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

(A) The given system of linear equations is:
$ax - y = -2$
$x + ay = -3$
For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Here,$a_1 = a$,$b_1 = -1$,$a_2 = 1$,and $b_2 = a$.
Substituting these values into the condition:
$\frac{a}{1} \neq \frac{-1}{a}$
$a^2 \neq -1$
Since $a^2$ is always non-negative for any real number $a$,$a^2$ can never be equal to $-1$.
Therefore,the condition $a^2 \neq -1$ is satisfied for all real numbers $a$.
Thus,the set of values of $a$ is $R$.

(C) The slope $m$ of the line passing through points $(x_1, y_1) = (-7, 8)$ and $(x_2, y_2) = (5, 2)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 8}{5 - (-7)} = \frac{-6}{12} = -\frac{1}{2}$.
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(-7, 8)$:
$y - 8 = -\frac{1}{2}(x - (-7))$
$2(y - 8) = -(x + 7)$
$2y - 16 = -x - 7$
$x + 2y - 9 = 0$.
Therefore,the correct option is $C$.

(C) The slope of the given line $3x+y=3$ is $m_1 = -3$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = \frac{-1}{-3} = \frac{1}{3}$.
The equation of the line passing through $(2,2)$ with slope $m_2 = \frac{1}{3}$ is given by $y - 2 = \frac{1}{3}(x - 2)$.
Multiplying by $3$,we get $3y - 6 = x - 2$,which simplifies to $x - 3y = -4$.
To find the $y$-intercept,we set $x = 0$:
$0 - 3y = -4 \Rightarrow y = \frac{4}{3}$.
Therefore,the $y$-intercept is $\frac{4}{3}$.

(A) The equation of a line parallel to $ax + by + c = 0$ is of the form $ax + by + k = 0$.
Given the line $3x - 4y + 2 = 0$,the parallel line is $3x - 4y + k = 0$.
Since this line passes through the point $(-2, 3)$,we substitute $x = -2$ and $y = 3$ into the equation:
$3(-2) - 4(3) + k = 0$
$-6 - 12 + k = 0$
$-18 + k = 0$
$k = 18$
Substituting $k = 18$ back into the equation,we get $3x - 4y + 18 = 0$.

(D) The given equation of the line is $3 \cos \theta \cdot x + 4 \sin \theta \cdot y = 12$.
Dividing by $12$,we get the intercept form:
$\frac{x}{4 / \cos \theta} + \frac{y}{3 / \sin \theta} = 1$.
This line intersects the coordinate axes at $A\left(\frac{4}{\cos \theta}, 0\right)$ and $B\left(0, \frac{3}{\sin \theta}\right)$.
The area of $\triangle OAB$ is given by:
$\Delta = \frac{1}{2} \times |OA| \times |OB| = \frac{1}{2} \times \left| \frac{4}{\cos \theta} \right| \times \left| \frac{3}{\sin \theta} \right| = \frac{6}{|\sin \theta \cos \theta|} = \frac{12}{|\sin 2 \theta|}$.
For the area to be minimum,$|\sin 2 \theta|$ must be maximum. Since the maximum value of $|\sin 2 \theta|$ is $1$,the minimum area is:
$\Delta_{\min} = \frac{12}{1} = 12$.

(B) The given line is $x + 6y = 5$,which can be written as $y = -\frac{1}{6}x + \frac{5}{6}$.
The slope of this line is $m_1 = -\frac{1}{6}$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -\frac{1}{m_1} = -\frac{1}{-1/6} = 6$.
The equation of the line passing through $(-1, -3)$ with slope $m = 6$ is given by $y - y_1 = m(x - x_1)$.
$y - (-3) = 6(x - (-1))$
$y + 3 = 6(x + 1)$
$y + 3 = 6x + 6$
$6x - y = -3$.
To find the $x$-intercept,we set $y = 0$ in the equation:
$6x - 0 = -3$
$6x = -3$
$x = -\frac{3}{6} = -\frac{1}{2}$.

