Position of the lines . The y-coordinate of every point on the line parallel to x-axis is 3, therefore, equation of the line parallel tox-axis and passing through (– 2, 3) is y = 3. Similarly, equation of the line parallel to y-axis and passing through (– 2, 3) is x = – 2.

(a) The slope of the line through (3, – 2) and (– 1, 4) is
\(m=\frac{4-(-2)}{-1-3}=\frac{6}{-4}=-\frac{3}{2}\)
(b) The slope of the line through the points (3, – 2) and (7, – 2) is
\(m=\frac{-2-(-2)}{7-3}=\frac{0}{4}=0\)
(c) The slope of the line through the points (3, – 2) and (3, 4) is
\(m=\frac{4-(-2)}{3-3}=\frac{6}{0}\), which is not defined.
(d) Here inclination of the line \(\alpha\) = 60°. Therefore, slope of the line is
m = tan 60° = \(\sqrt{3}\)

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 \(150^o\)

Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.
Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.
Hence, vertex A lies on the y-axis.
On applying Pythagoras theorem to ΔAOC, we obtain
(AC)² = (OA)² + (OC)²
⇒ (2a)² = (OA)² + a²
⇒ 4a² – a² = (OA)²
⇒ (OA)² = 3a²
⇒ OA = \(\sqrt{3}\)
\(\therefore\) coordinates of points \(A=(\pm \sqrt{3} a, 0)\)
thus, the vertics of the given equilateral triangle are \((0, a),(0,-a), \text { and }(\sqrt{3} a, 0) \text { or }(0, a),(0,-a)\) and \((-\sqrt{3} a, 0)\)

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).

Let m be the slope of required line
which passes through point (3, 2). Then equation of required line is
y - 2 = m(x - 3)... (i)
The equation of given line is x - 2y = 3
\(\Rightarrow y=\frac{x}{2}-\frac{3}{2}\)....(ii)
\(\therefore\)   Slope of given line is \(\frac{1}{2}\)
It is given that line (i) and (ii) make an angle of 45^o.
\(\therefore \tan 45^o=\left| \frac { m-\frac { 1 }{ 2 } }{ 1+\frac { m }{ 2 } } \right| \)
\(\therefore 1=|\frac{2m-1}{2+m}|\)
\(\Rightarrow \frac{2m-1}{2+m}=\pm 1\)
When \(\frac{2m-1}{2+m}=1\)
\(\Rightarrow\) 2m - 1 = 2 + m \(\Rightarrow\) m = 3
Then equation of required line is
y - 2 = 3(x - 3)
\(\Rightarrow\) y - 2 = 3x - 9 \(\Rightarrow\) 3x - y - 7 = 0
When \(\frac{2m-1}{2+m}=-1\Rightarrow\) 2m - 1 = -2 -m
\(\Rightarrow\)  3m = -1 \(\Rightarrow\) m = \(\frac{-1}{3}\)
Then the equation of required line is
\(y-2=\frac{-1}{3}(x-3)\)
\(\Rightarrow\)  3y - 6 = -x + 3 \(\Rightarrow\) x + 3y - 9 = 0.

(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.

(i) The given equation is x + 7y = 0.
It can be written as
\(\Rightarrow 7y=-x\Rightarrow y=\frac{-1}{7}x\)
\(\Rightarrow y = \frac{-1}{7}x+0\)
\(\text { This equation is of the form } y=m x+c, \text { where } m=-\frac{1}{7} \text { and } c=0 \text { . }\)
Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are \(-\frac{1}{7}\) and 0 respectively.
(ii) The given equation is 6x + 3y – 5 = 0
It can be written as
\(y=\frac{1}{3}(-6 x+5) \)
\(y=-2 x+\frac{5}{3}\)
\(\text { This equation is of the form } y=m x+c, \text { where } m=-2 \text { and } c=\frac{5}{3} \text { . }\)
Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are-2  and \(\frac{5}{3}\) respectively.
(iii) The given equation is y = 0.
It can be written as
y = 0.x + 0
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

