Slope of line, Equation of line in different forms Questions in English

(A) The given equation is $4x - 3y = 6$.
To reduce it to the intercept form $\frac{x}{a} + \frac{y}{b} = 1$,divide both sides by $6$:
$\frac{4x}{6} - \frac{3y}{6} = \frac{6}{6}$
$\frac{2x}{3} - \frac{y}{2} = 1$
This can be written as:
$\frac{x}{(\frac{3}{2})} + \frac{y}{(-2)} = 1$
Comparing this with the standard intercept form $\frac{x}{a} + \frac{y}{b} = 1$,we get the $x$-intercept $a = \frac{3}{2}$ and the $y$-intercept $b = -2$.

(A) The given equation is $3y + 2 = 0$.
This can be rewritten as $3y = -2$,or $y = -\frac{2}{3}$.
The intercept form of a line is given by $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ is the $x$-intercept and $b$ is the $y$-intercept.
Since the equation $y = -\frac{2}{3}$ represents a horizontal line parallel to the $x$-axis,it never intersects the $x$-axis.
Thus,the $x$-intercept does not exist,and the $y$-intercept is $-\frac{2}{3}$.

(A) The given equation is $y-2=0$.
This can be written as $0 \cdot x + 1 \cdot y = 2$.
To reduce this to the normal form $x \cos \omega + y \sin \omega = p$,we divide by $\sqrt{0^2 + 1^2} = 1$.
This gives $0 \cdot x + 1 \cdot y = 2$.
Comparing this with $x \cos \omega + y \sin \omega = p$,we have $\cos \omega = 0$ and $\sin \omega = 1$,which implies $\omega = 90^{\circ}$.
The perpendicular distance $p = 2$ and the angle $\omega = 90^{\circ}$.

(A) The equation of the given line is $3x - 4y + 2 = 0$.
The slope-intercept form of a line is $y = mx + c$.
Rewriting the given equation: $4y = 3x + 2$,which gives $y = \frac{3}{4}x + \frac{1}{2}$.
Thus,the slope $(m)$ of the given line is $\frac{3}{4}$.
Since parallel lines have the same slope,the slope of the required line is also $m = \frac{3}{4}$.
Using the point-slope form $(y - y_1) = m(x - x_1)$ for the point $(-2, 3)$:
$(y - 3) = \frac{3}{4}(x - (-2))$
$4(y - 3) = 3(x + 2)$
$4y - 12 = 3x + 6$
$3x - 4y + 18 = 0$.

(A) The given equation of the line is $x-7y+5=0$.
Rewriting in slope-intercept form $y=mx+c$,we get $7y=x+5$,or $y=\frac{1}{7}x+\frac{5}{7}$.
Therefore,the slope of the given line is $m_1 = \frac{1}{7}$.
The slope of the line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{1/7} = -7$.
The equation of a line with slope $m$ and $x$-intercept $d$ is given by $y=m(x-d)$.
Substituting $m=-7$ and $d=3$,we get $y=-7(x-3)$.
$y = -7x + 21$.
Rearranging the terms,we get $7x+y=21$.

(N/A) The given line is $Ax + By + C = 0$.
Rewriting it in slope-intercept form $y = mx + c$,we get $By = -Ax - C$,which implies $y = (\frac{-A}{B})x + (\frac{-C}{B})$.
The slope of this line is $m = \frac{-A}{B}$.
Since parallel lines have equal slopes,the slope of the required line is also $m = \frac{-A}{B}$.
Using the point-slope form of a line,the equation of the line passing through $(x_{1}, y_{1})$ with slope $m$ is $y - y_{1} = m(x - x_{1})$.
Substituting $m = \frac{-A}{B}$,we get $y - y_{1} = (\frac{-A}{B})(x - x_{1})$.
Multiplying both sides by $B$,we get $B(y - y_{1}) = -A(x - x_{1})$.
Rearranging the terms,we get $A(x - x_{1}) + B(y - y_{1}) = 0$.
Thus,the equation of the line is $A(x - x_{1}) + B(y - y_{1}) = 0$.

(C) The given equation of the line is $(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$.
If a line is parallel to the $x$-axis,its slope must be $0$.
Rewriting the equation in the form $y = mx + c$:
$(4-k^2)y = (k-3)x + (k^2 - 7k + 6)$
$y = \frac{k-3}{4-k^2}x + \frac{k^2 - 7k + 6}{4-k^2}$.
For the line to be parallel to the $x$-axis,the slope $m = \frac{k-3}{4-k^2}$ must be $0$.
This implies $k-3 = 0$,so $k = 3$.
However,we must ensure the coefficient of $y$ is not zero. If $k=3$,the coefficient of $y$ is $-(4-3^2) = -(4-9) = 5 \neq 0$.
Thus,the value of $k$ is $3$.

(C) The given equation of the line is $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$.
If a line is parallel to the $y$-axis,it is a vertical line of the form $x = c$.
For the equation to represent a vertical line,the coefficient of $y$ must be zero,and the coefficient of $x$ must be non-zero.
Setting the coefficient of $y$ to zero: $-(4-k^2) = 0$ $\Rightarrow 4-k^2 = 0$ $\Rightarrow k^2 = 4$ $\Rightarrow k = \pm 2$.
Now,check the coefficient of $x$ for these values:
If $k = 2$,the coefficient of $x$ is $(2-3) = -1 \neq 0$.
If $k = -2$,the coefficient of $x$ is $(-2-3) = -5 \neq 0$.
Since both values result in a valid vertical line,the value of $k$ is $\pm 2$.

