Given L₁ = 124.942, C₁ = 20 \(\Rightarrow\) (124.942,20)
and L₂ = 125.134, C₂ = 110 \(\Rightarrow\) (125.134,110)
Using two point form,
\(\frac{L-L_1}{L_2-L_1}=\frac{C-C_1}{C_2-C_1}\Rightarrow\frac{L-124.942}{125.134-124.942}\)
\(=\frac{C-20}{110-20}\)
\(\Rightarrow\frac{L-124.942}{0.192}=\frac{C-20}{90}\)
\(\Rightarrow L-124.942=\frac{0.192}{90}(C-20)\)
\(\Rightarrow\) L-124.942 = 0.0021 (C - 20)
\(\Rightarrow\) L-124.942 = 0.0021 (C - 20)
\(\Rightarrow\) L-124.942 = 0.0021C-.042
\(\Rightarrow\) L = 0.0021C - 0.042 + 124.942
\(\Rightarrow\) L = 0.0021C + 124.9

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

Since x-intercept is 3, A (3, 0) will be a point on the required line.
Any line parallel to 5x - 4y + 3 = 0 will be .of the form 5x - 4y + k = 0
Substituting the point (3, 0) we get
+15 - 0 + k = 0  
\(\Rightarrow \) k = -15
\(\therefore\) Required equation of the line is 5x - 4y + -15 = 0

(9 cos \(\alpha\), 9 sin \(\alpha\))
Let P (h, k) be any point on the required path
From the given information, we have
h = 9 cos \(\alpha\) and k = 9 sin \(\alpha\)
\(\Rightarrow\)  \({{h}\over{9}}\) = cos \(\alpha\) and \({{k}\over{9}}\) = sin \(\alpha\)
\({\left({{h}\over{9}} \right)}^{2}+{\left({{k}\over{9}} \right)}^{2}=cos^2\alpha+sin^2\alpha\)
\(\Rightarrow\)  \({{h^2}\over{81}}+{{k^2}\over{81}}=1\)     \([\because sin^2\ \alpha+cos^2\ \alpha=1]\)
\(\Rightarrow\)  h²+ k² = 81
\(\therefore\)  Locus of (h, k) is x² + y² = 81

Let y = 2x + c be the required line.
Given perpendicular distance from origin'to this line is \(\sqrt5\)
\(\frac{c}{\sqrt{1+4}}=\sqrt{5}\)
c  = 5
This required line is y - 2x + 5
2x - y + 5 = 0

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.

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

\((1,2) \text { lies on } 3 x^{2}+3 y^{2}-8 x-12 y+17=0\)
\(\text { Because, } 3(1)^{2}+3(2)^{2}-8(1)-12(2)+17\)
\(=3+12-8-24+17=0\)

Let the point be (x, y)
Its distance from origin is \(\sqrt{x^{2}+y^{2}}\)
Given \(x =\frac{1}{2} \sqrt{x^{2}+y^{2}} \)
\(\Rightarrow 2 x =\sqrt{x^{2}+y^{2}} \)
\(4 x^{2} =x^{2}+y^{2} \)
\(3 x^{2}-y^{2}=0 \)  is the required equation of the locus

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

Let P(h, k) be the locus of centroid of \(\Delta \) PQR.
Given P(6, 2 ) and Q(-2, 1) are the vertices of \(\Delta \)  PQR and R(\(\alpha \), \(\beta \)) be the third vertex.
Using the centroid formula,
\(\left( \frac { 6-2+\alpha }{ 3 } ,\frac { 2+1+\beta }{ 3 } \right) =(h,\ k)\)   \(\left[ \because \ centroid\ is=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \right] \)
\(\Rightarrow \left( \frac { 4+\alpha }{ 3 } ,\frac { 3+\beta }{ 3 } \right) =(h,k)\)
Equating the like co-ordinates both sides we get,
\(\Rightarrow \frac { 4+\alpha }{ 3 } =h\)
\(\Rightarrow 4+\alpha =3h\)
\(\Rightarrow \alpha =3h-4....(1)\)
\(\Rightarrow \frac { 3+\beta }{ 3 } =k\quad \)
\(\Rightarrow 3+\beta =3k \Rightarrow \beta =3k-3...(2)\)
Since R(\(\alpha\),  \(\beta\)) lies on the locus of y = x² - 3x + 4 \(\Rightarrow\) \(\beta\) = \(\alpha\)² - 3a + 4 ..(3)
Substituting 91) and (2) in (3) we get,
3k - 3 = (3h-4)²- 3(3h-4) + 4
\(\Rightarrow \)  3k-3 = 9h²+16-24h-9h+12 + 4
\(\Rightarrow \)  3k - 3 = 9h² - 33h + 32
\(\Rightarrow \) 9h²- 33h - 3k + 35 = 0
\(\therefore\) Locus of (h, k) is 9x²- 33x - 3y + 35 = 0

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 (h, k) be the mid-point on the required path.
Let the co-ordinates of A and B be A(a, 0) and B(0, b).
As the rod slides, the values of a and b change.
So a and b are two variables.
Then \(h=\frac { a+0 }{ 2 } ,and\quad k=\frac { 0+b }{ 2 } \)
\(\Rightarrow \ h=\frac { a }{ 2 } and\quad k=\frac { b }{ 2 } \)
\(\Rightarrow\)  a = 2h and b = 2k

From \(\Delta\)OAB we have
AB² = OA² + OB²
\(\Rightarrow\) a² + b² = 8²    [\(\because\) AB = 8 units]
\(\Rightarrow\)  (2h)² + (2k)² = 64
\(\Rightarrow\)  4h² + 4k² = 64
\(\Rightarrow\)    h² + k² = 16      [Dividing by 4]
\(\therefore\) Locus of (h, k) is x² + y² = 16