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

(D) The given equation is $3x + 3y + 7 = 0$.
To reduce this to the normal form $x \cos \alpha + y \sin \alpha = p$,we divide the entire equation by $\sqrt{A^2 + B^2}$,where $A=3$ and $B=3$.
$\sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$.
Dividing by $-3\sqrt{2}$ (to make the constant term positive),we get:
$\frac{3}{-3\sqrt{2}}x + \frac{3}{-3\sqrt{2}}y = \frac{-7}{-3\sqrt{2}}$
$-\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = \frac{7}{3\sqrt{2}}$.
Comparing this with $x \cos \alpha + y \sin \alpha = p$,we find $p = \frac{7}{3\sqrt{2}}$.

(A) The line $x = c$ represents a vertical line parallel to the $y$-axis.
Any line perpendicular to a vertical line must be a horizontal line.
$A$ horizontal line is represented by the equation $y = d$,where $d$ is a constant.
Therefore,the correct option is $A$.

(C) The given line $CD$ is $3x + 4y + 6 = 0$.
Rewriting in slope-intercept form $y = mx + c$,we get $4y = -3x - 6$,which implies $y = -\frac{3}{4}x - \frac{6}{4}$.
The slope of line $CD$ is $m_1 = -\frac{3}{4}$.
Since line $AB$ is perpendicular to line $CD$,the product of their slopes must be $-1$.
Let the slope of line $AB$ be $m_2$. Then $m_1 \times m_2 = -1$,so $(-\frac{3}{4}) \times m_2 = -1$,which gives $m_2 = \frac{4}{3}$.
Line $AB$ passes through the origin $(0, 0)$. Using the point-slope form $(y - y_1) = m(x - x_1)$,we get $(y - 0) = \frac{4}{3}(x - 0)$.
This simplifies to $3y = 4x$,or $4x - 3y = 0$.

(B) The slope of the given line $y = x$ is $m_1 = 1$.
Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Therefore,$m_2 = -1$.
Using the point-slope form of the equation of a line,$y - y_1 = m(x - x_1)$,where $(x_1, y_1) = (3, 2)$ and $m = -1$:
$y - 2 = -1(x - 3)$
$y - 2 = -x + 3$
$x + y = 5$.

(D) The equation of a line in slope-intercept form is given by $y = mx + c$.
Given that the line is inclined at $30^\circ$ to the positive direction of the $x$-axis,the slope $m = \tan(30^\circ) = \frac{1}{\sqrt{3}}$.
The line cuts off an intercept of $2$ from the negative direction of the $y$-axis,so $c = -2$.
Substituting these values into the equation,we get $y = \frac{1}{\sqrt{3}}x - 2$.
Multiplying both sides by $\sqrt{3}$,we get $\sqrt{3}y = x - 2\sqrt{3}$.
Rearranging the terms,we obtain $\sqrt{3}y - x + 2\sqrt{3} = 0$.

(A) The given line is $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$.
Any line perpendicular to $A \sin \theta + B \cos \theta = \frac{C}{r}$ is of the form $A \sin(\theta + \pi/2) + B \cos(\theta + \pi/2) = \frac{k}{r}$,which simplifies to $A \cos \theta - B \sin \theta = \frac{k}{r}$.
Substituting the values,we get $\sqrt{3} \cos \theta - 2 \sin \theta = \frac{k}{r}$.
Since the line passes through $(-1, \pi/2)$,we substitute $r = -1$ and $\theta = \pi/2$:
$\sqrt{3} \cos(\pi/2) - 2 \sin(\pi/2) = \frac{k}{-1}$.
$0 - 2(1) = -k \Rightarrow k = 2$.
Thus,the equation of the line is $\sqrt{3} \cos \theta - 2 \sin \theta = \frac{2}{r}$,which can be written as $2 = \sqrt{3} r \cos \theta - 2 r \sin \theta$.

(D) Given the line equation $\frac{x}{a} + \frac{y}{b} = 1$.
Since the line passes through $(2, -3)$,we have $\frac{2}{a} - \frac{3}{b} = 1$ --- $(1)$.
Since the line passes through $(4, -5)$,we have $\frac{4}{a} - \frac{5}{b} = 1$ --- $(2)$.
Let $X = \frac{1}{a}$ and $Y = \frac{1}{b}$.
The equations become $2X - 3Y = 1$ and $4X - 5Y = 1$.
Multiplying the first equation by $2$,we get $4X - 6Y = 2$.
Subtracting this from the second equation: $(4X - 5Y) - (4X - 6Y) = 1 - 2$,which gives $Y = -1$.
Substituting $Y = -1$ into $2X - 3Y = 1$: $2X - 3(-1) = 1 \implies 2X + 3 = 1 \implies 2X = -2 \implies X = -1$.
Thus,$\frac{1}{a} = -1 \implies a = -1$ and $\frac{1}{b} = -1 \implies b = -1$.
Therefore,$(a, b) = (-1, -1)$.