(C) The given equation of the line is $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$.
If the line passes through the origin $(0, 0)$,then the point must satisfy the equation.
Substituting $x = 0$ and $y = 0$ into the equation:
$(k-3)(0) - (4-k^2)(0) + k^2 - 7k + 6 = 0$
$k^2 - 7k + 6 = 0$
Factoring the quadratic equation:
$k^2 - 6k - k + 6 = 0$
$k(k-6) - 1(k-6) = 0$
$(k-6)(k-1) = 0$
Therefore,$k = 1$ or $k = 6$.

(A) The equation of the given line is $\sqrt{3} x + y + 2 = 0$.
This equation can be rewritten as $-\sqrt{3} x - y = 2$.
Dividing both sides by $\sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$,we get:
$-\frac{\sqrt{3}}{2} x - \frac{1}{2} y = 1$.
Comparing this with the normal form $x \cos \theta + y \sin \theta = p$,we have:
$\cos \theta = -\frac{\sqrt{3}}{2}$,$\sin \theta = -\frac{1}{2}$,and $p = 1$.
Since both $\sin \theta$ and $\cos \theta$ are negative,$\theta$ lies in the third quadrant.
$\theta = \pi + \frac{\pi}{6} = \frac{7 \pi}{6}$.
Thus,the values are $\theta = \frac{7 \pi}{6}$ and $p = 1$.

Let the intercepts cut by the given lines on the axes be $a$ and $b$.
It is given that
$a+b=1$ $(1)$
$ab=-6$ $(2)$
From $(1)$,$b = 1-a$. Substituting this into $(2)$:
$a(1-a) = -6$
$a - a^2 = -6$
$a^2 - a - 6 = 0$
$(a-3)(a+2) = 0$
So,$a=3$ or $a=-2$.
If $a=3$,then $b=1-3=-2$.
If $a=-2$,then $b=1-(-2)=3$.
The intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Case $I$: $a=3, b=-2$
$\frac{x}{3} + \frac{y}{-2} = 1 \implies -2x + 3y = -6 \implies 2x - 3y = 6$.
Case $II$: $a=-2, b=3$
$\frac{x}{-2} + \frac{y}{3} = 1 \implies -3x + 2y = 6 \implies 3x - 2y = -6$.
Thus,the required equations of the lines are $2x - 3y = 6$ and $3x - 2y = -6$.

(A) The equation of the given line is $\frac{x}{4} + \frac{y}{6} = 1$.
Multiplying by $12$,we get $3x + 2y = 12$,or $3x + 2y - 12 = 0$.
The slope of this line is $m_1 = -\frac{3}{2}$.
The slope of a line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{(-3/2)} = \frac{2}{3}$.
The given line meets the $y-$ axis where $x = 0$.
Substituting $x = 0$ in $\frac{x}{4} + \frac{y}{6} = 1$,we get $\frac{y}{6} = 1$,so $y = 6$.
The point of intersection is $(0, 6)$.
The equation of the line with slope $m_2 = \frac{2}{3}$ passing through $(0, 6)$ is given by $y - y_1 = m_2(x - x_1)$.
$y - 6 = \frac{2}{3}(x - 0)$.
$3(y - 6) = 2x$.
$3y - 18 = 2x$.
$2x - 3y + 18 = 0$.

(A) Let the equation of the line having equal intercepts on the axes be $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$ ......$(1)$.
To find the point of intersection,solve the system of equations:
$4x + 7y = 3$ ......$(2)$
$2x - 3y = -1$ ......$(3)$
Multiplying equation $(3)$ by $2$,we get $4x - 6y = -2$ ......$(4)$.
Subtracting $(4)$ from $(2)$ gives $(4x + 7y) - (4x - 6y) = 3 - (-2)$,which implies $13y = 5$,so $y = \frac{5}{13}$.
Substituting $y = \frac{5}{13}$ into $(3)$,$2x - 3(\frac{5}{13}) = -1 \Rightarrow 2x = \frac{15}{13} - 1 = \frac{2}{13}$,so $x = \frac{1}{13}$.
The point of intersection is $(\frac{1}{13}, \frac{5}{13})$.
Since the line $x + y = a$ passes through $(\frac{1}{13}, \frac{5}{13})$,we have $\frac{1}{13} + \frac{5}{13} = a$,which gives $a = \frac{6}{13}$.
Substituting $a$ into equation $(1)$,we get $x + y = \frac{6}{13}$,or $13x + 13y = 6$.

(A) Let the line $x + y = 4$ divide the line segment joining $A(-1, 1)$ and $B(5, 7)$ in the ratio $k: 1$ at point $P(x, y)$.
By the section formula,the coordinates of $P$ are:
$P = \left( \frac{k(5) + 1(-1)}{k + 1}, \frac{k(7) + 1(1)}{k + 1} \right) = \left( \frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1} \right)$
Since $P$ lies on the line $x + y = 4$,we substitute these coordinates into the equation:
$\frac{5k - 1}{k + 1} + \frac{7k + 1}{k + 1} = 4$
$\frac{5k - 1 + 7k + 1}{k + 1} = 4$
$\frac{12k}{k + 1} = 4$
$12k = 4(k + 1)$
$12k = 4k + 4$
$8k = 4$
$k = \frac{4}{8} = \frac{1}{2}$
Thus,the ratio $k: 1$ is $1: 2$.

