{a}+\frac{y}{b}=1$: $$\frac{x}{4}+\frac{y}{4}=1$$ Multiply by 4: $$x+y=4$$ $$x+y-4=0$$

(i) Rate of change of profit for Electronics:
Coordinates for Electronics: $(2,5)$ and $(8,17)$.
Slope $m_E=\frac{17-5}{8-2}=\frac{12}{6}=2$.
The rate of change is ` 2 lakhs per month.

(ii) Equation of the profit line for Fashion:
Coordinates for Fashion: $(3,8)$ and $(9,14)$.
Slope $m_F=\frac{14-8}{9-3}=\frac{6}{6}=1$.
Using point-slope form with $(3,8)$:
$$y-8=1(x-3)\Rightarrow y=x+5\Rightarrow x-y+5=0$$

(iii) Electronics equation & Equal profit:
Using slope $m_E=2$ and point $(2,5)$:
$$y-5=2(x-2)\Rightarrow y-5=2x-4\Rightarrow y=2x+1$$ For equal profit, equate the two $y$ values:
$$2x+1=x+5\Rightarrow x=4$$ Substituting $x=4$ into either equation gives $y=4+5=9$.
They will have equal profit in Month 4, and the profit amount will be ` 9 lakhs.

(iv) New product line:
Starts at `10 lakhs in month 1: Point $(1,10)$.
Grows parallel to Electronics: Slope $m=2$.
Equation: $y-10=2(x-1)\Rightarrow y=2x-2+10\Rightarrow y=2x+8$.
Profit in month 6 ($x=6$):
$$y=2\left(6\right)+8=12+8=20$$ The profit will be ` 20 lakhs.

(i) Slope of road AB:
$$m_{AB}=\frac{6-2}{4-1}=\frac{4}{3}$$

(ii) Equation of line BC:
Points: B$(4,6)$, C$(7,3)$.
Slope $m_{BC}=\frac{3-6}{7-4}=\frac{-3}{3}=-1$.
Equation using B$(4,6)$: $y-6=-1(x-4)\Rightarrow y-6=-x+4\Rightarrow x+y-10=0$.

(iii) Centroid and parallel line:
Centroid G is given by $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$.
$$G=(\frac{1+4+7}{3},\frac{2+6+3}{3})=(\frac{12}{3},\frac{11}{3})=(4,\frac{11}{3})$$ Line parallel to AB has slope $m=\frac{4}{3}$. Passing through $G(4,11/3)$:
$$y-\frac{11}{3}=\frac{4}{3}(x-4)$$ Multiply by 3: $$3y-11=4x-16\Rightarrow 4x-3y-5=0$$

(iv) Area of triangle / Collinear check:
$$\text{Area}=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$$ $$\text{Area}=\frac{1}{2}\left|1\right(6-3)+4(3-2)+7(2-6\left)\right|$$ $$\text{Area}=\frac{1}{2}\left|1\right(3)+4(1)+7(-4\left)\right|=\frac{1}{2}|3+4-28|=\frac{1}{2}|-21|=10.5$$ Since the area is 10.5 sq km (not zero), the points are not collinear.

(i) Rate of appreciation for Stock A:
Data points for Stock A: $(3,350)$ and $(7,550)$. Also purchased at month 0 for 200, i.e., $(0,200)$.
$$m_A=\frac{550-350}{7-3}=\frac{200}{4}=50$$ The rate of appreciation is ` 50 per month.

(ii) Equation for Stock B:
Data points for Stock B: $(2,400)$ and $(8,700)$.
$$m_B=\frac{700-400}{8-2}=\frac{300}{6}=50$$ Equation using point $(2,400)$: $$y-400=50(x-2)\Rightarrow y=50x-100+400\Rightarrow y=50x+300$$

(iii) Same price check:
Equation for Stock A: using y-intercept 200, $y=50x+200$.
Since the slopes for both Stock A and Stock B are identical ($m=50$) but their y-intercepts are different (200 vs 300), the lines are parallel. Therefore, the stocks will never have the same price.

(iv) Selling point and Stock C:
Investor wants to sell Stock A when $y=650$:
$$650=50x+200\Rightarrow 450=50x\Rightarrow x=9$$ Since he purchased at month 3, he should sell $9-3=6$ months after purchase.