(B) The slope of the line is $m = \frac{3}{4} = \tan \theta$. Thus,$\cos \theta = \frac{4}{5}$ and $\sin \theta = \frac{3}{5}$.
The coordinates of points at a distance $r = 5$ from $A(x_1, y_1) = (3, 2)$ are given by $(x_1 \pm r \cos \theta, y_1 \pm r \sin \theta)$.
For the positive direction: $(3 + 5 \times \frac{4}{5}, 2 + 5 \times \frac{3}{5}) = (3 + 4, 2 + 3) = (7, 5)$.
For the negative direction: $(3 - 5 \times \frac{4}{5}, 2 - 5 \times \frac{3}{5}) = (3 - 4, 2 - 3) = (-1, -1)$.
Thus,the required points are $(7, 5)$ and $(-1, -1)$.

(C) Let the diagonal be $BD$ along the line $8x - 15y = 0$. The slope of $BD$ is $m_1 = \frac{8}{15}$.
Let the vertex be $D(1, 2)$. The sides passing through $D$ are $AD$ and $CD$.
In a square,the diagonal makes an angle of $45^\circ$ with the sides.
Let the slope of the sides be $m$. Then,$\tan(45^\circ) = \left| \frac{m - \frac{8}{15}}{1 + m \cdot \frac{8}{15}} \right|$.
$1 = \left| \frac{15m - 8}{15 + 8m} \right|$.
This gives two cases:
Case $1$: $15m - 8 = 15 + 8m$ $\Rightarrow 7m = 23$ $\Rightarrow m = \frac{23}{7}$.
Case $2$: $15m - 8 = -(15 + 8m)$ $\Rightarrow 15m - 8 = -15 - 8m$ $\Rightarrow 23m = -7$ $\Rightarrow m = -\frac{7}{23}$.
The equations of the sides passing through $(1, 2)$ are:
$y - 2 = \frac{23}{7}(x - 1)$ $\Rightarrow 7y - 14 = 23x - 23$ $\Rightarrow 23x - 7y - 9 = 0$.
$y - 2 = -\frac{7}{23}(x - 1)$ $\Rightarrow 23y - 46 = -7x + 7$ $\Rightarrow 7x + 23y - 53 = 0$.

(A) The diagonal passes through the points $(1, 2)$ and $(3, 8)$.
The slope $m$ of the line passing through $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 8)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3$.
The equation of the line in point-slope form is $y - y_1 = m(x - x_1)$.
Substituting the values,we get:
$y - 2 = 3(x - 1)$
$y - 2 = 3x - 3$
$3x - y - 1 = 0$.

(D) Let the vertices of the triangle be $C(2a, 0)$,$B(0, a)$,and $A(2a, k)$.
Since the triangle is isosceles with base $BC$,we have $AB = AC$.
The length $AC = |k - 0| = |k|$.
The length $AB = \sqrt{(2a - 0)^2 + (k - a)^2} = \sqrt{4a^2 + (k - a)^2}$.
Equating $AB^2 = AC^2$,we get $4a^2 + (k - a)^2 = k^2$.
$4a^2 + k^2 - 2ak + a^2 = k^2$.
$5a^2 = 2ak$.
Since $a \neq 0$,we have $k = \frac{5a}{2}$.
Thus,the vertex $A$ is $(2a, \frac{5a}{2})$.
The other side is the line passing through $B(0, a)$ and $A(2a, \frac{5a}{2})$.
The slope $m = \frac{\frac{5a}{2} - a}{2a - 0} = \frac{\frac{3a}{2}}{2a} = \frac{3}{4}$.
The equation of the line is $y - a = \frac{3}{4}(x - 0)$.
$4y - 4a = 3x$,which simplifies to $3x - 4y + 4a = 0$.