(A) Let the line passing through $(-1, 2)$ make an angle $\theta$ with the positive $x$-axis. The coordinates of any point on this line at a distance $r = 3$ from $(-1, 2)$ are given by $(x, y) = (-1 + r \cos \theta, 2 + r \sin \theta)$.
Since $r = 3$,the point is $(-1 + 3 \cos \theta, 2 + 3 \sin \theta)$.
This point lies on the line $x + y = 4$,so:
$(-1 + 3 \cos \theta) + (2 + 3 \sin \theta) = 4$
$1 + 3(\cos \theta + \sin \theta) = 4$
$3(\cos \theta + \sin \theta) = 3$
$\cos \theta + \sin \theta = 1$
Dividing by $\sqrt{2}$:
$\frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta = \frac{1}{\sqrt{2}}$
$\cos(\theta - 45^{\circ}) = \cos 45^{\circ}$
$\theta - 45^{\circ} = \pm 45^{\circ}$
Case $1$: $\theta - 45^{\circ} = 45^{\circ} \Rightarrow \theta = 90^{\circ}$.
Case $2$: $\theta - 45^{\circ} = -45^{\circ} \Rightarrow \theta = 0^{\circ}$.
Thus,the required directions are $0^{\circ}$ or $90^{\circ}$ with the $x$-axis.

(B) Let the point $A$ be $(1, k)$. The incident ray makes an angle of $30^{\circ}$ with the normal to the line $x=1$. The normal to $x=1$ is a horizontal line. Thus,the angle of the incident ray with the horizontal is $30^{\circ}$.
From the geometry,the slope of the incident ray is $\tan(180^{\circ} - 30^{\circ}) = -\tan 30^{\circ} = -\frac{1}{\sqrt{3}}$.
The equation of the incident ray passing through $(2, 2\sqrt{3})$ is $y - 2\sqrt{3} = -\frac{1}{\sqrt{3}}(x - 2)$.
At $x=1$,$y - 2\sqrt{3} = -\frac{1}{\sqrt{3}}(1 - 2) = \frac{1}{\sqrt{3}}$,so $y = 2\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{7}{\sqrt{3}}$. Thus $A = (1, \frac{7}{\sqrt{3}})$.
However,based on the provided diagram,the angle with the vertical line $x=1$ is $30^{\circ}$,meaning the angle with the horizontal is $60^{\circ}$.
Slope of incident ray $m_1 = \tan(180^{\circ} - 60^{\circ}) = -\sqrt{3}$.
Equation: $y - 2\sqrt{3} = -\sqrt{3}(x - 2) \implies y = -\sqrt{3}x + 4\sqrt{3}$.
At $x=1$,$y = -\sqrt{3} + 4\sqrt{3} = 3\sqrt{3}$. So $A = (1, 3\sqrt{3})$.
The reflected ray makes an angle of $60^{\circ}$ with the horizontal,so its slope is $m_2 = \tan(-60^{\circ}) = -\sqrt{3}$.
The equation of line $AB$ is $y - 3\sqrt{3} = -\sqrt{3}(x - 1) \implies y = -\sqrt{3}x + 4\sqrt{3}$.
Checking the options for $(x, y)$ satisfying $y = -\sqrt{3}x + 4\sqrt{3}$:
For $x=4$,$y = -\sqrt{3}(4) + 4\sqrt{3} = 0$. This is not in the options.
Re-evaluating the diagram: The angle with the horizontal is $60^{\circ}$. The reflected ray makes an angle of $-60^{\circ}$ with the horizontal. The line $AB$ passes through $A(1, k)$ and has slope $-\sqrt{3}$.
If $A=(1, \sqrt{3})$,then $y - \sqrt{3} = -\sqrt{3}(x - 1) \implies y = -\sqrt{3}x + 2\sqrt{3}$.
For $x=3$,$y = -3\sqrt{3} + 2\sqrt{3} = -\sqrt{3}$. This matches option $B$.

(B) Given the lines $3x + 4y = 9$ and $y = mx + 1$.
Substitute $y$ from the second equation into the first:
$3x + 4(mx + 1) = 9$
$3x + 4mx + 4 = 9$
$(3 + 4m)x = 5$
$x = \frac{5}{3 + 4m}$
For $x$ to be an integer,$(3 + 4m)$ must be a divisor of $5$. The divisors of $5$ are $\{1, -1, 5, -5\}$.
Case $1$: $3 + 4m = 1$ $\Rightarrow 4m = -2$ $\Rightarrow m = -0.5$ (Not an integer)
Case $2$: $3 + 4m = -1$ $\Rightarrow 4m = -4$ $\Rightarrow m = -1$ (Integer)
Case $3$: $3 + 4m = 5$ $\Rightarrow 4m = 2$ $\Rightarrow m = 0.5$ (Not an integer)
Case $4$: $3 + 4m = -5$ $\Rightarrow 4m = -8$ $\Rightarrow m = -2$ (Integer)
The integral values of $m$ are $\{-1, -2\}$.
Thus,the number of integral values of $m$ is $2$.

(D) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the $x$ and $y$ intercepts respectively.
The arithmetic mean of the reciprocals of the intercepts is given by $\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{1}{4}$.
This implies $\frac{1}{a} + \frac{1}{b} = \frac{1}{2}$.
Since the line passes through $(x, y)$,we have $\frac{x}{a} + \frac{y}{b} = 1$.
If we test the point $(2, 2)$,we get $\frac{2}{a} + \frac{2}{b} = 2(\frac{1}{a} + \frac{1}{b}) = 2(\frac{1}{2}) = 1$.
Thus,the line always passes through the point $(2, 2)$.
Since stone $B$ is at $(2, 2)$,only stone $B$ is on the path of the man.

