Given equation is \(\frac{x}{a}+\frac{y}{b}=1\)
\(\Rightarrow \frac{bx+ay}{ab}=1\)
\(\Rightarrow\)bx + ay - ab = 0....(1)
Given that p= Length of the perpendicular from the origin to the line (1)
\(\Rightarrow p=\left|\frac{b(0)+a(0)-ab}{\sqrt{b^2+a^2}}\right|\)
\(\Rightarrow p^2=\left(\frac{ab}{\sqrt{a^2+b^2}}\right)^2\Rightarrow P^2=\frac{a^2b^2}{a^2+b^2}\)
\(\Rightarrow\frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2}\)
\(\Rightarrow\frac{1}{p^2}=\frac{a^2}{a^2b^2}+\frac{b^2}{a^2b^2}\)
\(\Rightarrow\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2}\)
Hence proved.

Given (x₁, y₁) = (2, 3) and \(\theta=30^o\)
Equation of the line in parametric form is
\(\frac{x-x_1}{\cos\theta}=\frac{y-y_1}{\sin\theta}\Rightarrow\frac{x-2}{\cos30^\circ}=\frac{y-3}{\sin30^\circ}\)
\(\frac{x-2}{\frac{\sqrt3}2}=\frac{y-3}{\frac{1}{2}}\Rightarrow x-2=\sqrt3(y-3)\)
\(\Rightarrow x-\sqrt3y=2-3\sqrt3\)
Points on the line at a distance 4 from P(2, 3) are \((x_1\pm r\cos\theta,y_1\pm r\sin\theta)\)
\(\Rightarrow\left(2\pm4\cos30^o,3\pm4.\frac{1}{2}\right)\)
\(\Rightarrow(2\pm3\sqrt3,3\pm2)\)
\(\Rightarrow(2\pm2\sqrt3,5)\) and \((2-2\sqrt3,1)\)

Given equation is \(\sqrt{3x}+y+2=0\)
\(\Rightarrow-\sqrt{3x}-y=2\)
Dividing by 2 we get,
\(\left(-\frac{\sqrt3}{2}\right)x+\left(-\frac{1}{2}\right)y=1\)....(1)
Comparing this with \(x\cos\alpha+y\sin\alpha=p\)  we get
\(\cos\alpha=-\frac{\sqrt3}{2}\sin\alpha=\frac{-1}{2}\) and p = 1
\(\Rightarrow \cos \alpha=-\cos\frac{\pi}{6}\)
\(\Rightarrow \cos\alpha=\cos\left(\pi+\frac{\pi}{6}\right)\) [the angle is in III quadrant, both \(\cos\alpha\) and \(\sin\alpha\) are negative]
\(\Rightarrow\cos\alpha=\cos\left(\frac{7\pi}{6}\right)\)
\(\Rightarrow\alpha=\frac{7\pi}{6}\) and p = 1.

It is given that 
\(\angle XAB=120^\circ\)
\(\angle PAO=180^\circ-120^\circ=60^\circ\)
In \(\triangle OAP, 90^\circ+60^\circ+\angle AOP=80^\circ\)
\(\Rightarrow\angle AOP=30^\circ\)
\(\therefore p=4\) and \(\alpha=30^o\)
Equation of the line is normal form is
\(x\cos\alpha+y\sin\alpha=p\)
\(\Rightarrow x\cos\ 30^o+y\sin\ 30^o=4\)
\(\Rightarrow x.\frac{\sqrt{3}}{2}+y.\frac12=4\)
\(\Rightarrow\sqrt3x+y=8\)

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.

Since the line is perpendicular to y-axis, it is parallel to x-axis
\(\Rightarrow\) Slope = m = 0.
Now, equation of the line is y- y₁ = m(x -x₁)
\(\Rightarrow\) y + 3 = 0(x + 4)
\(\Rightarrow\) y = 3 = 0

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

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

(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