(B) Let $p$ be the length of the perpendicular from the origin to the line. The equation of the line in normal form is $x \cos 30^\circ + y \sin 30^\circ = p$,which simplifies to $\frac{\sqrt{3}}{2}x + \frac{1}{2}y = p$ or $\sqrt{3}x + y = 2p$.
This line intersects the coordinate axes at points $A\left(\frac{2p}{\sqrt{3}}, 0\right)$ and $B(0, 2p)$.
The area of the triangle $\Delta OAB$ formed with the axes is given by $\text{Area} = \frac{1}{2} \times |\text{base}| \times |\text{height}| = \frac{1}{2} \times \left|\frac{2p}{\sqrt{3}}\right| \times |2p| = \frac{2p^2}{\sqrt{3}}$.
Given that the area is $\frac{50}{\sqrt{3}}$,we have $\frac{2p^2}{\sqrt{3}} = \frac{50}{\sqrt{3}}$,which implies $p^2 = 25$,so $p = \pm 5$.
Substituting $p = \pm 5$ into the equation $\sqrt{3}x + y = 2p$,we get $\sqrt{3}x + y = \pm 10$,or $\sqrt{3}x + y \mp 10 = 0$. Thus,the lines are $\sqrt{3}x + y \pm 10 = 0$.

(A) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the intercepts on the axes.
The area of the triangle formed with the axes is given by $\frac{1}{2} |ab| = 6$,which implies $|ab| = 12$ $(i)$.
The length of the hypotenuse is given by $\sqrt{a^2 + b^2} = 5$,which implies $a^2 + b^2 = 25$ $(ii)$.
From $(i)$,$b = \frac{12}{a}$. Substituting this into $(ii)$:
$a^2 + (\frac{12}{a})^2 = 25$
$a^4 - 25a^2 + 144 = 0$
$(a^2 - 16)(a^2 - 9) = 0$
So,$a^2 = 16$ or $a^2 = 9$,which gives $a = \pm 4$ or $a = \pm 3$.
If $a = \pm 4$,then $b = \pm 3$. If $a = \pm 3$,then $b = \pm 4$.
The equations of the lines are $\frac{x}{\pm 4} + \frac{y}{\pm 3} = 1$ or $\frac{x}{\pm 3} + \frac{y}{\pm 4} = 1$,which can be written as $\frac{x}{4} + \frac{y}{3} = \pm 1$ (and other variations).
Checking the options,$\frac{x}{4} + \frac{y}{3} = \pm 1$ satisfies the given conditions.

(B) Let the slope of line $AB$ be $m_1 = 2$ and the slope of line $BC$ be $m_{BC} = \frac{1 - 2}{2 - 1} = -1$.
Let the slope of line $AC$ be $m_2$.
The angle $\theta$ between $AB$ and $BC$ is given by $\tan \theta = \left| \frac{m_1 - m_{BC}}{1 + m_1 m_{BC}} \right| = \left| \frac{2 - (-1)}{1 + 2(-1)} \right| = \left| \frac{3}{-1} \right| = 3$.
Since $\triangle ABC$ is isosceles with $AB = AC$,the angle between $AC$ and $BC$ is also $\theta$.
Thus,$\tan \theta = \left| \frac{m_{BC} - m_2}{1 + m_{BC} m_2} \right| = 3$.
$\left| \frac{-1 - m_2}{1 - m_2} \right| = 3$.
Case $1$: $\frac{-(1 + m_2)}{1 - m_2} = 3 \implies -1 - m_2 = 3 - 3m_2 \implies 2m_2 = 4 \implies m_2 = 2$ (This corresponds to line $AB$).
Case $2$: $\frac{-(1 + m_2)}{1 - m_2} = -3 \implies 1 + m_2 = 3 - 3m_2 \implies 4m_2 = 2 \implies m_2 = \frac{1}{2}$.
Since $AC$ passes through $C(2, 1)$ with slope $m_2 = \frac{1}{2}$,its equation is $y - 1 = \frac{1}{2}(x - 2) \implies y - 1 = \frac{x}{2} - 1 \implies y = \frac{x}{2}$.

(A) The vertices of the square are formed by the intersection of the given lines:
$A(0, 0)$ (intersection of $x=0$ and $y=0$),
$B(0, 1)$ (intersection of $x=0$ and $y=1$),
$C(1, 1)$ (intersection of $x=1$ and $y=1$),
$D(1, 0)$ (intersection of $x=1$ and $y=0$).
The diagonals are the lines joining opposite vertices $A(0, 0)$ to $C(1, 1)$ and $B(0, 1)$ to $D(1, 0)$.
$1$. For diagonal $AC$: The slope $m = \frac{1-0}{1-0} = 1$. The equation is $y - 0 = 1(x - 0) \Rightarrow y = x$.
$2$. For diagonal $BD$: The slope $m = \frac{0-1}{1-0} = -1$. The equation is $y - 1 = -1(x - 0)$ $\Rightarrow y - 1 = -x$ $\Rightarrow x + y = 1$.