(D) Let $B = (0, a)$. Since $\angle OBA = 60^{\circ}$ and $\triangle OAB$ is a right-angled triangle at $O$,we have $\tan(60^{\circ}) = \frac{OA}{OB} = \frac{OA}{a}$. Thus,$OA = a\sqrt{3}$. So,$A = (a\sqrt{3}, 0)$.
Since $\triangle OAD$ is an equilateral triangle with vertices $O(0,0)$ and $A(a\sqrt{3}, 0)$,the third vertex $D$ is located at $(\frac{a\sqrt{3}}{2}, \frac{a\sqrt{3} \cdot \sqrt{3}}{2}) = (\frac{a\sqrt{3}}{2}, \frac{3a}{2})$.
The slope of line $DB$ passing through $D(\frac{a\sqrt{3}}{2}, \frac{3a}{2})$ and $B(0, a)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{3a}{2} - a}{\frac{a\sqrt{3}}{2} - 0} = \frac{\frac{a}{2}}{\frac{a\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$.

(B) Given that the lines $ax + 9y = 5$ and $4x + by = 3$ are parallel,their slopes must be equal.
For the line $ax + 9y = 5$,the slope is $m_1 = -\frac{a}{9}$.
For the line $4x + by = 3$,the slope is $m_2 = -\frac{4}{b}$.
Since the lines are parallel,$m_1 = m_2$,which implies $-\frac{a}{9} = -\frac{4}{b}$,so $ab = 36$.
We need to find the minimum value of $a + b$ where $a, b > 0$.
Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality:
$\frac{a + b}{2} \geq \sqrt{ab}$
Substituting $ab = 36$:
$\frac{a + b}{2} \geq \sqrt{36}$
$\frac{a + b}{2} \geq 6$
$a + b \geq 12$.
Thus,the least possible value of $a + b$ is $12$.

(B) Given that $AHKF$,$FKDE$,and $HBCK$ are unit squares. Let the side length of each square be $1$.
We can set up a coordinate system with $K$ at the origin $(0,0)$.
Then the coordinates are: $K(0,0)$,$H(-1,0)$,$F(0,1)$,$A(-1,1)$,$B(-1,-1)$,$D(1,0)$.
The line $AD$ passes through $A(-1,1)$ and $D(1,0)$. The slope $m_1 = \frac{0-1}{1-(-1)} = -\frac{1}{2}$. The equation is $y - 0 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{1}{2}$.
The line $BF$ passes through $B(-1,-1)$ and $F(0,1)$. The slope $m_2 = \frac{1-(-1)}{0-(-1)} = 2$. The equation is $y - 1 = 2(x - 0) \implies y = 2x + 1$.
To find $X$,set $- \frac{1}{2}x + \frac{1}{2} = 2x + 1 \implies -x + 1 = 4x + 2 \implies 5x = -1 \implies x = -\frac{1}{5}$.
Then $y = 2(-\frac{1}{5}) + 1 = \frac{3}{5}$. So $X(-\frac{1}{5}, \frac{3}{5})$.
In $\triangle ABF$,the base $AF = 1$ and height $AB = 2$. Area $= \frac{1}{2} \times 1 \times 2 = 1$.
In $\triangle AXF$,the base $AF = 1$ and the height is the perpendicular distance from $X$ to $AF$. Since $AF$ lies on the line $y=1$,the height is $|1 - \frac{3}{5}| = \frac{2}{5}$.
Area of $\triangle AXF = \frac{1}{2} \times 1 \times \frac{2}{5} = \frac{1}{5}$.
The ratio of the areas is $\frac{\text{Area}(\triangle AXF)}{\text{Area}(\triangle ABF)} = \frac{1/5}{1} = \frac{1}{5}$.

(A) The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
The normal form of the line is $x \cos \alpha + y \sin \alpha = p$,where $\alpha$ is the angle the perpendicular from the origin makes with the positive $x$-axis.
Given the perpendicular makes an angle of $\frac{\pi}{6}$ with the positive $y$-axis,it makes an angle of $\alpha = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$ with the positive $x$-axis.
Thus,the equation is $x \cos \frac{\pi}{3} + y \sin \frac{\pi}{3} = p$,which simplifies to $\frac{x}{2} + \frac{y \sqrt{3}}{2} = p$,or $\frac{x}{2p} + \frac{y}{2p/\sqrt{3}} = 1$.
Comparing this with $\frac{x}{a} + \frac{y}{b} = 1$,we get $a = 2p$ and $b = \frac{2p}{\sqrt{3}}$.
The area of $\triangle OAB$ is $\frac{1}{2} ab = \frac{98}{3} \sqrt{3}$.
Substituting $a$ and $b$: $\frac{1}{2} (2p) \left( \frac{2p}{\sqrt{3}} \right) = \frac{98}{3} \sqrt{3} \implies \frac{2p^2}{\sqrt{3}} = \frac{98\sqrt{3}}{3} \implies 2p^2 = \frac{98 \cdot 3}{3} = 98 \implies p^2 = 49$.
Now,$a^2 - b^2 = (2p)^2 - \left( \frac{2p}{\sqrt{3}} \right)^2 = 4p^2 - \frac{4p^2}{3} = \frac{8p^2}{3}$.
Substituting $p^2 = 49$: $a^2 - b^2 = \frac{8 \cdot 49}{3} = \frac{392}{3}$.