Stock C depreciates perpendicular to Stock B. $m_C=-\frac{1}{m_B}=-\frac{1}{50}$.
Starts at 800 at month 0, so y-intercept is 800.
Equation: $y=-\frac{1}{50}x+800\Rightarrow x+50y=40000$.

(i) Length of the diagonal AC:
Using distance formula between A$(0,0)$ and C$(12,8)$: $$AC=\sqrt{{(12-0)}^2+{(8-0)}^2}=\sqrt{144+64}=\sqrt{208}=4\sqrt{13}\text{m}$$

(ii) Equation of the line AC:
Slope $m=\frac{8-0}{12-0}=\frac{8}{12}=\frac{2}{3}$.
Since it passes through the origin, y-intercept is 0. $$y=\frac{2}{3}x\Rightarrow 2x-3y=0$$

(iii) Coordinates of P and perpendicular line:
The distance AP = 5. Let the angle of inclination of AC be $\theta$.
$\cos \theta =\frac{12}{4\sqrt{13}}=\frac{3}{\sqrt{13}}$ and $\sin \theta =\frac{8}{4\sqrt{13}}=\frac{2}{\sqrt{13}}$.
Coordinates of P $(x_P,y_P)$:
$$x_P=0+5\cos \theta =\frac{15}{\sqrt{13}}, y_P=0+5\sin \theta =\frac{10}{\sqrt{13}}$$ Line perpendicular to AC has slope $m_{\perp }=-\frac{3}{2}$. Passing through P:
$$y-\frac{10}{\sqrt{13}}=-\frac{3}{2}(x-\frac{15}{\sqrt{13}})$$ $$2y-\frac{20}{\sqrt{13}}=-3x+\frac{45}{\sqrt{13}}$$ $$3x+2y=\frac{65}{\sqrt{13}}=5\sqrt{13}$$

(iv) Coordinates of D and intersection of diagonals:
For a rectangle with A$(0,0)$, B$(12,0)$, C$(12,8)$, the opposite corner D must be $(0,8)$.
Midpoint of diagonal AC: $(\frac{0+12}{2},\frac{0+8}{2})=(6,4)$.
Midpoint of diagonal BD: $(\frac{12+0}{2},\frac{0+8}{2})=(6,4)$.
Since the midpoints are the same, the diagonals intersect at and bisect each other at (6, 4).

Check the Assertion:
For the lines $3x+4y-7=0$ and $6x+8y+5=0$:
$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{4}{8}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{-7}{5}$
Since $\frac{1}{2}=\frac{1}{2}\ne \frac{-7}{5}$, the lines are indeed parallel. Thus, Assertion (A) is true.
Check the Reason:
The reason states the correct mathematical condition for two lines to be parallel, which is exactly the logic we used. Reason (R) is true and perfectly explains the Assertion.

Check the Assertion:
Using the point-slope form given in the reason: $$y-(-5)=\frac{3}{4}(x-2)$$ $$y+5=\frac{3}{4}(x-2)$$ $$4y+20=3x-6$$ $$3x-4y-26=0$$ The Assertion states the equation is $3x-4y-23=0$, which is incorrect. Thus, Assertion (A) is false.
Check the Reason:
The reason states the correct point-slope formula, which is absolutely true.

Check the Assertion:
The slope of $4x-5y+7=0$ is $m_1=\frac{4}{5}$.
The slope of a line perpendicular to it is $m_2=-\frac{5}{4}$.
Passing through $(2,-3)$: $$y-(-3)=-\frac{5}{4}(x-2)$$ $$4y+12=-5x+10\Rightarrow 5x+4y+2=0$$ Thus, Assertion (A) is true.
Check the Reason:
For lines $4x-5y+7=0$ and $5x+4y+2=0$, checking the product sum:
$a_1{a}_2+b_1{b}_2=\left(4\right)\left(5\right)+(-5)\left(4\right)=20-20=0$.
Reason (R) is true and perfectly explains the Assertion.

Check the Assertion:
Break-even point is where Total Revenue equals Total Cost ($R=C$). $$100x=50x+2000$$ $$50x=2000\Rightarrow x=40$$ Thus, Assertion (A) is true.
Check the Reason:
The reason states that the break-even point occurs when total revenue equals zero. This is false. The break-even point occurs when total revenue equals total cost (or profit equals zero).