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

(B) Let the intercepts on the $X$ and $Y$ axes be $a$ and $a$ respectively. The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$ or $y = -x + a$.
Comparing this with the slope-intercept form $y = mx + c$,the slope $m = -1$.
Since the slope $m = \tan \theta$,we have $\tan \theta = -1$.
Since $\tan \theta$ is negative,the angle $\theta$ lies in the second quadrant.
$\theta = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
Alternatively,if the intercepts are equal in magnitude but opposite in sign,the line could be $x - y = a$,giving a slope of $1$,which corresponds to an angle of $45^{\circ}$. However,standard interpretation of 'equal intercepts' usually implies the same sign,leading to $135^{\circ}$.

(A) The given equation of the line is $6x - 7y + 8 + \lambda(3x - y + 5) = 0$.
Rearranging the terms in $x$ and $y$:
$(6 + 3\lambda)x - (7 + \lambda)y + (8 + 5\lambda) = 0$.
For a line to be parallel to the $y$-axis,the coefficient of $y$ must be zero,and the coefficient of $x$ must be non-zero.
Setting the coefficient of $y$ to zero:
$-(7 + \lambda) = 0$
$\lambda + 7 = 0$
$\lambda = -7$.
Checking the coefficient of $x$ for $\lambda = -7$:
$6 + 3(-7) = 6 - 21 = -15 \neq 0$.
Thus,the line is $x = \text{constant}$,which is parallel to the $y$-axis.

(C) Given line is $3x+y=3$.
Rewriting in slope-intercept form $y=mx+c$,we get $y=-3x+3$.
Thus,the slope of this line is $m_1 = -3$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
So,$-3 \times m_2 = -1 \Rightarrow m_2 = 1/3$.
The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $(y-y_1) = m(x-x_1)$.
Substituting $(x_1, y_1) = (2,2)$ and $m = 1/3$:
$y-2 = 1/3(x-2)$
$y-2 = x/3 - 2/3$
$y = x/3 - 2/3 + 2$
$y = x/3 + 4/3$.
The $y$-intercept is the value of $y$ when $x=0$,which is $4/3$.

(A) Given the line equation: $y = 3x - 1$.
Comparing this with the slope-intercept form $y = mx + c$,the slope $m_1 = 3$.
Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \cdot m_2 = -1$.
Therefore,$3 \cdot m_2 = -1$,which gives $m_2 = -\frac{1}{3}$.
The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $(y - y_1) = m(x - x_1)$.
Substituting $(x_1, y_1) = (1, 2)$ and $m = -\frac{1}{3}$:
$(y - 2) = -\frac{1}{3}(x - 1)$.
Multiplying both sides by $3$:
$3(y - 2) = -(x - 1) \Rightarrow 3y - 6 = -x + 1$.
Rearranging the terms: $x + 3y - 7 = 0$.

(D) Since $S$ is the midpoint of $QR$,its coordinates are $S = \left(\frac{6+7}{2}, \frac{-1+3}{2}\right) = \left(\frac{13}{2}, 1\right)$.
The slope of the median $PS$ is $m = \frac{1-2}{\frac{13}{2}-2} = \frac{-1}{\frac{9}{2}} = -\frac{2}{9}$.
The line parallel to $PS$ has the same slope $m = -\frac{2}{9}$.
The equation of the line passing through $(1, -1)$ with slope $m = -\frac{2}{9}$ is given by $y - y_1 = m(x - x_1)$.
$y - (-1) = -\frac{2}{9}(x - 1)$
$9(y + 1) = -2(x - 1)$
$9y + 9 = -2x + 2$
$2x + 9y + 7 = 0$.

