Distance between two parallel lines a₁x + b₁y + c₁ = 0 and a₁x + b₁y + c₂ = 0 is
D = \(\frac{|c_1-c_2|}{\sqrt{a_1^2+b_1^2}}\)
Given lines can be written as 3x + 4y - 12 = 0 and 3x + 4y+\(\frac{1}{2}\)=0
Here a₁ = 3, b₁ = 4, c₁ = \(\frac{1}{2}\)
D = \(\frac{|c_1-c_2|}{\sqrt{a_2^1+b_2^1}}=\left|\frac{-12-\frac{1}{2}}{\sqrt{3^2+4^2}}\right|=\frac{25}{2\times5}\) = 2.5 units

Intercept form √3x - y + 4 = 0 ⇒ √3x - y = -4
\(\frac{-\sqrt{3}}{4}x+\frac{y}{4}=1\)
That is \(\frac{x}{(-\frac{4}{\sqrt{3}})}+\frac{y}{4}=1\)
Comparing the above equation with the equation \(\frac{x}{a}+\frac{y}{b}=1\)
We get, x-intercept = -\(\frac{4}{\sqrt{3}}\) and y-intercept = 4

By unrolling the hollow cylinder and flattening it into a rectangle, and with a single reflection allows us to determine the ant's path, as shown the figure. Let the baseline x-axis in cm and the vertical line through A (initial position of the ant) be the y-axis. Let H be the position of honey drop and E be the entry point of ant inside the vessel. From the given information we have
Let A(x₁, y₁) and H(x₂, y₂) be (0, 4) and (12, 13) respectively.
(i) The shortest distance between A and H is
\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{12^2+9^2}=15\)
the ant would need to crawl 15 cm to get the honey.
(ii) The equation of the path AH is \(\frac{y-4}{13-4}=\frac{x-0}{12-0}\)
y = 0.75x + 4
(iii) At the entry point E, y = 10 ⇒ x = 8
E = (8, 10)

Given that up to 1.8 kilometers the fixed rent is Rs. 25.
The corresponding equation is
y = 25, 0  ≤  x  ≤ 1.8......(1)
Also Rs. 12 for every additional kilometer after 1.8 kilometers.

The corresponding equation is
y = 25 + 12 (x - 1.8), x ≥ 1.8 ... (2)
The combined equation of (1) and (2), we get
y = \(\begin{cases} 25,\quad 0\le x\le 1.8 \\ 25+12\left( x-1.8 \right) ,\ x>1.8 \end{cases}\)
When x = 15, from (2), we get cost to travel 15 kilometers is Rs. 183.40.

Take conveniently PQ as x-axis, PA as y-axis and P is the origin (instead of conventional origin 0).
Therefore, the coordinates are P(0, 0), A(0, 3) and B(6, 5)
lf the image of A about the x-axis is \(\bar A\), then \(\bar A\) is (0, -3).
The required R is the point of intersection of the line \(\bar A\)B and x-axis.
AR and BR are the path of the cable (road)

The shortest length of the cable is \(AR+BR=BR+R\bar A=B\bar A\)
\(B\bar A=\sqrt{(6-0)^2+(5+3)^2}=10\ km\)
Equation of the line  \(\bar A\)B is \(y-(-3)=\frac{5-(-3)}{6-0}(x-0)\)
4x - 3y = 9
When y = 0, R is \((\frac{9}{4},0)\)
That is the substation should be located at a distance of 2.25 km from P.
The equation of AR is 4x + 3y = 9
The equations of the cable lines (roads) of RA and RB are 4x - 3y = 9 and 4x + 3y = 9.

Let P be the mid point of the line segment joining points A (1, 1) and B (2, 3). Then 
P is \(\left(\frac{1+2}{2},\frac{3+1}{2}\right)\)
\(\Rightarrow\) P is \(\left(\frac{3}{2},2\right)\)
Let m be the slope of perpendicular bisector of AB
Then m x slope of AB =-1
\(\Rightarrow m\left(\frac{3-1}{2-1}\right)=-1\)
\(\Rightarrow m\left(\frac21\right)=-1\Rightarrow m=\frac{-1}{2}\)
Since the perpendicular bisector of J\B passes through \(P\left(\frac{3}{2},2\right)\) and has slope \(m=\frac{-1}{2}\), its equation is
\(\Rightarrow y-2=-\frac{1}{2}\left(x-\frac{3}{2}\right)\)
\(\Rightarrow 2y-4=-x+\frac{3}{2}\)
\(\Rightarrow\)4y-8=-2x+3\(\Rightarrow\)2x+4y-11=0.

Shape of the hall in the shopping mall is cuboid. When you open out the cuboid, the not of the cuboid will be as shown in the following diagram.
 
The path of the escalator is from OA to AB to BC to CD
In OAE, OA² = AE² + OE²
\(\Rightarrow{ OA }^{ 2 }={ \left[ \frac { 1 }{ 4 } (720) \right] }^{ 2 }+{ OE }^{ 2 }\)
\(\Rightarrow\)   OA² = (180)² + (800)²
\(\Rightarrow\)  OA² = (20\(\times\)9)² + (20 \(\times\) 40)²
\(\Rightarrow\)   OA² = 20² + (9²+ 40²)
\(\Rightarrow\)  OA² = 20² \(\times\) 1681
\(\Rightarrow\)  OA² = 20² \(\times\) 41²
\(\Rightarrow\)  OA² = 20 \(\times\) 41 = 820
\(\therefore\)  Total length of the escalator = OA + A + BC + CD
= 4 \(\times\) OA    (since \(\Delta\)OAE \(\equiv \) \(\Delta\)ABB' \(\equiv \) \(\Delta\)BCC' \(\equiv \) \(\Delta\)CDD')
= 4 \(\times\) 820
The minimum length = 3280 units
(ii) The height at which the escalator changes its direction.
\(AE=\frac { 1 }{ 4 } (720)=180\quad units\)
\(BE=\frac { 1 }{ 2 } (720)=360\quad units\quad and\quad GE=\frac { 3 }{ 4 } (7200=540\quad units\)
(iii) Slope of the escalator at the turning points
Let \(\angle \)AOE = \(\theta\)
\(In\ \Delta OAE,\ tan\theta =\frac { opp }{ adj } =\frac { AE }{ DE } =\frac { 180 }{ 800 } =\frac { 9 }{ 40 } \)
\(\therefore \ Slope\ at\ the\ point\ A=\frac { 9 }{ 40 } \)
\(Since\ \Delta OAE\equiv \Delta ABB'\equiv \Delta BCC'\equiv \Delta CDD'\)
Slope at the points B, C will be \(\frac{9}{40}\)

