Perpendicular distance from origin to the line is
4x + 3y - 20 = 0 is
\(\left(\frac{-20}{\sqrt{16+9}}\right)=\frac{20}{\sqrt{25}}=\frac{20}{5}=4 \text { units }\)
Area of the square = 4 \(\times\) 4 = 16 sq. units.

\(5 x-y=0 \text { perpendicular line is } x+5 y+k=0\)
\(x \text { intercept is }-\mathrm{k} \text { and } \mathrm{y} \text { intercepts } \frac{-\mathrm{k}}{5}\)
\(\text { Area of the } \Delta=\frac{1}{2}(-\mathrm{k})\left(\frac{-\mathrm{k}}{5}\right)\)
\(5 =\frac{\mathrm{k}^{2}}{10} \text { (given) } \)
\(\mathrm{k}^{2} =50 \)
\(\mathrm{k} =\pm 5 \sqrt{2}\)
\(\text {Equation of the line is } x+5 y \pm 5 \sqrt{2}=0\)

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

\({\tan}^{-1}\left( {2 \over 11} \right)\)

(3x +y) (x - 3y) = 0 \(\Rightarrow\) 3x²- 8xy - 3y² = 0

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.

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

 (-2, 4)
Distance from the point (x₁,  y₁) to the line ax + by + c = 0 is \(\pm \frac { a{ x }_{ 1 }+b{ y }_{ 1 }+c }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
Distance from the point (-2, 4) to the line 4x + 3y + 4 = 0 is
\(\pm \left| \frac { 4(-2)+3(4)+4 }{ \sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } } \right| \)
\(=\pm \frac { (-8+12+4) }{ \sqrt { 25 } } =\frac { 8 }{ 5 } \) Unit

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.

12x²+7xy+ky²+13x-y+3=0
a=12, h=\(\frac { 7 }{ 2 } \), b=k, g=\(\frac { 13 }{ 2 } \), f =\(\frac { 1 }{ 2 } \), c=3
af²+bg²+ch²-abc-2fgh=0
12\(\left( -\frac { 1 }{ 2 } \right) ^{ 2 }+k\left( \frac { 13 }{ 2 } \right) ^{ 2 }+3\left( \frac { 7 }{ 2 } \right) ^{ 2 }\)-12(k)(3)-2\(\left( -\frac { 1 }{ 2 } \right) \left( \frac { 13 }{ 2 } \right) \left( \frac { 7 }{ 2 } \right) \)=0
\(\frac { 12 }{ 4 } +\frac { 169k }{ 4 } +\frac { 147 }{ 4 } -36k+\frac { 91 }{ 4 } \)=0
⇒ 12 + 169k+ 147 -144k+ 91 = 0
25k = -250 ⇒ k = -10
The equation is 12x²+7xy-10y²+13x-y+3=0
To find separate equations: 12x²+7xy-10y=(3x-2y)(4x+5y)
Let 12x²+7xy-10y²+13x-y+3=0 (3x-2y+l)(4x+5y+m)
Equating the coefficient of x ⇒ 4l+ 3m = 13.....(1)
Equating the coefficient of y ⇒ 5l-2m = -1 .....(2)
(1) x 2 ⇒ 8l+6m=26
(2) x 3 ⇒ 15l-6m=-3
23l=23
l=1
4+3=13
3m=9 ⇒ m=3
The separate equations are 3x - 2y + 1 = 0 and 4x + 5y + 3 = 0

Given that:
2x+y=5.....(i)
x+3y+8=0...(ii)
3x+4y=7 ....(iii)
Equation of any line passing through the point of intersection of equation (i) and (ii) is
(2x+y-5)+λ(x+3y+8)=0 ...(iv)     (λ=constant)
⇒ 2x+y-5+λx+3λy+8λ=0
⇒ (2+λ)x+(1+3λ)y-5+8λ=0
Slope of line m₁ (say) = \(\frac { -(2+\lambda ) }{ 1+3\lambda } \)   \(\left[ \because m=\frac { -a }{ b } \right] \)
Now slope of line 3x + 4y = 7 is
m₂(say) = -\(\frac { 3 }{ 4 } \)
If equation (iii) is parallel to equation (iv) then m₁ = m₂
⇒ \(\frac { -(2+\lambda ) }{ 1+3\lambda } =-\frac { 3 }{ 4 } \)
⇒ \(\frac { 2+\lambda }{ 1+3\lambda } =\frac { 3 }{ 4 } \) ⇒ 8+4λ=3+9λ
⇒ 9λ-4λ=5  ⇒ 5λ=5  ⇒  λ=1
On putting the value of A.in equation (iv) we get
(2x+y-5)+1(x+3y+8)=0
⇒ 2x+y-5+x+3y+8=0  ⇒ 3x+4y+3=0
Hence, the required equation is 3x+4y+3=0

Equation of the first line is
3x - 4y + 4 = 0
-3x + 4y = 4
Dividing throughout by \(\sqrt { { \left( -3 \right) }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 25 } =5\)
We get, \(\frac { -3 }{ 5 } x+\frac { 4 }{ 5 } y=\frac { 4 }{ 5 } \)
This is in normal form \(x\cos { \alpha } +y\sin { \alpha } =p\)
\(\Rightarrow { p }_{ 1 }=\frac { 4 }{ 5 } \)
Equation of the second line is
4x - 3y + 12 = 0
\(\Rightarrow\) -4x + 3y = 12
Dividing throughout by \(\sqrt { { \left( -4 \right) }^{ 2 }+{ 3 }^{ 2 } } =5\) we get,
\(\frac { -4 }{ 5 } x+\frac { 3 }{ 5 } y=\frac { 12 }{ 5 } \)
This is also in normal form \(x\cos { \theta } +y\sin { \theta } ={ p }_{ 2 }\)
\(\Rightarrow { p }_{ 2 }=\frac { 12 }{ 5 } \)
Clearly p₂ > p₁ (using (1) and (2))
\(\therefore\) The line 4x- 3y+ 12 = 0 is nearer to the origin.

3x²+10xy+8y²+14x+22y+15=0
a= 3, h = 5, b = 8, g = 7, f= 11, c = 15
The condition is af²+bg²+ch² abc-2fgh = 0
3(11)² +8(7)² +15(5)² -(3)(8)(15)-2(11)(7)(5)=363+392+375-360-770=0
Hence the equation represents a pair of straight lines
tanθ=\(\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\pm \frac { 2 }{ 11 } \)
tanθ=\(\frac { 2 }{ 11 } \)
⇒ θ=tan-1\(\left( \frac { 2 }{ 11 } \right) \).

If the line 2x - y = 5 makes an angle \(\theta\) with x - axis. Then, tan \(\theta\) = 2. Let P (a, a) be a point on the line 2x - y = 5. Then, 2 a - a = 5 \(\Rightarrow\) a = 5

So, the coordinates of Pare (5, 5). If the line 2x - y - 5 = 0 is rotated about point P through 45° in anti-clockwise direction, then the line in its new position makes angle 8 + 45° with x -axis. Let m be the slope of the line in its new position. Then,
 \(m'=tan(\theta+45^o)=\frac{\tan\theta+\tan45^o}{1-\tan\theta\tan45^o}=\frac{2+1}{1-2\times 1}=-3\)
Thus, the line in its new position passes through P (5, 5) and has slope m' = -3
So, its equationy -5 = m' (x - 5) or, y -5 = -3 (x - 5) or, 3x + y - 20 = 0.