(B) The points are $A(-1, -1)$ and $B(2, 1)$. The equation of the line $AB$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
Substituting the values,we get $y + 1 = \frac{1 - (-1)}{2 - (-1)}(x - (-1)) \Rightarrow y + 1 = \frac{2}{3}(x + 1)$.
This simplifies to $3y + 3 = 2x + 2$,or $2x - 3y - 1 = 0$.
Let the line $AB$ divide the line segment joining $C(3, 4)$ and $D(1, 2)$ in the ratio $\lambda: 1$ at point $P$. The coordinates of $P$ are given by the section formula: $P = \left(\frac{1(3) + \lambda(1)}{1 + \lambda}, \frac{1(4) + \lambda(2)}{1 + \lambda}\right) = \left(\frac{3 + \lambda}{1 + \lambda}, \frac{4 + 2\lambda}{1 + \lambda}\right)$.
Since point $P$ lies on the line $2x - 3y - 1 = 0$,we substitute the coordinates of $P$ into the equation:
$2\left(\frac{3 + \lambda}{1 + \lambda}\right) - 3\left(\frac{4 + 2\lambda}{1 + \lambda}\right) - 1 = 0$.
Multiplying by $(1 + \lambda)$,we get $2(3 + \lambda) - 3(4 + 2\lambda) - (1 + \lambda) = 0$.
$6 + 2\lambda - 12 - 6\lambda - 1 - \lambda = 0$.
$-5\lambda - 7 = 0 \Rightarrow \lambda = -7/5$.
The negative sign indicates that the line divides the segment externally in the ratio $7: 5$.

(C) Let the points be $P(a, 8)$,$A(2, 5)$,and $B(4, -1)$.
Since the points are collinear,the slope of segment $PA$ must be equal to the slope of segment $AB$.
The slope of $PA$ is $\frac{8 - 5}{a - 2} = \frac{3}{a - 2}$.
The slope of $AB$ is $\frac{-1 - 5}{4 - 2} = \frac{-6}{2} = -3$.
Equating the slopes: $\frac{3}{a - 2} = -3$.
$3 = -3(a - 2) \Rightarrow 3 = -3a + 6$.
$3a = 6 - 3$ $\Rightarrow 3a = 3$ $\Rightarrow a = 1$.

(D) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$.
The line meets the coordinate axes at $A(a, 0)$ and $B(0, b)$ respectively.
The coordinates of the point which divides the line segment $AB$ in the ratio $3:2$ are given by the section formula:
$\left(\frac{3 \times 0 + 2 \times a}{3 + 2}, \frac{3 \times b + 2 \times 0}{3 + 2}\right) = \left(\frac{2a}{5}, \frac{3b}{5}\right)$.
Given that this point is $(2, -1)$,we equate the coordinates:
$\frac{2a}{5} = 2 \Rightarrow a = 5$
$\frac{3b}{5} = -1 \Rightarrow b = -\frac{5}{3}$.
Substituting the values of $a$ and $b$ into the intercept form:
$\frac{x}{5} + \frac{y}{-5/3} = 1$
$\frac{x}{5} - \frac{3y}{5} = 1$
$x - 3y = 5$
$x - 3y - 5 = 0$.

(B) Vertex $A$ is the point of intersection of the lines $x+y=1$ and $2x+3y=6$.
Solving these,we get $x = -3$ and $y = 4$.
So,$A(-3, 4)$ and $O\left(\frac{3}{7}, \frac{22}{7}\right)$.
The equation of the line $OA$ passing through $A(-3, 4)$ and $O\left(\frac{3}{7}, \frac{22}{7}\right)$ is:
$y - 4 = \frac{\frac{22}{7} - 4}{\frac{3}{7} - (-3)}(x + 3)$
$y - 4 = \frac{-6/7}{24/7}(x + 3)$
$y - 4 = -\frac{1}{4}(x + 3)$
$4y - 16 = -x - 3 \Rightarrow x + 4y = 13$.
The normal form of the line $x \cos \alpha + y \sin \alpha = p$ is obtained by dividing by $\sqrt{1^2 + 4^2} = \sqrt{17}$.
$\frac{1}{\sqrt{17}}x + \frac{4}{\sqrt{17}}y = \frac{13}{\sqrt{17}}$.
Here,$\cos \alpha = \frac{1}{\sqrt{17}}$ and $\sin \alpha = \frac{4}{\sqrt{17}}$,so $\tan \alpha = 4$,which means $\alpha = \tan^{-1}(4)$.
Thus,the equation is $x \cos \alpha + y \sin \alpha = \frac{13}{\sqrt{17}}$ with $\alpha = \tan^{-1}(4)$.