(A) The equation of the straight line is $ax + by + c = 0$.
Since the line always passes through the point $(1, -2),$ we substitute $x = 1$ and $y = -2$ into the equation:
$a(1) + b(-2) + c = 0$
$a - 2b + c = 0$
Rearranging the terms,we get:
$a + c = 2b$
This condition $a + c = 2b$ implies that $a, b,$ and $c$ are in Arithmetic Progression $(A.P.)$.

(A) Given $u = a_1x + b_1y + c_1 = 0$ and $v = a_2x + b_2y + c_2 = 0$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \lambda$ (where $\lambda$ is a constant),we have $a_1 = \lambda a_2$,$b_1 = \lambda b_2$,and $c_1 = \lambda c_2$.
Substituting these into the equation $u + kv = 0$:
$(a_1x + b_1y + c_1) + k(a_2x + b_2y + c_2) = 0$
$(\lambda a_2x + \lambda b_2y + \lambda c_2) + k(a_2x + b_2y + c_2) = 0$
$(\lambda + k)(a_2x + b_2y + c_2) = 0$
Since $a_2x + b_2y + c_2 = 0$ represents the line $v$,the equation $u + kv = 0$ simplifies to a multiple of the original line $v$ (or $u$),which is the same straight line.

(D) The equation of the line $ax + by + 8 = 0$ can be written in intercept form as $\frac{x}{-8/a} + \frac{y}{-8/b} = 1$.
The equation of the line $2x - 3y + 6 = 0$ can be written as $2x - 3y = -6$,which simplifies to $\frac{x}{-3} + \frac{y}{2} = 1$.
According to the problem,the intercepts of the first line are equal in length but opposite in sign to the intercepts of the second line.
Therefore,$-\frac{8}{a} = -(-3) = 3 \Rightarrow a = -\frac{8}{3}$.
And $-\frac{8}{b} = -(2) = -2 \Rightarrow b = 4$.

(C) The equation of a line in slope-intercept form is given by $y = mx + c$.
Given that the line makes an angle of $\theta = 135^\circ$ with the $x$-axis,the slope $m$ is:
$m = \tan(135^\circ) = -1$.
The line cuts the $y$-axis at a distance of $-5$ from the origin,so the $y$-intercept $c = -5$.
Substituting these values into the slope-intercept form:
$y = (-1)x + (-5)$
$y = -x - 5$
Rearranging the terms to the general form:
$x + y + 5 = 0$.
Thus,the correct option is $C$.

(C) The intercept form of a line is $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the line makes equal intercepts on the axes,we have $a = b$.
Thus,the equation becomes $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$ .....$(i)$.
Since the line passes through the point $(2, 4)$,we substitute these coordinates into equation $(i)$:
$2 + 4 = a \implies a = 6$.
Substituting $a = 6$ back into equation $(i)$,we get $x + y = 6$,or $x + y - 6 = 0$.

(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively.
Given that $a + b = -1$,so $b = -1 - a$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting $b$,we get $\frac{x}{a} + \frac{y}{-(1+a)} = 1$,which is $\frac{x}{a} - \frac{y}{1+a} = 1$.
Since the line passes through $(4, 3)$,we have $\frac{4}{a} - \frac{3}{1+a} = 1$.
Multiplying by $a(1+a)$,we get $4(1+a) - 3a = a(1+a)$.
$4 + 4a - 3a = a + a^2$.
$4 + a = a + a^2$,which simplifies to $a^2 = 4$,so $a = 2$ or $a = -2$.
If $a = 2$,then $b = -1 - 2 = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1$,or $\frac{x}{2} - \frac{y}{3} = 1$.
If $a = -2$,then $b = -1 - (-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$.

(B) Let $ABCD$ be a rectangle with opposite vertices $A(1, 3)$ and $C(5, 1)$.
Since the diagonals of a rectangle bisect each other,the intersection point of the diagonals is the midpoint of $AC$.
Midpoint of $AC = \left(\frac{1+5}{2}, \frac{3+1}{2}\right) = (3, 2)$.
The other two vertices $B$ and $D$ lie on the line $y = 2x + c$. Since the diagonal $BD$ passes through the intersection point of the diagonals,the point $(3, 2)$ must lie on the line $y = 2x + c$.
Substituting $x = 3$ and $y = 2$ into the equation:
$2 = 2(3) + c$
$2 = 6 + c$
$c = 2 - 6 = -4$.
Thus,the value of $c$ is $-4$.