The equation of the given line is
\(3 x-4 y+2=0 \)
\(\text { or } y=\frac{3 x}{4}+\frac{2}{4} \)
\(\text { or } y=\frac{3}{4} x+\frac{1}{2},\)
\(\text { which is of the form } y=m x+c\)
\(\text { Slope of the given line }=\frac{3}{4}\)
\(\text { It is known that parallel lines have the same slope. }\)
\(\text { Slope of the other line }=m=\frac{3}{4}\)
\(\text { Now, the equation of the line that has a slope of } \frac{3}{4} \text { and passes through the point }(-2,3) \text { is }\)
\((y-3)=\frac{3}{4}\{x-(-2)\} \)
\(4 y-12=3 x+6 \)
\(\text { i.e., } 3 x-4 y+18=0\)

Here \(m=\tan\frac{2\pi}{3}=\tan 120^o\)
\(=\tan(90^o+30^o)=-\cot 30^o=-\sqrt{3}\)
Now, the Equation of the line passing through point (0, 2) having slope \(-\sqrt{3}\) is
\(y-2=-\sqrt{3}(x-0) \quad\left[\because y-y_{1}=m\left(x-x_{1}\right)\right] \)
\(\Rightarrow y-2=-\sqrt{3} x \Rightarrow \sqrt{3} x+y-2=0\)
Since, the line BD, parallel to AC, intersect below Y-axis. It means that the y-coordinate of intersection point is negative and x-coordinate is 0, Thus, intersection point is (0, - 2).


Hence, equation of line through the point B (0, - 2) and having slope  \(-\sqrt{3}\) is
\(y+2=-\sqrt{3}(x-0)\left[\because y-y_{1}=m\left(x-x_{1}\right)\right] \)
\(\Rightarrow \sqrt{3} x+y+2=0\)

Here p = 5 and \(\alpha\) = 30°
Putting these values in x cos \(\alpha\) + y sin \(\alpha\) =p, we have
x cos 30° + Y sin 30° = 5
\(\Rightarrow \frac{\sqrt{3}}{2}x+\frac{1}{2}y=5\Rightarrow\sqrt{3}x+y=10\)

Let P(x, 0) be any point on the x-axis which is equidistant from Q(7, 6) and R(3,4).
Then  \(PQ=\sqrt{(x-7)^2+(0-6)^2}\)
\(=\sqrt{x^2-14x+49+36}\)
\(=\sqrt{x^2-14x+85}\)
\(PR = \sqrt{(x-3)^2+(0-4)^2}\)
\(=\sqrt{x^2-6x+9+16}\)
\(=\sqrt{x^2-6x+25}\)
Since PQ = PR,
\(\therefore \sqrt{x^2-14x+85} = \sqrt{x^2-6x+25}\)
Squaring both sides, we have
x² - 14x + 85 = x² - 6x + 25
\(\Rightarrow\)  -14x + 6x = 25 - 85  \(\Rightarrow\)  -8x = -60
\(\Rightarrow\)   \(x=\frac{15}{2}\)
Thus coordinates of point on the x-axis is \((\frac{15}{2},0)\)

Here P(x₁, Y₁) and Q(x₂, Y₂) are two points.
Then  \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
(i) PQ is parallel to the y-axis then x₂ - x₁ = 0
Then \(PQ=\sqrt{(y_2-y_1)^2}=|y_2-y_1|\)
(ii) PQ is parallel to the x-axis then y₂ - y₁ = 0
Then \(PQ=\sqrt{(x_2-x_1)^2}=|x_2-x_1|\)

Let the coordinates of the point A be ( x, 0 ) .From the figure, the slope of the reflected ray is given by
tan \(\theta \) = \(\frac{3}{5 - x} \) .... ( i )
 

Again , the slope of the incident ray is given by 
tan ( \(\pi\) - \(\theta\) ) = \(\frac{-2}{ x-1 }\)   [ \(\because\) m = \(\frac{y_2-y_1}{x_2-x_1}\) ]
\(\Rightarrow\) - tan \(\theta\)  = \( \frac{-2}{ x-1 }\) [ \(\because\) tan ( \(\pi\) - \(\theta\) = - tan \(\theta\)  )  ]
\(\Rightarrow\)  tan \(\theta\) =  \( \frac{2}{ x - 1}\)
From Eqs.(i) and (ii), we get
\(\frac{3}{ 5 - x }\) = \(\frac{2}{ x - 1}\)
\(\Rightarrow\) 3x  - 3 = 10 -2x
\(\Rightarrow\)  x = \( \frac{13}{5}\)
Therefore , the required coordinates of the point A are ( \(\frac{13}{5}\),0 ).