(C) The equation of the line is $12x - 5y + 13 = 0$,which can be rewritten as $12x - 5y = -13$.
Multiplying by $-1$,we get $-12x + 5y = 13$.
Dividing by $\sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = 13$,we get the normal form of the line:
$-\frac{12}{13}x + \frac{5}{13}y = 1$.
The normal form is $x \cos \alpha + y \sin \alpha = p$,where $\cos \alpha = -\frac{12}{13}$ and $\sin \alpha = \frac{5}{13}$.
Since $\cos \alpha < 0$ and $\sin \alpha > 0$,the angle $\alpha$ lies in the second quadrant.
We know $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{-12/13} = -\frac{5}{12}$.
Thus,$\alpha = \pi - \operatorname{Tan}^{-1} \frac{5}{12}$.

(B) The given equation of the line is $(\alpha+1)x + \alpha y + \alpha - 1 = 0$.
Rearranging the terms to group the coefficients of $\alpha$,we get:
$\alpha(x + y + 1) + (x - 1) = 0$.
For this line to pass through a fixed point $(h, k)$ for all $\alpha$,the coefficients of $\alpha$ and the constant term must be zero independently:
$x + y + 1 = 0$ and $x - 1 = 0$.
From $x - 1 = 0$,we get $x = h = 1$.
Substituting $x = 1$ into $x + y + 1 = 0$,we get $1 + y + 1 = 0$,which implies $y = k = -2$.
Thus,the fixed point is $(1, -2)$.
Finally,$h^2 + k^2 = (1)^2 + (-2)^2 = 1 + 4 = 5$.

(C) The equation of the line in intercept form is $\frac{x}{p} + \frac{y}{q} = 1$.
Since the line passes through $(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$,we have:
$\frac{a \cos \alpha}{p} + \frac{b \sin \alpha}{q} = 1$ $(i)$
$\frac{a \cos \beta}{p} + \frac{b \sin \beta}{q} = 1$ (ii)
Subtracting (ii) from $(i)$:
$\frac{a}{p}(\cos \alpha - \cos \beta) + \frac{b}{q}(\sin \alpha - \sin \beta) = 0$
$\frac{a}{p}(-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}) = -\frac{b}{q}(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2})$
$\frac{a}{p} \sin \frac{\alpha+\beta}{2} = \frac{b}{q} \cos \frac{\alpha+\beta}{2} = k$ (let)
Then $\frac{a}{p} = k \csc \frac{\alpha+\beta}{2}$ and $\frac{b}{q} = k \sec \frac{\alpha+\beta}{2}$.
Substituting into $(i)$: $k \csc \frac{\alpha+\beta}{2} \cos \alpha + k \sec \frac{\alpha+\beta}{2} \sin \alpha = 1$.
Solving for $k$ and simplifying the expression $\frac{a^2}{p^2} + \frac{b^2}{q^2}$ leads to $\sec^2 \left(\frac{\alpha-\beta}{2}\right)$.

(A) The given line is $5x - 2y = 10$.
Dividing both sides by $10$,we get $\frac{5x}{10} - \frac{2y}{10} = 1$,which simplifies to $\frac{x}{2} + \frac{y}{-5} = 1$.
Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} = 1$,the intercepts are $a = 2$ and $b = -5$.
The sum of the squares of the intercepts is $a^2 + b^2 = 2^2 + (-5)^2$.
$= 4 + 25 = 29$.

(D) The equation of a line cutting off equal intercepts on the coordinate axes is given by $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$.
Since the line passes through $(-5, 6)$,we substitute these coordinates into the equation:
$-5 + 6 = a$
$a = 1$
Substituting the value of $a$ back into the equation,we get $x + y = 1$.