(A) The initial line passes through $A(9,0)$ and makes an angle of $30^{\circ}$ with the positive $x$-axis.
When the line is rotated clockwise by $15^{\circ}$ about $A$,the new angle it makes with the positive $x$-axis is $\theta = 30^{\circ} - 15^{\circ} = 15^{\circ}$.
The slope of the new line is $m = \tan 15^{\circ}$.
Using the value $\tan 15^{\circ} = \tan(45^{\circ}-30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$.
The equation of the line passing through $(x_1, y_1) = (9,0)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
Substituting the values: $y - 0 = (2-\sqrt{3})(x - 9)$.
Rearranging the equation: $y = (2-\sqrt{3})(x - 9)$ $\Rightarrow \frac{y}{2-\sqrt{3}} = x - 9$ $\Rightarrow x - \frac{y}{2-\sqrt{3}} = 9$.
Since $\frac{1}{2-\sqrt{3}} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3}$,this does not match the options directly.
Let's re-examine the options. The equation $y = (2-\sqrt{3})(x-9)$ can be written as $x-9 = \frac{y}{2-\sqrt{3}}$.
Alternatively,$y = (2-\sqrt{3})(x-9)$ $\Rightarrow \frac{y}{\sqrt{3}-2} = -(x-9) = 9-x$ $\Rightarrow x + \frac{y}{\sqrt{3}-2} = 9$.

(A) Let the initial line be $PQ$ with slope $\tan \alpha$. After rotating clockwise by $\frac{\alpha}{2}$,the new line $PR$ has an angle of inclination $\alpha - \frac{\alpha}{2} = \frac{\alpha}{2}$.
Given the slope of the new line $PR$ is $2-\sqrt{3}$,we have $\tan(\frac{\alpha}{2}) = 2-\sqrt{3} = \tan 15^{\circ}$.
Thus,$\frac{\alpha}{2} = 15^{\circ}$,which implies $\alpha = 30^{\circ}$.
The equation of line $PR$ passing through $P(a, 0)$ with slope $m = 2-\sqrt{3}$ is $y - 0 = (2-\sqrt{3})(x - a)$,which simplifies to $(2-\sqrt{3})x - y - a(2-\sqrt{3}) = 0$.
The perpendicular distance from the origin $(0, 0)$ to this line is given as $\frac{1}{\sqrt{2}}$.
Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$,we get $\frac{|-a(2-\sqrt{3})|}{\sqrt{(2-\sqrt{3})^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$.
Simplifying the denominator: $\sqrt{4 + 3 - 4\sqrt{3} + 1} = \sqrt{8 - 4\sqrt{3}} = \sqrt{2(4 - 2\sqrt{3})} = \sqrt{2}(\sqrt{3}-1)$.
So,$\frac{|a|(2-\sqrt{3})}{\sqrt{2}(\sqrt{3}-1)} = \frac{1}{\sqrt{2}}$.
Since $2-\sqrt{3} = \frac{(\sqrt{3}-1)^2}{2}$ is not quite right,let's use $\sqrt{8-4\sqrt{3}} = \sqrt{2}\sqrt{4-2\sqrt{3}} = \sqrt{2}(\sqrt{3}-1)$.
$|a| = \frac{\sqrt{2}(\sqrt{3}-1)}{\sqrt{2}(2-\sqrt{3})} = \frac{\sqrt{3}-1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} = 2\sqrt{3} + 3 - 2 - \sqrt{3} = \sqrt{3}+1$.
$a^2 = (\sqrt{3}+1)^2 = 3+1+2\sqrt{3} = 4+2\sqrt{3}$.
Now,calculate $3a^2 \tan^2 \alpha - 2\sqrt{3} = 3(4+2\sqrt{3}) \tan^2 30^{\circ} - 2\sqrt{3} = 3(4+2\sqrt{3}) \cdot \frac{1}{3} - 2\sqrt{3} = 4+2\sqrt{3} - 2\sqrt{3} = 4$.

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.
Since $P(-4, 1)$ divides $AB$ in the ratio $1:2$,using the section formula:
$-4 = \frac{1(0) + 2(a)}{1+2} \implies -4 = \frac{2a}{3} \implies a = -6$.
$1 = \frac{1(b) + 2(0)}{1+2} \implies 1 = \frac{b}{3} \implies b = 3$.
Thus,the intercepts are $a = -6$ and $b = 3$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
$\frac{x}{-6} + \frac{y}{3} = 1$.
Multiplying by $-6$,we get $x - 2y = -6$,or $x - 2y + 6 = 0$.

(D) Step $1$: Find the midpoint $M$ of the line segment joining $(4, -5)$ and $(-2, 9)$.
$M = (\frac{4 + (-2)}{2}, \frac{-5 + 9}{2}) = (\frac{2}{2}, \frac{4}{2}) = (1, 2)$.
Step $2$: Find the slope $m$ of the line passing through $(-3, 6)$ and $(1, 2)$.
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 6}{1 - (-3)} = \frac{-4}{4} = -1$.
Step $3$: Find the inclination $\theta$.
Since $m = \tan(\theta) = -1$,and $0 \le \theta < \pi$,we have $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

(B) Let the intercepts of the line on the $X$-axis and $Y$-axis be $b$ and $a$ respectively. The points are $A(0, a)$ and $B(b, 0)$.
Area of $\triangle OAB = \frac{1}{2} |ab| = 12 \implies |ab| = 24$.
Assuming the line has a negative slope,we can write the intercept form of the line as $\frac{x}{b} + \frac{y}{a} = 1$.
Since the line passes through $(2, 3)$,we have $\frac{2}{b} + \frac{3}{a} = 1$.
From $ab = 24$,we have $b = \frac{24}{a}$.
Substituting $b$ in the equation: $\frac{2}{24/a} + \frac{3}{a} = 1 \implies \frac{2a}{24} + \frac{3}{a} = 1 \implies \frac{a}{12} + \frac{3}{a} = 1$.
Multiplying by $12a$: $a^2 + 36 = 12a \implies a^2 - 12a + 36 = 0 \implies (a-6)^2 = 0 \implies a = 6$.
Then $b = \frac{24}{6} = 4$.
The equation of the line is $\frac{x}{4} + \frac{y}{6} = 1$.
Multiplying by $12$: $3x + 2y = 12$.