(D) First,find the slope $m$ of the line passing through $(-1, 3)$ and $(4, -2)$:
$m = \frac{-2 - 3}{4 - (-1)} = \frac{-5}{5} = -1$
Using the point-slope form $(y - y_1) = m(x - x_1)$ with point $(-1, 3)$:
$y - 3 = -1(x - (-1))$
$y - 3 = -1(x + 1)$
$y - 3 = -x - 1$
$x + y = 2$
If the line passes through the point $(p, q)$,then the coordinates must satisfy the equation:
$p + q = 2$

(A) The equation of the line passing through the points $(a, 0)$ and $(0, b)$ is given by the intercept form: $\frac{x}{a} + \frac{y}{b} = 1$.
To check which point lies on the line,we substitute the coordinates $(x, y)$ of each option into the equation.
For option $A$: $(3a, -2b)$,we have $\frac{3a}{a} + \frac{-2b}{b} = 3 - 2 = 1$.
Since the equation is satisfied,the point $(3a, -2b)$ lies on the line.

(C) Let the intercepts of the line on the $x$-axis and $y$-axis be $2a$ and $a$ respectively.
The equation of the line in intercept form is $\frac{x}{2a} + \frac{y}{a} = 1$,which simplifies to $x + 2y = 2a$.....$(i)$
The line bisects the join of $(3, -4)$ and $(5, 2)$. The midpoint of these points is $(\frac{3+5}{2}, \frac{-4+2}{2}) = (4, -1)$.
Since the line passes through $(4, -1)$,we substitute these coordinates into equation $(i)$:
$4 + 2(-1) = 2a$
$4 - 2 = 2a$
$2 = 2a \Rightarrow a = 1$
Substituting $a = 1$ into equation $(i)$,we get $x + 2y = 2(1)$,which is $x + 2y = 2$.

(C) The slope $m$ of the line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \tan \theta = \frac{y_2 - y_1}{x_2 - x_1}$.
Substituting the given coordinates $(1, 0)$ and $(2, \sqrt{3})$:
$m = \frac{\sqrt{3} - 0}{2 - 1} = \frac{\sqrt{3}}{1} = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = \tan^{-1}(\sqrt{3}) = 60^o$.

(D) The general equation of a line is $lx + my + n = 0$.
For a line to be parallel to the $x$-axis,its slope must be $0$.
The slope of the line $lx + my + n = 0$ is given by $m_{slope} = -\frac{l}{m}$.
Setting the slope to $0$,we get $-\frac{l}{m} = 0$,which implies $l = 0$ (provided $m \neq 0$).
Thus,the equation becomes $my + n = 0$,or $y = -\frac{n}{m}$,which represents a line parallel to the $x$-axis.

(C) The slope $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Substituting the given points $(1, 0)$ and $(-2, \sqrt{3})$:
$m = \frac{\sqrt{3} - 0}{-2 - 1} = \frac{\sqrt{3}}{-3} = -\frac{1}{\sqrt{3}}$.
Since $m = \tan(\theta)$,we have $\tan(\theta) = -\frac{1}{\sqrt{3}}$.
Since the slope is negative,the angle $\theta$ lies in the second quadrant.
$\theta = 180^o - 30^o = 150^o$.

(D) The slope of the line passing through $(4, 3)$ and $(2, k)$ is given by $m_1 = \frac{k - 3}{2 - 4} = \frac{k - 3}{-2}$.
The slope of the given line $y = 2x + 3$ is $m_2 = 2$.
Since the two lines are perpendicular,the product of their slopes must be $-1$,so $m_1 \times m_2 = -1$.
Substituting the values: $\left( \frac{k - 3}{-2} \right) \times 2 = -1$.
This simplifies to: $-(k - 3) = -1$,which means $k - 3 = 1$.
Therefore,$k = 4$.

(B) line equally inclined to both axes makes an angle of $45^{\circ}$ or $135^{\circ}$ with the positive $x$-axis.
Therefore,the slope $m = \tan(45^{\circ}) = 1$ or $m = \tan(135^{\circ}) = -1$.
The equations of such lines are $y = x$ and $y = -x$.
Thus,there are $2$ such straight lines.

(C) Let the line $y - x + 2 = 0$ divide the line segment joining $A(3, -1)$ and $B(8, 9)$ in the ratio $\lambda : 1$.
Using the section formula,the coordinates of the point of division are $\left( \frac{8\lambda + 3}{\lambda + 1}, \frac{9\lambda - 1}{\lambda + 1} \right)$.
Since this point lies on the line $y - x + 2 = 0$,we substitute the coordinates into the equation:
$\frac{9\lambda - 1}{\lambda + 1} - \frac{8\lambda + 3}{\lambda + 1} + 2 = 0$
Multiplying by $(\lambda + 1)$,we get:
$(9\lambda - 1) - (8\lambda + 3) + 2(\lambda + 1) = 0$
$9\lambda - 1 - 8\lambda - 3 + 2\lambda + 2 = 0$
$3\lambda - 2 = 0$
$\lambda = \frac{2}{3}$
Thus,the required ratio is $2 : 3$.

