We have \(\sqrt{3}x+y=1\)
\(\Rightarrow y=-\sqrt{3}x+1\)
\(\therefore m_1=-\sqrt{3}\)
Also  \(x+\sqrt{3}y=1\)
\(\Rightarrow \sqrt{3}y=-x+1\)
\(\Rightarrow y=\frac{-1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}\)
\(\therefore m_2=\frac{-1}{\sqrt{3}}\)
Let \(\theta\) be the angle between the lines. Then
\(\tan\theta=|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3}(\frac{-1}{\sqrt{3}})}|\)
\(=|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}|=|\frac{-2}{\sqrt{3}}\times \frac{1}{2}|\)
\(=|\frac{-1}{\sqrt{3}}|=\frac{1}{\sqrt{3}}\)
\(\tan\theta=\tan 30^o\) and \(\tan(180^o-30^o)\)
\(\theta=30^o\) and \(150^o\)

(i) Let (x', y') the coordinates of the given point (x, y0 in the new system of translation of axes.
\(\therefore\) x'=x-h \(\Rightarrow\) x=x'+h=x'+1 and y=y'+k=y'+1
Substituting these values of x and y in the given equation x²+xy-3y²-y+2 = 0, we get
(x'+1)²+(x'+1)(y'+1)-3(y'+1)²-(y'+1)+2=0
\(\Rightarrow\) x'²+1+2x'+x'y'+x'+y'+1-3y'²-3-6y'-y'-1+2=0
\(\Rightarrow\)  x'²+x'y'-3y'²+3x'-6y' = 0
Hence the equation of the given pair of straight lines in a new system is
x²+xy-3y²+3x-6y=0
(ii) The orgin is shifted to a point (1, 1) and Let (x',y') be the new coordinates of the given point (x, y) in the line.
\(\therefore\) x = x' + h
\(\Rightarrow\) x = x' + 1 and y = y' + 1
Substituting these values of x and y in the given equation xy - y² - x+y=0 we get,
(x'+1) (y'+1)-(y'+1)²-(x'+1)+(y'+1) = 0
\(\Rightarrow\)  x'y' +x' +y' +1-y'²-1-2y'-x'-1+y'+1 = 0
\(\Rightarrow\)  x'y'-y'² = 0
Hence the equation of the given pair of straight lines in a new system is xy-y² = 0
(iii) Here h = 1 and k = 1
Let (x', y' ) be the coordinates of the new point
\(\therefore\)  x = x'+h, y=y' +k
\(\Rightarrow\)  x = x'+1, \(\Rightarrow\) y = y' +1
Now substituting the values of x and y is given equation xy-x-y+1 = 0, we get
(x' +1) (y'+1) - (x'+1)-(y'+1)+1=0
\(\Rightarrow\)  x'y' +x' +y' +1-x'-1-y'-1+1 =0
\(\Rightarrow\) x' y' = 0
Hence the equation of the given pair of straight lines in a new system of translation of axes is xy = 0.

The given lines are 5x - 3y - 1 = 0 and 2x + 3y - 23 = 0.
Equation of any line passing through the intersection of the given lines is in the form
(5x - 3y - 1) + k (2x + 3y - 23) = 0 ...(i)
⇒ (5 + 2k) x + (-3 + 3k) y - 1 - 23k = 0
Slope of the line = \(-(5+2k)\over -3+3k\)
=m₁ (Say)
and the slope of the line 5x - 3y - 1 = 0...(ii)
\(={{-5\over -3}}=m_2\)(say)
\(∴\ m_2={5\over 3}\)
as the equation (i) is perpendicular to equation (ii)
∴ m₁ x m₂ =-1
\({-(5+2k)\over -3+3k}\times{5\over 3}=-1\)
\(⇒\ {5+2k\over -3+3k}={3\over 5}\)
⇒ 25 + 10k = -9 + 9k
⇒ k=-34
Now putting the value of k is equation (i) we get
(5x - 3y - 1) - 34 (2x + 3y - 23) = O·
⇒ 5x - 3y - 1 - 68x - 102y + 782 = 0
⇒ - 63x - 105y + 781 = 0
⇒ 63x + 105y -781 = 0

The given equation are 2x-5y+9=0 and x+2y+3=0
Equation of any line passing through the point of intersection of the given lines is in the form
(2x-5y+9)+k(x+2y+3) = 0....(i)
\(\Rightarrow\) 2x-5y+9+kx+2ky+3k=0 \(\Rightarrow\) (2+k)x+(2k-5)y+(3k=9)=0
Slope of this line (i) is \(\frac { -(2+k) }{ (2k-5) } ={ m }_{ 1 }\) (Say)
Slope of the given line 3x + 4y + 7 = 0 is =\(\frac { -3 }{ 4 } ={ m }_{ 2 }\) (Say)
If these two lines are parallel to each other
then m₁ = m₂
\(\Rightarrow -\left( \frac { 2+k }{ 2k-5 } \right) =\frac { -3 }{ 4 } \Rightarrow \) 6k-15=8+4k \(\Rightarrow\) 2k = 23
\(\therefore k=\frac { 23 }{ 2 } \)
Now substituting the value of k in Eq. (1)
(2x-5y+9)+\(\frac { 23 }{ 2 } \)(x+2y+3) = 0
\(\Rightarrow\) 4x-10y+18+23x+46y+69=0
\(\Rightarrow\) 27x + 36y + 87 = 0
which is the required equation.