(i) Given points are A (1, 3) B (2, 1) and C\((\frac{1}{2},4)\)
Slope of AB = \(\frac{1-3}{2-1}=\frac{-2}{2}=-2\)
Slope of BC = \(\frac{4-1}{\frac{1}{2}-2}=\frac{3}{\frac{1-4}{2}}=\frac{6}{-3}=-2\)
∴ Slope of AB = Slope of BC
ஃ AB II BC and B is a common point
ஃ Points A, B, C are collinear. 
(ii) Equation of AB is \(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \quad \frac { y-3 }{ -2 } =\frac { x-1 }{ 1 } \)
⇒ 2x + y - 5 = 0
Substitute the point C\((\frac{1}{2},4)\) we get,
= 2\((\frac{1}{2})\) + 4 - 5
= 1 + 4 - 5 = 5 - 5 = 0
ஃ Points A, B, C are collinear.
(iii) 
Area of ΔABC = \(\frac{1}{2}[6+\frac{1}{2}+4-(1+8+\frac{3}{2})]=\frac{1}{2}[(10+\frac{1}{2})-(9+\frac{3}{2})]\)
= \(\frac{1}{2}[10+\frac{1}{2}-9-\frac{3}{2})=\frac{1}{2}(1-1)=\frac{0}{2}=0\)
Since Area of ΔABC = 0, the given points A, B, C are collinear.

Any line perpendicular to 5x - y + 7 will be of the form x + 5y + k = 0 .....(1)
Let A and B are the meeting point of the line (1) with the co-ordinate axis.
Putting y = 0 in (1) we get, x + k = 0   \(\Rightarrow\) x = - k
\(\therefore\)  A is (-k, 0)      \(\Rightarrow\) OA = -k
Putting x = 0 in (1) we get, 5y + k = 0

\(\Rightarrow \quad y=-\frac { k }{ 5 } \)
\(\therefore \ B\ is\ \left( 0,-\frac { k }{ 5 } \right) \Rightarrow \ OB=-\frac { k }{ 5 } \)
\(\therefore\)  Area of the right angled \(\Delta\) OAB,
\(=\frac { 1 }{ 2 } \times OA\times OB\)
\(=\frac { 1 }{ 2 } \times (-k)\left( -\frac { k }{ 5 } \right) \)
Given that \(\frac { 1 }{ 2 } \times (-k)\left( -\frac { k }{ 5 } \right) =10\ sq.units.\)
\(\Rightarrow \quad \frac { { k }^{ 2 } }{ 10 } =\pm 10\ \Rightarrow \ { k }^{ 2 }=\pm 100\)
\(\Rightarrow \ k=\pm 10\)
\(\therefore \) The required equation is x + 5y \(\pm \) 10 = 0   
\(\Rightarrow\) x + 5y = \(\pm \)10

Let P(1, 2) and B(5, 3) are the given points.
By the property of reflector \(\angle \) XAB = \(\angle \) OAP = \(\theta\)
Let m₁ be the slope of the x-axis, m₂ and m₃ be the slopes of the lines AP and AB.
To find XAB,
Clearly m₁ = 0 Since it represents slope of x-axis.
\({ m }_{ 2 }=\frac { 2-0 }{ 1-x } =\frac { 2 }{ 1-x } and\left[ m=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\({ m }_{ 3 }=\frac { 3-0 }{ 5-x } =\frac { 3 }{ 5-x } \)

\(tan\ \theta =\left| \frac { { m }_{ 1 }-{ m }_{ 3 } }{ 1+{ m }_{ 1 }.{ m }_{ 3 } } \right| \)
\(tan\ \theta =\left| \frac { 0-\frac { 3 }{ 5-x } }{ 1+0\left( \frac { 3 }{ 5 } -x \right) } \right| =\frac { 3 }{ 5-x } \)
To find \(\angle \)OAP,
\(tan(-\theta )=\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\left| \frac { 0-\frac { 2 }{ 1-x } }{ 1+0\left( \frac { 2 }{ 1-x } \right) } \right| =\frac { 2 }{ 1-x } \)  [Since OAP is in the clockwise direction]
\(\Rightarrow \ -tan\theta =\frac { 2 }{ 1-x } \ \quad [\because tan\theta (-\theta )=tan\quad \theta ]\)
\(\Rightarrow \quad tan\quad \theta =\frac { -2 }{ 1-x } \)
From (1) and (2),
\(\frac { 3 }{ 5-x } =\frac { -2 }{ 1-x } \Rightarrow 3-3x=-10+2x\)
\(\Rightarrow\)  3 + 10 = 2x + 3x 
\(\Rightarrow\)  13 = 5x 
\(\Rightarrow \ x=\frac { 13 }{ 5 } \)
The required co-ordinates of A is \( \left( \frac { 13 }{ 5 } ,0 \right) \)