(A) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively.
Then the coordinates of the points where the line intersects the axes are $A(a, 0)$ and $B(0, b)$.
The midpoint of the segment $AB$ is given by $(\frac{a+0}{2}, \frac{0+b}{2}) = (\frac{a}{2}, \frac{b}{2})$.
Given that the midpoint is $(3, 2)$,we have:
$\frac{a}{2} = 3 \Rightarrow a = 6$
$\frac{b}{2} = 2 \Rightarrow b = 4$
The intercept form of the equation of a line is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting $a=6$ and $b=4$,we get $\frac{x}{6} + \frac{y}{4} = 1$.
Multiplying by $12$,we get $2x + 3y = 12$.

(A) Let the line passing through $A(-5, -4)$ make an angle $\theta$ with the $x$-axis. The equation of the line is $\frac{x + 5}{\cos \theta} = \frac{y + 4}{\sin \theta} = r$.
For point $B$ on $x + 3y + 2 = 0$,we have $(r_1 \cos \theta - 5) + 3(r_1 \sin \theta - 4) + 2 = 0$,which gives $r_1(\cos \theta + 3 \sin \theta) = 15$,so $\frac{15}{AB} = \cos \theta + 3 \sin \theta$.
For point $C$ on $2x + y + 4 = 0$,we have $2(r_2 \cos \theta - 5) + (r_2 \sin \theta - 4) + 4 = 0$,which gives $r_2(2 \cos \theta + \sin \theta) = 10$,so $\frac{10}{AC} = 2 \cos \theta + \sin \theta$.
For point $D$ on $x - y - 5 = 0$,we have $(r_3 \cos \theta - 5) - (r_3 \sin \theta - 4) - 5 = 0$,which gives $r_3(\cos \theta - \sin \theta) = 6$,so $\frac{6}{AD} = \cos \theta - \sin \theta$.
Substituting these into the given relation $(\frac{15}{AB})^2 + (\frac{10}{AC})^2 = (\frac{6}{AD})^2$,we get $(\cos \theta + 3 \sin \theta)^2 + (2 \cos \theta + \sin \theta)^2 = (\cos \theta - \sin \theta)^2$.
Expanding this,$\cos^2 \theta + 9 \sin^2 \theta + 6 \sin \theta \cos \theta + 4 \cos^2 \theta + \sin^2 \theta + 4 \sin \theta \cos \theta = \cos^2 \theta + \sin^2 \theta - 2 \sin \theta \cos \theta$.
Simplifying,$4 \cos^2 \theta + 9 \sin^2 \theta + 12 \sin \theta \cos \theta = 0$,which is $(2 \cos \theta + 3 \sin \theta)^2 = 0$.
Thus,$2 \cos \theta + 3 \sin \theta = 0$,so $\tan \theta = -\frac{2}{3}$.
The equation of the line is $y + 4 = -\frac{2}{3}(x + 5)$,which simplifies to $2x + 3y + 22 = 0$.

(D) The median $PS$ connects vertex $P(2, 2)$ to the midpoint $S$ of side $QR$.
$S = \left( \frac{6 + 7}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{13}{2}, 1 \right)$.
The slope $m$ of $PS$ is given by $m = \frac{1 - 2}{\frac{13}{2} - 2} = \frac{-1}{\frac{9}{2}} = -\frac{2}{9}$.
Since the required line is parallel to $PS$,its slope is also $-\frac{2}{9}$.
The equation of the line passing through $(1, -1)$ with slope $-\frac{2}{9}$ is:
$y - (-1) = -\frac{2}{9}(x - 1)$
$9(y + 1) = -2(x - 1)$
$9y + 9 = -2x + 2$
$2x + 9y + 7 = 0$.

(B) Let the line intersect the $x$-axis at $A(-a, 0)$ and the $y$-axis at $B(0, b)$.
The area of the triangle formed by the axes is given by $\frac{1}{2} \times |OA| \times |OB| = T$.
Since $OA = |-a| = a$,we have $\frac{1}{2} \times a \times |b| = T$,which implies $|b| = \frac{2T}{a}$.
Taking $b = \frac{2T}{a}$ (assuming the intercept is positive),the intercept form of the line is $\frac{x}{-a} + \frac{y}{2T/a} = 1$.
Multiplying by $2T$,we get $-\frac{2Tx}{a} + ay = 2T$.
Rearranging gives $2Tx - a^2y + 2aT = 0$.