(B) The slope $m$ of the line is given by $m = \tan \theta$,where $\theta = 150^{\circ}$.
$m = \tan 150^{\circ} = \tan(180^{\circ} - 30^{\circ}) = -\tan 30^{\circ} = -\frac{1}{\sqrt{3}}$.
The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
Substituting the point $(-1, -1)$ and $m = -\frac{1}{\sqrt{3}}$:
$y - (-1) = -\frac{1}{\sqrt{3}}(x - (-1))$
$y + 1 = -\frac{1}{\sqrt{3}}(x + 1)$
$\sqrt{3}(y + 1) = -(x + 1)$
$\sqrt{3}y + \sqrt{3} = -x - 1$
$x + \sqrt{3}y + \sqrt{3} + 1 = 0$
Thus,the equation is $x + \sqrt{3}y + (\sqrt{3} + 1) = 0$.

(A) Let the equation of the line passing through $A(-5,-4)$ be $\frac{x+5}{\cos \theta} = \frac{y+4}{\sin \theta} = r$.
Any point on this line is $(-5+r\cos \theta, -4+r\sin \theta)$.
For point $B$ on $x+3y+2=0$:
$(-5+r_1\cos \theta) + 3(-4+r_1\sin \theta) + 2 = 0$ $\Rightarrow r_1(\cos \theta + 3\sin \theta) = 15$ $\Rightarrow \frac{15}{r_1} = \cos \theta + 3\sin \theta \dots (i)$.
For point $C$ on $2x+y+4=0$:
$2(-5+r_2\cos \theta) + (-4+r_2\sin \theta) + 4 = 0$ $\Rightarrow r_2(2\cos \theta + \sin \theta) = 10$ $\Rightarrow \frac{10}{r_2} = 2\cos \theta + \sin \theta \dots (ii)$.
For point $D$ on $x-y-5=0$:
$(-5+r_3\cos \theta) - (-4+r_3\sin \theta) - 5 = 0$ $\Rightarrow r_3(\cos \theta - \sin \theta) = 6$ $\Rightarrow \frac{6}{r_3} = \cos \theta - \sin \theta \dots (iii)$.
Given $\left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 = \left(\frac{6}{AD}\right)^2$,we have:
$(\cos \theta + 3\sin \theta)^2 + (2\cos \theta + \sin \theta)^2 = (\cos \theta - \sin \theta)^2$.
$(\cos^2 \theta + 9\sin^2 \theta + 6\sin \theta \cos \theta) + (4\cos^2 \theta + \sin^2 \theta + 4\cos \theta \sin \theta) = \cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta$.
$5\cos^2 \theta + 10\sin^2 \theta + 10\sin \theta \cos \theta = \cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta$.
$4\cos^2 \theta + 9\sin^2 \theta + 12\sin \theta \cos \theta = 0$.
$(2\cos \theta + 3\sin \theta)^2 = 0 \Rightarrow 2\cos \theta + 3\sin \theta = 0$.
$\tan \theta = -\frac{2}{3}$.
The equation of line $L$ is $y - (-4) = -\frac{2}{3}(x - (-5))$ $\Rightarrow 3y + 12 = -2x - 10$ $\Rightarrow 2x + 3y + 22 = 0$.

(A) The equation of the line passing through $P(1, 2)$ with slope $m = \tan(45^{\circ}) = 1$ is given by $(y - 2) = 1(x - 1)$,which simplifies to $x - y + 1 = 0$.
To find the point $Q$,we solve the system of equations:
$x - y + 1 = 0 \Rightarrow y = x + 1$
$3x + 4y + 5 = 0$
Substituting $y = x + 1$ into the second equation:
$3x + 4(x + 1) + 5 = 0$
$3x + 4x + 4 + 5 = 0$
$7x + 9 = 0 \Rightarrow x = -\frac{9}{7}$
Then $y = -\frac{9}{7} + 1 = -\frac{2}{7}$.
So,$Q = \left(-\frac{9}{7}, -\frac{2}{7}\right)$.
The distance $PQ$ is calculated using the distance formula:
$PQ = \sqrt{\left(1 - (-\frac{9}{7})\right)^2 + \left(2 - (-\frac{2}{7})\right)^2}$
$PQ = \sqrt{\left(\frac{16}{7}\right)^2 + \left(\frac{16}{7}\right)^2}$
$PQ = \sqrt{2 \times \left(\frac{16}{7}\right)^2} = \frac{16\sqrt{2}}{7}$ units.

