Equation of line joining (1, 2) and (3, 4) is
\(\frac{y-2}{4-2} =\frac{x-1}{3-1} \)
\(x-y+1 =0 \)
Any line perpendicular to this x + y + k = 0
This passes through midpoint of (1, 2) and (3, 4)
That is (2, 3)
k = -5, x+y-5 = 0
x intercept is 5, y intercept is 5.

The point M is (0, 3)
The equation of line joining ('-1, 2) and (0, 3) is
\(\frac{y-2}{3-2} =\frac{x+1}{0+1} \)
\(x-y+3 =0\)

\(\text {Perimeter }=4+2 \sqrt{2}\)
\(a+a+\sqrt{2} =4+2 \sqrt{2} \)
\(2 a+\sqrt{2} a =4+2 \sqrt{2} \)
\(\therefore a =2 \)
\(\text {Equation of line is } \frac{x}{2}+\frac{y}{2}=1\)
\(x+y-2=0\)

\(\text { Slope of } 3 x-y+5=0 \text { is } \frac{-3}{-1}=3=m_{1}\)
Let m₂ be the slope of the second line
\(\text {Given } \tan \theta=\tan 45^{\circ}=1 \Rightarrow \frac{m_{1}-m_{2}}{1+m_{1} m_{2}}=\pm 1\)
\(\frac{3-m_{2}}{1+3 m_{2}}=1 \quad \frac{m_{2}-3}{1+3 m_{2}}=1\)
\(3-m_{2}=1+3 m_{2} \quad m_{2}-3=1+3 m_{2}\)
\(2=4 m_{2} \quad-2 m_{2}=4\)
\(\mathrm{m}_{2}=\frac{1}{2} \quad \mathrm{~m}_{2}=-2\)
\(\left(\frac{1}{2},-2\right)\)

Since A and B are the points on the axes, its co-ordinate are A(x, 0) and B(0, y)
Given that p(r, c) is the mid-point of Ab.
\(\therefore\) Using mid-point formula,
\((r,c)=\left( \frac { x+0 }{ 2 } ,\frac { 0+y }{ 2 } \right) \)   

\(\Rightarrow \quad r=\frac { x }{ 2 } and\quad c=\frac { y }{ 2 } \)
\(\Rightarrow \quad x=2r\ and\ y=2x\)
\(\therefore \ The\ point\ A\ and\ B\ are\left( \begin{matrix} { x }_{ 2 } & { y }_{ 2 } \\ 0 & 2c \end{matrix} \right) and\left( \begin{matrix} { x }_{ 1 } & { y }_{ 1 } \\ 2r & 0 \end{matrix} \right) \)
\(\therefore \quad Equation\ of\ AB\ is\frac { y-0 }{ 2c-0 } =\frac { x-2r }{ 0-2r } \)
\(\Rightarrow \quad \frac { y }{ 2c } =\frac { x-2r }{ -2r } \ \ \Rightarrow \ \frac { y }{ c } =\frac { x-2r }{ -r } \)
\(\Rightarrow -ry=cx-2rc\ \Rightarrow \ cr+xy=-2rc\)
\(\Rightarrow \ cx+ry=2rc\)
Dividing by rc, we get, \(\frac { cx }{ rc } +\frac { ry }{ rc } =\frac { 2rc }{ rc } \) \(\Rightarrow \frac { x }{ r } +\frac { y }{ c } =2\)  Hence proved.

Let P(x₁, y₁) be the moving point and A(a, 0) B(-a, 0) are the fixed points
Given PA² + PB² = 2C²
\(\Rightarrow\) (x₁ - a)²+ (y₁ - 0)² + (x₁ + a)² + (y₁ - 0)² = 2c² [using distance formula]

\(\Rightarrow2x^2_1+2y^2_1+2a^2=2c^2\) 
\(\Rightarrow x^2_1+y^2_1+a^2=c^2\)
\(\Rightarrow x^2_1+y^2_1=c^2-a^2\)
\(\therefore\) Locus of (x₁, y₁) is x²+ y² = c²- a²

If the equation of two lines are in general form as a₁ x + b₁ y₁ + c = 0 and a₂x + b₂y + c₂ = 0
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } }\ or\ { a }_{ 1 }{ b }_{ 2 }={ a }_{ 2 }{ b }_{ 1 }\)
Given lines are 3x + 2y + 9 = 0 and 12x + 8y - 15 = 0
\(\frac { 3 }{ 12 } =\frac { 2 }{ 8 } \)
\(\Rightarrow \frac { 1 }{ 4 } =\frac { 1 }{ 4 } \)
Hence the given lines are parallel.