(C) $S$ is the midpoint of $QR = \left(\frac{6+7}{2}, \frac{-1+3}{2}\right) = \left(\frac{13}{2}, 1\right)$.
Slope of $PS = \frac{2-1}{2-\frac{13}{2}} = \frac{1}{-\frac{9}{2}} = -\frac{2}{9}$.
The line is parallel to $PS$,so its slope is $m = -\frac{2}{9}$.
The equation of the line passing through $(1, -1)$ with slope $m = -\frac{2}{9}$ is $y - (-1) = -\frac{2}{9}(x - 1)$.
$9(y + 1) = -2(x - 1)$ $\Rightarrow 9y + 9 = -2x + 2$ $\Rightarrow 2x + 9y + 7 = 0$.
To find the $X$-intercept,set $y = 0$: $2x + 7 = 0 \Rightarrow x = -\frac{7}{2}$.
To find the $Y$-intercept,set $x = 0$: $9y + 7 = 0 \Rightarrow y = -\frac{7}{9}$.
Thus,the intercepts are $-\frac{7}{2}$ and $-\frac{7}{9}$.

(A) Since the points $A(5, k)$,$B(-3, 1)$,and $C(-7, -2)$ are collinear,the slope of $AB$ must be equal to the slope of $BC$.
Slope of $AB = \frac{1 - k}{-3 - 5} = \frac{1 - k}{-8}$.
Slope of $BC = \frac{-2 - 1}{-7 - (-3)} = \frac{-3}{-4} = \frac{3}{4}$.
Equating the slopes: $\frac{1 - k}{-8} = \frac{3}{4}$.
Multiplying both sides by $-8$: $1 - k = \frac{3}{4} \times (-8)$.
$1 - k = -6$.
$k = 1 + 6 = 7$.

(A) To check if the points are collinear,we can check the slopes of the line segments formed by these points.
Slope of $AB = \frac{0 - (-b)}{0 - (-a)} = \frac{b}{a}$.
Slope of $BC = \frac{b - 0}{a - 0} = \frac{b}{a}$.
Since the slope of $AB$ is equal to the slope of $BC$,the points $A, B,$ and $C$ are collinear.
Now,check the slope of $CD = \frac{ab - b}{a^2 - a} = \frac{b(a - 1)}{a(a - 1)} = \frac{b}{a}$ (for $a \neq 1, a \neq 0$).
Since the slope of $BC$ is equal to the slope of $CD$,the points $B, C,$ and $D$ are also collinear.
Since all points lie on the same line with slope $\frac{b}{a}$,the points $A, B, C,$ and $D$ are collinear.

(B) The intercept form of the equation of a line is $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the line passes through $(1, 3)$,we substitute these coordinates into the equation:
$\frac{1}{a} + \frac{3}{b} = 1$.
We are also given $b = 3a$. Substituting this into the equation:
$\frac{1}{a} + \frac{3}{3a} = 1$
$\frac{1}{a} + \frac{1}{a} = 1$
$\frac{2}{a} = 1 \Rightarrow a = 2$.
Since $b = 3a$,we have $b = 3(2) = 6$.
Substituting $a = 2$ and $b = 6$ into the intercept form:
$\frac{x}{2} + \frac{y}{6} = 1$.
Multiplying by $6$ to clear the denominators:
$3x + y = 6$.

(B) The line $L$ passes through the origin and makes an angle of $30^{\circ}$ with the positive $Y$-axis in the anticlockwise direction.
This means the angle $\theta$ that the line makes with the positive $X$-axis is $90^{\circ} + 30^{\circ} = 120^{\circ}$.
The slope $m$ of the line is given by $m = \tan(\theta)$.
Therefore,$m = \tan(120^{\circ}) = \tan(180^{\circ} - 60^{\circ}) = -\tan(60^{\circ}) = -\sqrt{3}$.

(B) The slope $m$ of the line passing through $(x_1, y_1) = \left(-\frac{1}{2}, 1\right)$ and $(x_2, y_2) = (1, 3)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{1 - (-1/2)} = \frac{2}{3/2} = \frac{4}{3}$
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(1, 3)$:
$y - 3 = \frac{4}{3}(x - 1)$
$3y - 9 = 4x - 4$
$4x - 3y + 5 = 0$
To find the $x$-intercept,set $y = 0$:
$4x - 3(0) + 5 = 0$
$4x = -5$
$x = -\frac{5}{4}$
Thus,the $x$-intercept is $-\frac{5}{4}$.

(C) The slope of the line $m = \tan(90^\circ + \alpha) = -\cot \alpha = -\frac{\cos \alpha}{\sin \alpha}$.
Using the point-slope form $(y - y_1) = m(x - x_1)$:
$(y - p \sin \alpha) = -\frac{\cos \alpha}{\sin \alpha}(x - p \cos \alpha)$
$y \sin \alpha - p \sin^2 \alpha = -x \cos \alpha + p \cos^2 \alpha$
$x \cos \alpha + y \sin \alpha = p(\cos^2 \alpha + \sin^2 \alpha)$
Since $\cos^2 \alpha + \sin^2 \alpha = 1$,the equation is $x \cos \alpha + y \sin \alpha = p$.