\(\text { The equation of the given line is }\)
\(\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1 \)
\(\text { Or, bx } \cos \theta+a y \sin \theta-a b=0\)
\(\text { Length of the perpendicular from point }\left(\sqrt{a^{2}-b^{2}}, 0\right) \text { to line }(1) \text { is }\)
\(∴\ p_1=\left|{\sqrt{a^2-b^2}cos\theta\over a}+{0\times sin\theta\over b}-1\over \sqrt{ \left(cos\theta\over a\right)^2+\left(sin\theta\over b\right) ^2 } \right|\)
\(=\left| {\sqrt{a^2-b^2}cos\theta\over a}-1\over \sqrt{ {cos^2\theta\over a^2}+{sin^2\theta\over b^2} } \right|\)
\(\text { Length of the perpendicular from point }\left(-\sqrt{a^{2}-b^{2}}, 0\right) \text { to line }(2) \text { is }\)
\(p_2=\left|-{\sqrt{a^2-b^2}cos\theta\over a}+{0\times sin\theta\over b}-1\over \sqrt{ \left(cos\theta\over a\right)^2+\left(sin\theta\over b\right) ^2 } \right|\)
\(=\left| -{\sqrt{a^2-b^2}cos\theta\over a}-1\over \sqrt{ {cos^2\theta\over a^2}+{sin^2\theta\over b^2} } \right|\)
\(\text { On multiplying equations }(2) \text { and }(3), \text { we obtain }\)
now p₁p₂
\(=\left|{\sqrt{a^2-b^2}cos\theta\over a}+{0\times sin\theta\over b}-1\over \sqrt{ \left(cos\theta\over a\right)^2+\left(sin\theta\over b\right) ^2 } \right|\)\(\left|-{\sqrt{a^2-b^2}cos\theta\over a}+{0\times sin\theta\over b}-1\over \sqrt{ \left(cos\theta\over a\right)^2+\left(sin\theta\over b\right) ^2 } \right|\)

\({=\left|\left[{\sqrt{a^2-b^2}cos\theta\over a}-1\right]\left[{\sqrt{a^2-b^2}cos\theta\over a}+1 \right] \right|\over {cos^2\theta\over a^2}+{sin^2\theta\over b^2}}\)
= \(\left|\left[ {(a^2-b^2cos^2\theta\over a^2}-1\right]\right|\over {cos^2\theta\over a^2}+{1-cos^2\theta\over b^2}\)
\(={\left|\left[ {(a^2-b^2)cos^2\theta\over a^2}-1\right]\right|\over {a^2-(a^2-b^2)cos^2\theta\over b^2}}\)
\(=a^2 - (a^2 - b^2) cos^2θ\times{b^2\over a^2-(a^2-b^2)cos^2\theta}\)
= b²

Let the image of the point A(3, 8) in the line mirror DE be \(C(\alpha, \beta)\) . Then AC is perpendicular bisector of DE.
The coordinates of point B are \((\frac{\alpha+3}{2},\frac{\beta+8}{2})\)
Since point B lies on the line x + 3y = 7,

\(\therefore \frac{\alpha +3}{2}+\frac{3(\beta+8)}{2}=7\)
\(\Rightarrow \alpha +3+3\beta+24 =14\)
\(\therefore \alpha+3\beta+13=0\).....(i)
Since AC is perpendicular on DE
\(\therefore\)  Slope of AC x Slope of DE = -1
\(\Rightarrow \frac{\beta-8}{\alpha-3}\times \frac{-1}{3}=-1\Rightarrow\beta-8=3 \alpha -9\)
\(\Rightarrow 3\alpha - \beta - 1 = 0\)
Solving (i) and (ii) we get
\(\alpha=-1\) and \(\beta =-4\)
Thus image of point (3, 8)is (-1, -4).

\(\text { Let } y=m x+c \text { be the line through point }(-1,2) \text { . }\)
\(\text { Accordingly, } 2=m(-1)+c\)
\(\Rightarrow 2=-m+c \)
\(\Rightarrow c=m+2\)
\(y=m x+m+2 \ldots(1)\)
\(\text { The given line is }\)
\(x+y=4 \ldots(2)\)
\(\text { On solya equations ili and } 121 \text { , we obtain }\)
\(x=\frac{2-m}{m+1} \text { and } y=\frac{5 m+2}{m+1}\)
\(\therefore\left(\frac{2-m}{m+1}, \frac{5 m+2}{m+1}\right) \text { is the point of intersection of lines (1) and (2). }\)
\(\text { Since this point is at a distance of } 3 \text { units from point }(-1,2), \text { according to distance formula } \)
\(\sqrt{\left(\frac{2-m}{m+1}+1\right)^{2}+\left(\frac{5 m+2}{m+1}-2\right)^{2}}=3 \)
\(\Rightarrow\left(\frac{2-m+m+1}{m+1}\right)^{2}+\left(\frac{5 m+2-2 m-2}{m+1}\right)^{2}=3^{2} \)
\(\Rightarrow \frac{9}{(m+1)^{2}}+\frac{9 m^{2}}{(m+1)^{2}}=9 \)
\(\Rightarrow \frac{1+m^{2}}{(m+1)^{2}}=1 \)
\(\Rightarrow 1+m^{2}=m^{2}+1+2 m \)
\(\Rightarrow 2 m=0 \)
\(\Rightarrow m=0\)
\(\text { Thus, the slope of the required line must be zero i.e., the line must be parallel to the x- axis }\)