(D) Let the required line passing through $(1, 2)$ be inclined at an angle $\theta$ to the $x$-axis. The equation of the line is $\frac{x-1}{\cos \theta} = \frac{y-2}{\sin \theta} = r$,where $r$ is the distance from $(1, 2)$.
Any point on this line is $(1 + r \cos \theta, 2 + r \sin \theta)$.
Given $r = \frac{\sqrt{6}}{3}$,the point is $(1 + \frac{\sqrt{6}}{3} \cos \theta, 2 + \frac{\sqrt{6}}{3} \sin \theta)$.
Since this point lies on $x + y = 4$,we have $(1 + \frac{\sqrt{6}}{3} \cos \theta) + (2 + \frac{\sqrt{6}}{3} \sin \theta) = 4$.
$\frac{\sqrt{6}}{3} (\cos \theta + \sin \theta) = 1 \implies \sin \theta + \cos \theta = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}$.
Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2}$.
$\sin(\theta + 45^\circ) = \sin 60^\circ$ or $\sin 120^\circ$.
$\theta + 45^\circ = 60^\circ \implies \theta = 15^\circ$ or $\theta + 45^\circ = 120^\circ \implies \theta = 75^\circ$.
Since $75^\circ$ is one of the options,the correct answer is $75^\circ$.

(A) The equation of a line passing through $P(3, 4)$ with an inclination of $\theta = \frac{\pi}{6}$ is given by the parametric form: $\frac{x - 3}{\cos(\pi/6)} = \frac{y - 4}{\sin(\pi/6)} = r$.
Any point $Q$ on this line is $(3 + r\cos(30^\circ), 4 + r\sin(30^\circ)) = (3 + \frac{r\sqrt{3}}{2}, 4 + \frac{r}{2})$.
Since $Q$ lies on the line $12x + 5y + 10 = 0$,we substitute the coordinates:
$12(3 + \frac{r\sqrt{3}}{2}) + 5(4 + \frac{r}{2}) + 10 = 0$.
$36 + 6r\sqrt{3} + 20 + 2.5r + 10 = 0$.
$66 + r(6\sqrt{3} + 2.5) = 0$.
Since $PQ$ is a length,we take the magnitude: $r = |\frac{-66}{6\sqrt{3} + 2.5}| = \frac{66}{6\sqrt{3} + 2.5} = \frac{132}{12\sqrt{3} + 5}$.

(B) To find the intersection point with the $Y$-axis,we set $x = 0$.
Substituting $x = 0$ into the equation $3x - 2y - 6 = 0$:
$3(0) - 2y - 6 = 0$
$-2y = 6$
$y = -3$
Thus,the coordinates of the intersection point on the $Y$-axis are $(0, -3)$.

(D) line perpendicular to the $y$-axis is parallel to the $x$-axis and is of the form $y = k$.
Since the line passes through the point $(1, -2)$,the $y$-coordinate must be $-2$.
Therefore,the equation of the line is $y = -2$,which can be written as $y + 2 = 0$.

(C) The equation of a line parallel to $ax + by + c = 0$ is of the form $ax + by + k = 0$ ... $(1)$.
Since the line passes through the point $(c, d)$,we substitute these coordinates into equation $(1)$:
$ac + bd + k = 0$,which gives $k = -(ac + bd)$.
Substituting the value of $k$ back into equation $(1)$:
$ax + by - (ac + bd) = 0$
$a(x - c) + b(y - d) = 0$.

(D) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively.
Given $a + b = -1$,so $b = -1 - a$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting $b$,we get $\frac{x}{a} + \frac{y}{-1 - a} = 1$.
Since the line passes through $(4, 3)$,we have $\frac{4}{a} + \frac{3}{-1 - a} = 1$.
$\frac{4}{a} - \frac{3}{1 + a} = 1$.
$4(1 + a) - 3a = a(1 + a)$.
$4 + 4a - 3a = a + a^2$.
$4 + a = a + a^2$.
$a^2 = 4$,which gives $a = 2$ or $a = -2$.
If $a = 2$,then $b = -1 - 2 = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1$,i.e.,$\frac{x}{2} - \frac{y}{3} = 1$.
If $a = -2$,then $b = -1 - (-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$.

(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $a$ respectively.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{a} = 1$.
This simplifies to $x + y = a$,or $x + y - a = 0$.
The slope $m$ of a line $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$.
Here,$A = 1$ and $B = 1$,so $m = -\frac{1}{1} = -1$.
Alternatively,if the line makes equal intercepts on the axes,it makes an angle of $135^{\circ}$ with the positive $x$-axis,so the slope is $\tan(135^{\circ}) = -1$.