(D) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the length of the perpendicular from the origin to the line and $\alpha$ is the angle that the perpendicular makes with the positive $X$-axis.
Given $p = 4$ and $\alpha = 30^{\circ}$.
Substituting these values into the formula:
$x \cos 30^{\circ} + y \sin 30^{\circ} = 4$
$x \left(\frac{\sqrt{3}}{2}\right) + y \left(\frac{1}{2}\right) = 4$
Multiplying both sides by $2$,we get:
$\sqrt{3} x + y = 8$

(A) The slope of the line is $m = -\frac{1}{\sqrt{2}}$.
Since the line makes an intercept of $2 \sqrt{2}$ units on the negative direction of the $y$-axis,the $y$-intercept is $c = -2 \sqrt{2}$.
Using the slope-intercept form $y = mx + c$:
$y = -\frac{1}{\sqrt{2}}x - 2 \sqrt{2}$
Multiply the entire equation by $\sqrt{2}$:
$\sqrt{2}y = -x - 4$
Rearranging the terms,we get:
$x + \sqrt{2}y + 4 = 0$.

(C) The equation of the line passing through points $(x_1, y_1) = (1, 4)$ and $(x_2, y_2) = (-5, 1)$ is given by the two-point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$.
Substituting the values: $\frac{y - 4}{1 - 4} = \frac{x - 1}{-5 - 1}$.
$\frac{y - 4}{-3} = \frac{x - 1}{-6}$.
$2(y - 4) = x - 1$ $\Rightarrow 2y - 8 = x - 1$ $\Rightarrow x - 2y + 7 = 0$ ...$(1)$.
We are given the second line: $4x + 3y - 5 = 0$ ...$(2)$.
From $(1)$,$x = 2y - 7$. Substituting this into $(2)$:
$4(2y - 7) + 3y - 5 = 0$.
$8y - 28 + 3y - 5 = 0$.
$11y - 33 = 0 \Rightarrow y = 3$.
Substituting $y = 3$ into $x = 2y - 7$: $x = 2(3) - 7 = 6 - 7 = -1$.
Thus,the point of intersection is $(-1, 3)$.

(B) Let $A \equiv (a, 0)$ and $B \equiv (0, b)$.
Let $P \equiv (5, 6)$ be the point that divides the line segment $AB$ internally in the ratio $3: 1$.
Using the section formula,the coordinates of $P$ are given by:
$5 = \frac{3 \times 0 + 1 \times a}{3 + 1}$ $\Rightarrow 5 = \frac{a}{4}$ $\Rightarrow a = 20$.
$6 = \frac{3 \times b + 1 \times 0}{3 + 1}$ $\Rightarrow 6 = \frac{3b}{4}$ $\Rightarrow 3b = 24$ $\Rightarrow b = 8$.
Thus,the $X$-intercept is $20$ and the $Y$-intercept is $8$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values,we get $\frac{x}{20} + \frac{y}{8} = 1$.
Multiplying by $40$,we get $2x + 5y = 40$.

(C) Let $a$ and $b$ be the intercepts made by the line on the axes.
We are given $a + b = 8$ and $ab = 15$.
Solving the quadratic equation $t^2 - 8t + 15 = 0$,we get $(t - 3)(t - 5) = 0$,so $(a, b) = (3, 5)$ or $(5, 3)$.
The intercept form of the equation of a line is $\frac{x}{a} + \frac{y}{b} = 1$.
Case $1$: When $a = 3$ and $b = 5$,the equation is $\frac{x}{3} + \frac{y}{5} = 1$,which simplifies to $5x + 3y - 15 = 0$.
Case $2$: When $a = 5$ and $b = 3$,the equation is $\frac{x}{5} + \frac{y}{3} = 1$,which simplifies to $3x + 5y - 15 = 0$.
Thus,the equations are $5x + 3y - 15 = 0$ and $3x + 5y - 15 = 0$.

(B) Let the line intersect the $x$-axis at $(h, 0)$ and the $y$-axis at $(0, k)$.
Since $(a, -2a)$ is the midpoint of the line segment,we have:
$\frac{h + 0}{2} = a \Rightarrow h = 2a$
$\frac{0 + k}{2} = -2a \Rightarrow k = -4a$
Using the intercept form of the line equation,$\frac{x}{h} + \frac{y}{k} = 1$:
$\frac{x}{2a} + \frac{y}{-4a} = 1$
Multiplying by $4a$,we get:
$2x - y = 4a$
Thus,the equation of the line is $2x - y = 4a$.

(A) The slope $m_1$ of the line passing through $(2, 3)$ and $(1, -2)$ is given by $m_1 = \frac{-2-3}{1-2} = \frac{-5}{-1} = 5$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -\frac{1}{5}$.
Using the point-slope form $(y - y_1) = m(x - x_1)$ for the point $(7, -4)$:
$y - (-4) = -\frac{1}{5}(x - 7)$
$5(y + 4) = -(x - 7)$
$5y + 20 = -x + 7$
$x + 5y + 13 = 0$.

(B) The given line is $x - 2y = 4$,which can be written as $2y = x - 4$ or $y = \frac{1}{2}x - 2$. The slope of this line is $m_1 = \frac{1}{2}$.
Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$. Thus,$m_2 = -2$.
The equation of the line passing through $A(6, 1)$ with slope $m_2 = -2$ is given by $y - y_1 = m_2(x - x_1)$.
Substituting the values,we get $y - 1 = -2(x - 6)$.
$y - 1 = -2x + 12$.
$2x + y = 13$.
To find the $y$-intercept,we set $x = 0$ in the equation $2x + y = 13$.
$2(0) + y = 13 \Rightarrow y = 13$.
Therefore,the $y$-intercept of the line is $13$.