The equation of given lines are
4x + 7y - 3 = 0 and 2x - 3y + 1 = O.
Now the equation of any line through intersection of these lines is
4x + 7y - 3 + k (2x - 3y + 1) = 0....(i)
\(\Rightarrow\) (4 + 2k)x + (7 - 3k)y = 3 - k
\(\Rightarrow\)  \(\frac{(4+2k)x}{3-k}+\frac{(7-3k)y}{3-k}=1\)
\(\Rightarrow\)  \(\frac{x}{\frac{3-k}{4+2k}}+\frac{y}{\frac{3-k}{7-3k}}=1\)
It is given that \(\frac{3-k}{4+2k}=\frac{3-k}{7-3k}\)
\(\Rightarrow\)  \((3-k)[\frac{1}{4+2k}-\frac{1}{7-3k}]=0\)
\(\Rightarrow\)   3 - k = 0
or \(\frac{1}{4+2k}-\frac{1}{7-3k}=0\Rightarrow 3=k\)
or 7 - 3k - 4 - 2k = 0
\(\Rightarrow\)  k = 3 or -5k = -3
\(\Rightarrow\)  k = 3 or \(k=\frac{3}{5}\)
Putting k = 3 in (i), we have
4x + 7y - 3 + 3(2x - 3y + 1) = 0
\(\Rightarrow\) 4x + 7y - 3 + 6x - 9y + 3 = 0
\(\Rightarrow\) 10x - 2y = 0 \(\Rightarrow\) 5x - y = 0
Putting \(k=\frac{3}{5}\) in equ (i), we have
4x + 7y - 3 + 3(2x - 3y + 1) = 0
\(\Rightarrow\) 20x + 35y - 15 + 6x - 9y + 3 = 0
\(\Rightarrow\) 13x + 13y - 6 = 0.

The equation of lines are
y - x = 0.....(i)
x + y = 0....(ii)
x - k = 0....(iii)

By solving (i) and (ii), we get the coordinates of point C.
\(\therefore\) Coordinate of Care (0, 0).
By solving (ii) and (iii), we get the coordinates of point A.
\(\therefore\) Coordinate of A are (k, - k).
By solving (i) and (iii), we get the coordinates of point B.
\(\therefore\) Coordinates of B are (k, k).
\(\therefore\) Area of \(\Delta ABC=\frac { 1 }{ 2 } \left| \begin{matrix} k & -k & 1 \\ k & k & 1 \\ 0 & 0 & 1 \end{matrix} \right| \)
= \(\frac{1}{2}\)[(k² + k²) + (0 - 0) + (0 - 0)]
= \(\frac{1}{2}\times\) 2k² = k² sq.units.

(i) The given equation is 3x + 2y – 12 = 0.
It can be written as
\(\Rightarrow\)  3x + 2y = 12
\(\Rightarrow \frac{3x}{12}+\frac{2y}{12}=1\Rightarrow\frac{x}{4}+\frac{y}{6}=1\)
\(\text { This equation is of the form } \frac{x}{a}+\frac{y}{b}=1 \text { , where } a=4 \text { and } b=6 \text { . }\)
Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.
(ii) The given equation is 4x – 3y = 6.
It can be written as
\(\Rightarrow \frac{4x}{6}-\frac{3y}{6}=1\Rightarrow \frac{2x}{3}-\frac{y}{2}=1\)
\(\Rightarrow \frac{x}{\frac{3}{2}}+\frac{y}{-2}=1\)
\(\text { This equation is of the form } \frac{x}{a}+\frac{y}{b}=1 \text { , where } a=\frac{3}{2} \text { and } b=-2 \text { . }\)
Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are \(\frac {3}{2}\) and -2 respectively.
(iii) The given equation is 3y + 2 = 0.
It can be written as
\(\Rightarrow 3y=-2\Rightarrow\frac{3y}{2}=1\)
\(\Rightarrow{0x\over -2}+{3y\over -2}=1\Rightarrow {0x\over -2}+{y\over \frac{-2}{3}}=1\)
\(\text { This equation is of the form } \frac{x}{a}+\frac{y}{b}=1 \text { , where } a=0 \text { and } b=-\frac{2}{3} \text { . }\)
Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 0 and \(-\frac {2}{3}\) respectively.