(i) with y-intercept (-4)
Equation of the line passing through (x, y) with y - intercept c is y = mx + c
Since the y - intercept is c = - 4, (0, -4) is also a point on the line.
Slope of line joining (1, 1) and (0, 4) is
\(m={y_2-y_1\over x_2-x_3}={-4-1\over 0-1}={-5\over -1}=5\)
∴ Required equation is y = 5x - 4. [ m = 5, c = -4]

(ii) with slope 3
Equation of the line passing through (1, 1) with slope 3 is
y -1  = 3 (x - x₁)    [∵ y - y₁ = m(x - x₁)]
⇒ y - 1 = 3x - 3
⇒ 3x - y = -1 + 3
⇒ 3x - y = 2
(iii) and (-2, 3)
Equation of the line passing through (1, 1) and (-2, 3)
\(\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\)
⇒ \(\frac{y-1}{3-1}=\frac{x-1}{-2-1}\)
⇒ \(\frac{y-1}{2}=\frac{x-1}{-3}\)
⇒ -3y + 3 = 2x - 2
⇒ 2x + 3y = 3 + 2
⇒ 2x + 3y = 5
(iv) and the perpendicular from the origin makes an angle 60° with x- axis .
Given \(\alpha\) = 60°
Perpendicular distance p = distance between op
= \(\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2}\)

ஃ Required equation in normal form is x cos \(\alpha\) + y sin \(\alpha\) = p
⇒ x cos 60° +y sin 60° = √2
⇒ \(x(\frac{1}{2})+y\frac{\sqrt{3}}{2}=\sqrt{2}\)
⇒ \(\frac{x+\sqrt{3}y}{2}=\sqrt{2}\)
⇒ x + √3y = 2√2

Equation of the straight line in intercept form is \(\frac { x }{ a } +\frac { y }{ b } =1\)
Given a + b = 1 \(\Rightarrow\) b = 1 - a
\(\Rightarrow \quad \frac { x }{ a } +\frac { y }{ 1-a } =1\)
Since (8, 3) lies on this line, we get,
\(\Rightarrow \frac { 8 }{ a } +\frac { 3 }{ 1-a } =1 \)
\(\Rightarrow \frac { 8-8a+3a }{ a(1-a) } =1\)
\(\Rightarrow \frac { 8-5a }{ a-{ a }^{ 2 } } =1\)
\(\Rightarrow \quad 8-5a=a-{ a }^{ 2 }\)
\(\Rightarrow\)  8 - 5a - a + a² = 0   \(\Rightarrow\)  a²- 6a + 8 = 0
\(\Rightarrow\)  (a-4)(a-2) = 0   \(\Rightarrow\)  a = 4 or 2
If a = 4, b = 1-4 = -3
if a = 2, b = 1-2 = -1
When a = 4, b = -3, equation of straight line is \(\frac { x }{ 4 } +\frac { y }{ -3 } =1\)
\(\Rightarrow\)   -3x + 4y = -12    [Form (1)]
\(\Rightarrow\)  3x - 4y = 12
If a = 2 and b = -1, equation of the other straight line is \(\frac { x }{ 2 } +\frac { y }{ -1 } =1\)  [From (1)]
\(\Rightarrow\)  -x + 2y = -2   \(\Rightarrow\) x - 2y = 2

Equation of the line in intercept form is \(\frac{x}{a}+\frac{y}{b}=1\)
⇒ \(\frac{x}{a}+\frac{y}{b}-1=0\)
p=Length of perpendicular from (0, 0) to (1)
= \(\frac { \left| 0+0-1 \right| }{ \sqrt { { \left( \frac { 1 }{ a } \right) }^{ 2 }+{ \left( \frac { 1 }{ b } \right) }^{ 2 } } } \)
= \(\frac { 1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } \)
⇒ p\(\frac { 1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } \) = 1
Squaring on both sides we get,
\({ p }^{ 2 }\left( \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \right) =1\)
⇒ \(\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } =\frac { 1 }{ { p }^{ 2 } } \)
Hence proved