(D) The given line is $3x + y = 3$,which can be written as $y = -3x + 3$. The slope of this line is $m_1 = -3$.
Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$-3 \times m_2 = -1$,which gives $m_2 = 1/3$.
The equation of the line passing through $(2, 2)$ with slope $m_2 = 1/3$ is given by $y - y_1 = m_2(x - x_1)$.
Substituting the values: $y - 2 = \frac{1}{3}(x - 2)$.
Multiplying by $3$: $3y - 6 = x - 2$,which simplifies to $x - 3y + 4 = 0$.
To find the $y$-intercept,set $x = 0$: $0 - 3y + 4 = 0$,which gives $3y = 4$,so $y = 4/3$.
Therefore,the $y$-intercept is $4/3$.

(A) The slope $m$ of the line is given by $m = \tan(60^{\circ}) = \sqrt{3}$.
The $y$-intercept $c$ is given as $-5$ because it is on the negative direction of the $y$-axis.
Using the slope-intercept form of the line equation,$y = mx + c$,we substitute the values:
$y = \sqrt{3}x + (-5)$
$y = \sqrt{3}x - 5$.

(A) The given line is $x + y + 4 = 0$.
Any line perpendicular to $ax + by + c = 0$ is of the form $bx - ay + k = 0$.
Thus,the equation of a line perpendicular to $x + y + 4 = 0$ is $x - y + k = 0$.
Since this line passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ into the equation:
$1 - 2 + k = 0$
$-1 + k = 0$
$k = 1$.
Substituting $k = 1$ back into the equation,we get $x - y + 1 = 0$.

(C) The slope of the line is $m = \tan(45^{\circ}) = 1$.
The midpoint of the line segment joining $(5, 3)$ and $(4, 4)$ is given by:
$M = \left( \frac{5+4}{2}, \frac{3+4}{2} \right) = \left( \frac{9}{2}, \frac{7}{2} \right)$.
The equation of the line passing through $(x_1, y_1) = \left( \frac{9}{2}, \frac{7}{2} \right)$ with slope $m = 1$ is:
$y - y_1 = m(x - x_1)$
$y - \frac{7}{2} = 1 \left( x - \frac{9}{2} \right)$
$y - \frac{7}{2} = x - \frac{9}{2}$
$x - y - \frac{9}{2} + \frac{7}{2} = 0$
$x - y - 1 = 0$.

(A) Given that the line makes an intercept of $3$ units on the negative direction of the $y$-axis,so $c = -3$.
The angle of inclination is $\theta = {\tan ^{ - 1}}\left( {\frac{3}{5}} \right)$,which implies the slope $m = \tan \theta = \frac{3}{5}$.
The slope-intercept form of the equation of a line is $y = mx + c$.
Substituting the values of $m$ and $c$:
$y = \frac{3}{5}x - 3$
Multiply by $5$ to clear the fraction:
$5y = 3x - 15$
Rearranging the terms,we get:
$3x - 5y - 15 = 0$.

(C) The given equation of the line is $x + \sqrt{3}y = 4$.
Divide both sides by $\sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.
We get $\frac{1}{2}x + \frac{\sqrt{3}}{2}y = \frac{4}{2}$.
This simplifies to $x \cos(\pi/3) + y \sin(\pi/3) = 2$.

(B) Let the equation of the line parallel to $\frac{x}{a} + \frac{y}{b} = 1$ be $\frac{x}{a} + \frac{y}{b} = K$ $(1)$.
Since the line passes through $(a, b)$,we substitute $x = a$ and $y = b$ into equation $(1)$:
$\frac{a}{a} + \frac{b}{b} = K$
$1 + 1 = K$
$K = 2$.
Therefore,the required equation is $\frac{x}{a} + \frac{y}{b} = 2$.

(A) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.
The coordinates of the points where the line intersects the $x$-axis and $y$-axis are $(a, 0)$ and $(0, b)$ respectively.
The midpoint of the segment intercepted between the axes is $(\frac{a}{2}, \frac{b}{2})$.
Given that the line is bisected at $(x_1, y_1)$,we have $\frac{a}{2} = x_1 \implies a = 2x_1$ and $\frac{b}{2} = y_1 \implies b = 2y_1$.
Substituting these values into the intercept form equation:
$\frac{x}{2x_1} + \frac{y}{2y_1} = 1$
Multiplying by $2$,we get $\frac{x}{x_1} + \frac{y}{y_1} = 2$.