(D) $1$. Find the coordinates of $M$: The line $x-y-2=0$ cuts the $X$-axis where $y=0$. Substituting $y=0$ into the equation,we get $x-0-2=0$,so $x=2$. Thus,$M = (2,0)$.
$2$. Find the slope of the original line $MN$: The equation $x-y-2=0$ can be written as $y=x-2$. Comparing this with $y=mx+c$,the slope $m_1 = 1$. This corresponds to an angle of $\theta_1 = 45^{\circ}$.
$3$. Find the new slope: The line is rotated by $45^{\circ}$ anticlockwise. The new angle $\theta_2 = 45^{\circ} + 45^{\circ} = 90^{\circ}$.
$4$. Find the equation of the new line: $A$ line passing through $M(2,0)$ with an angle of $90^{\circ}$ is a vertical line. The equation of a vertical line passing through $(2,0)$ is $x=2$.

(A) The equation of the given line is $2x - 3y + 17 = 0$,which can be written as $3y = 2x + 17$ or $y = \frac{2}{3}x + \frac{17}{3}$.
Thus,the slope of this line is $m_1 = \frac{2}{3}$.
The slope of the line passing through $(7, 17)$ and $(15, \beta)$ is $m_2 = \frac{\beta - 17}{15 - 7} = \frac{\beta - 17}{8}$.
Since the two lines are perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 \times m_2 = -1$.
$\frac{2}{3} \times \frac{\beta - 17}{8} = -1$.
$\frac{\beta - 17}{12} = -1$.
$\beta - 17 = -12$.
$\beta = 17 - 12 = 5$.

(C) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle made by the perpendicular with the positive $X$-axis.
Given $p = 5$ and $\alpha = 210^{\circ}$.
Substituting these values into the equation:
$x \cos(210^{\circ}) + y \sin(210^{\circ}) = 5$
Since $\cos(210^{\circ}) = \cos(180^{\circ} + 30^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2}$ and $\sin(210^{\circ}) = \sin(180^{\circ} + 30^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2}$,
$x(-\frac{\sqrt{3}}{2}) + y(-\frac{1}{2}) = 5$
Multiplying both sides by $-2$:
$\sqrt{3}x + y = -10$
$\sqrt{3}x + y + 10 = 0$

(A) The given line is $2x - 3y + 5 = 0$.
Any line perpendicular to this line is of the form $3x + 2y + \lambda = 0$.
Since the line makes an intercept of $3$ with the positive $Y$-axis,it passes through the point $(0, 3)$.
Substituting $x = 0$ and $y = 3$ into the equation $3x + 2y + \lambda = 0$:
$3(0) + 2(3) + \lambda = 0$
$6 + \lambda = 0$
$\lambda = -6$.
Thus,the required equation is $3x + 2y - 6 = 0$.

(B) The slope of the line $AB$ is given by $m = \frac{1-0}{3-2} = 1$.
Since $m = \tan \theta = 1$,the angle of inclination is $\theta = 45^{\circ}$.
When the line is rotated about point $A$ in the anticlockwise direction by $15^{\circ}$,the new angle of inclination becomes $\theta' = 45^{\circ} + 15^{\circ} = 60^{\circ}$.
The slope of the new line is $m' = \tan 60^{\circ} = \sqrt{3}$.
Since the line passes through $A(2,0)$,the equation of the line in point-slope form is $(y - 0) = \sqrt{3}(x - 2)$.
This simplifies to $y = \sqrt{3}x - 2\sqrt{3}$.

(A) The line bisects the angle between the coordinate axes in the second quadrant. The angle made by this line with the positive $x$-axis is $135^{\circ}$.
Therefore,the slope of the line is $m = \tan(135^{\circ}) = -1$.
The line passes through the point $(-3, 1)$.
Using the point-slope form,the equation of the line is:
$(y - y_1) = m(x - x_1)$
$(y - 1) = -1(x - (-3))$
$y - 1 = -1(x + 3)$
$y - 1 = -x - 3$
$x + y + 2 = 0$

(C) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$.
Here,the perpendicular distance from the origin is $p = 7$ and the angle $\alpha = 120^{\circ}$.
Substituting these values into the equation:
$x \cos 120^{\circ} + y \sin 120^{\circ} = 7$
Since $\cos 120^{\circ} = -\frac{1}{2}$ and $\sin 120^{\circ} = \frac{\sqrt{3}}{2}$,we get:
$x(-\frac{1}{2}) + y(\frac{\sqrt{3}}{2}) = 7$
Multiplying both sides by $2$:
$-x + \sqrt{3}y = 14$
Rearranging the terms to the standard form:
$x - \sqrt{3}y + 14 = 0$

(B) The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the length of the perpendicular from the origin and $\alpha$ is the angle the perpendicular makes with the positive $X$-axis.
Given $p = 2 \sqrt{2}$ and $\alpha = 135^{\circ}$.
The equation of the line is $x \cos(135^{\circ}) + y \sin(135^{\circ}) = 2 \sqrt{2}$.
Since $\cos(135^{\circ}) = -\frac{1}{\sqrt{2}}$ and $\sin(135^{\circ}) = \frac{1}{\sqrt{2}}$,we have:
$x \left(-\frac{1}{\sqrt{2}}\right) + y \left(\frac{1}{\sqrt{2}}\right) = 2 \sqrt{2}$.
Multiplying both sides by $\sqrt{2}$,we get:
$-x + y = 4$
or $x - y + 4 = 0$.