Given (0,0) is the vertex of y² = 4x and P (h, k) be the mid-point on the locus and the co-ordinates of R be (a, b)
Now P is the mid-point of OR
ஃ (h, k) = \((\frac{0+a}{2},\frac{0+b}{2})=(\frac{a}{2},\frac{b}{2})\)
⇒ h = \(\frac{a}{2}\) and k = \(\frac{b}{2}\)
⇒ a = 2h and b = 2k

Here a and b are two variable which are to be eliminated.
Since (a, b) lies on y² = 4x
⇒ b² = 4a
⇒ (2k)² = 4(2h)
⇒ 4k² = 8h
⇒ k² = 2h
ஃ Locus of (h, k) in y² = 2x

Given that P (-3,1) lie on the locus of x² - 5x + ky = 0.
⇒ (-3)²-5 (-3)+k(1) = 0
⇒ 9 +15 + k = 0
⇒ k = -24
Also, it is given that (2, b) lie on the locus of x² - 5x + ky = 0.
⇒ 2² - 5 (2) + kb = 0
⇒ 4 - 10 - 24 (b) = 0
⇒ - 6 - 24b = 0
⇒ -24b = 6
⇒ b = \(\frac{-6}{24}=\frac{-1}{4}\)

Given x = a cos³ \(\theta\) , y = a sin³ \(\theta\)
\(\Rightarrow \quad \frac { x }{ a } ={ cos }^{ 3 }\theta \quad and\quad \frac { y }{ a } ={ sin }^{ 3 }\quad \theta \)
Taking power \(\left( \frac { 2 }{ 3 } \right) \)for both the equations, we get
\({ \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }={ \left( { cos }^{ 3 }\theta \right) }^{ \frac { 2 }{ 3 } }and\)
\({ \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }={ { (sin }^{ 3 }\theta })^{ \frac { 2 }{ 3 } }\)

\({ \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }={ cos }^{ 2 }\theta \quad and\quad { \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }={ sin }^{ 2 }\theta \)
We know that cos² \(\theta\) + sin² \(\theta\) = 1
\(\therefore { \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }+{ \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }=1\)
\(\Rightarrow \quad \frac { { x }^{ \frac { 2 }{ 3 } } }{ { x }^{ \frac { 2 }{ 3 } } } +\frac { { x }^{ \frac { 2 }{ 3 } } }{ { x }^{ \frac { 2 }{ 3 } } } =1\)
\(\Rightarrow \quad \frac { { x }^{ \frac { 2 }{ 3 } }+{ y }^{ \frac { 2 }{ 3 } } }{ { a }^{ \frac { 2 }{ 3 } } } =1\)
\(\Rightarrow \quad { x }^{ \frac { 2 }{ 3 } }+{ y }^{ \frac { 2 }{ 3 } }={ a }^{ \frac { 2 }{ 3 } }\)
\(\therefore \) The required point is (0, 12)

Let the vertex B is (-4, 7) and the equation of the diagonal AC is 5x -y + 7 = 0
In rhombus, the diagonals are perpendicular to each other.

\(\therefore\)  Equation of the diagonal BD is x + 5y + k = 0 which is \(\bot \) to AC
Since BD passes through the point B (-4, 7) we get
-4 + 5 (7) + k = 0
\(\Rightarrow\)  31 + k = 0
\(\Rightarrow\)  k = -31
\(\therefore\)   Equation of the all another diagonal is x + 5y - 31 = 0.

(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.

(i) Find the linear relationship between C and F.
By the given data
x₁ (100°C)  y₁(212°F)
x₂(0°C) y₂(32°F)
Using two point form, the linear relationship between C and F is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\frac { y-212 }{ 32-212 } =\frac { x-100 }{ 0-100 } \)
\(\Rightarrow \quad \frac { y-212 }{ -180 } =\frac { x-100 }{ -100 } \)
\(\Rightarrow \quad \frac { y-212 }{ 9 } =\frac { x-100 }{ 5 } =\frac { 5 }{ 9 } (y-212)=x-100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-212)+100\quad \Rightarrow \quad x=\frac { 5 }{ 9 } y-\frac { 5 }{ 9 } \times 212+100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } y-118+100\quad \Rightarrow x=\frac { 5 }{ 9 } y-18\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-32) \Rightarrow C=\frac { 5 }{ 9 } (F-32)\quad .....(1)\)
[\(\because \)  x represents Celsius and y represents Fahrenheit]
Which is the required relationship between C and F.
(ii) Find the value of C for 98.6°F and
Find C when F = 98.6° F
Substituting F = 98.6° in (1) we get,
C = \(\frac{5}{9}(98.6-32)=\frac{5}{9}(66.6)=\frac{333}{9}=37°\)
(iii) Find the value of F for 38°C.
Substituting C = 38° in (1) we get,
38=\(\frac{5}{9}(F-32)\)
⇒ \(\frac{342}{5}+32\) = F
⇒ F = 100.4°C

Let P(h, k) be any point on the locus
By the given condition,
\(h=\frac { a }{ 2 } (cosec\theta +sin\theta ),k=\frac { b }{ 2 } (cosec\quad \theta -sin\theta )\)
\(\Rightarrow \ \frac { 2h }{ a } =cosec\ \theta +sin\theta \ and\ \frac { 2k }{ b } =cosec\ \theta -sin\ \theta \)
Squaring and subtracting we get,
\(\Rightarrow \ { \left( \frac { 2h }{ a } \right) }^{ 2 }-{ \left( \frac { 2k }{ b } \right) }^{ 2 }.(cosec\ \theta +sin\ \theta { ) }^{ 2 }-(cosec\ \theta -sin\ \theta )^{ 2 }\)
\(\Rightarrow \ \frac { { 4h }^{ 2 } }{ { a }^{ 2 } } -\frac { { 4k }^{ 2 } }{ { b }^{ 2 } } ={ cosec }^{ 2 }\theta +{ sin }^{ 2 }\theta +2cosec\theta \ sin\ \theta -({ cosec }^{ 2 }\theta +{ sin }^{ 2 }\theta -2cosec\ sin\ \theta )\)
\(\Rightarrow \ \frac { { 4h }^{ 2 } }{ { a }^{ 2 } } -\frac { { 4k }^{ 2 } }{ { b }^{ 2 } } ={ cosec }^{ 2 }\theta +{ sin }^{ 2 }\theta +2\ cosec\theta \ sin\ \theta -{ cosec }^{ 2 }\theta -{ sin }^{ 2 }\theta +2\ cosec\ sin\ \theta \)
\(\Rightarrow \ \frac { { 4h }^{ 2 } }{ { a }^{ 2 } } -\frac { { 4k }^{ 2 } }{ { b }^{ 2 } } =4\ cosec\ \theta \ sin\ \theta \)
\(\Rightarrow \quad \frac { 4{ h }^{ 2 } }{ { a }^{ 2 } } -\frac { { 4k }^{ 2 } }{ { b }^{ 2 } } =4\)     \(\left[ \because \ cosec\ sin\theta =\frac { 1 }{ sin\theta } .sin\ \theta =1 \right] \)
Dividing by 4 we get,
\(\Rightarrow \frac { { h }^{ 2 } }{ { a }^{ 2 } } -\frac { { k }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow \frac { { b }^{ 2 }{ h }^{ 2 }-{ a }^{ 2 }{ k }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =1\)
\(\Rightarrow \)  b²h²-a²k² = a²b²
\(\therefore \)  Locus of (h, k) is
b²x²- a²y² = a²b² 
Hence proved

Let P(h, k) be the point on the locus and A(3, 5) B (1, -1) be the given points.
By the given condition, PA² + PB² = 20

Given PA² + PB² = 20
⇒ (h - 3)² + (k - 5)² + (h - 1)²+ (k + 1)² = 20
h² - 6h + 9 + k² - 10k + 25 + h² - 2h + 1 + k² + 2k + 1 = 20
⇒ 2h² + 2k² - 8h - 8k + 36 = 20
⇒ 2h² + 2k² - 8h - 8k + 36 - 20 = 0
Dividing by 2 we get,
h² + k² - 4h - 4k + 8 = 0
ஃ Locus of (h, k) is x² + y²- 4x - 4y + 8 = 0