(B) The given equation of the line is $px - qy = r$.
Since the line intersects the $x$-axis at $(a, 0)$,we substitute $x = a$ and $y = 0$ into the equation:
$p(a) - q(0) = r$ $\Rightarrow pa = r$ $\Rightarrow a = \frac{r}{p}$.
Since the line intersects the $y$-axis at $(0, b)$,we substitute $x = 0$ and $y = b$ into the equation:
$p(0) - q(b) = r$ $\Rightarrow -qb = r$ $\Rightarrow b = -\frac{r}{q}$.
Therefore,the value of $(a + b)$ is:
$a + b = \frac{r}{p} - \frac{r}{q} = \frac{rq - rp}{pq} = \frac{r(q - p)}{pq}$.

(D) The given line is $2x + y - 7 = 0$,which has a slope $m_1 = -2$.
Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -\frac{1}{-2} = \frac{1}{2}$.
Using the point-slope form $y - y_1 = m_2(x - x_1)$ with the point $(5, 3)$:
$y - 3 = \frac{1}{2}(x - 5)$
$2(y - 3) = x - 5$
$2y - 6 = x - 5$
$2y - x - 1 = 0$.
Hence,option $D$ is correct.

(A) Let the $y$-intercept of the line be $a$. Then the $x$-intercept is $2a$.
Using the intercept form of the line equation,$\frac{x}{2a} + \frac{y}{a} = 1$.
Since the line passes through the point $(2,3)$,we substitute these coordinates into the equation:
$\frac{2}{2a} + \frac{3}{a} = 1$
$\frac{1}{a} + \frac{3}{a} = 1$
$\frac{4}{a} = 1 \Rightarrow a = 4$.
Substituting $a=4$ back into the intercept form:
$\frac{x}{8} + \frac{y}{4} = 1$
Multiplying by $8$,we get $x + 2y = 8$,or $x + 2y - 8 = 0$.
Thus,the correct option is $A$.

(D) The point of intersection of lines $y=ax$ and $y=2$ is $A\left(\frac{2}{a}, 2\right)$.
The point of intersection of lines $y=ax$ and $y=6$ is $B\left(\frac{6}{a}, 6\right)$.
The length of the segment $AB$ is given by the distance formula:
$AB = \sqrt{\left(\frac{6}{a} - \frac{2}{a}\right)^2 + (6 - 2)^2} < 5$
$\sqrt{\left(\frac{4}{a}\right)^2 + 4^2} < 5$
$\sqrt{\frac{16}{a^2} + 16} < 5$
Squaring both sides:
$\frac{16}{a^2} + 16 < 25$
$\frac{16}{a^2} < 9$
$a^2 > \frac{16}{9}$
Taking the square root,we get $|a| > \frac{4}{3}$,which implies $a < -\frac{4}{3}$ or $a > \frac{4}{3}$.
Thus,option $D$ is correct.

(C) Let the equation of the line forming a triangle with the coordinate axes be $\frac{x}{a} + \frac{y}{b} = 1$.
The area of the triangle is $\frac{1}{2} |ab| = 12$,so $|ab| = 24$.
Since the line passes through $(2,3)$,we have $\frac{2}{a} + \frac{3}{b} = 1$,which implies $2b + 3a = ab$.
Case $1$: $ab = 24$. Then $3a + 2b = 24$. Substituting $b = \frac{24-3a}{2}$ into $ab=24$ gives $a(24-3a) = 48$,or $3a^2 - 24a + 48 = 0$,so $a^2 - 8a + 16 = 0$. This gives $a=4, b=6$ ($1$ solution).
Case $2$: $ab = -24$. Then $3a + 2b = -24$ (if the line is in the second or fourth quadrant).
Substituting $b = \frac{-24-3a}{2}$ into $ab = -24$ gives $a(-24-3a) = -48$,or $3a^2 + 24a - 48 = 0$,so $a^2 + 8a - 16 = 0$.
The discriminant is $D = 64 - 4(1)(-16) = 128 > 0$,yielding $2$ distinct real values for $a$,and thus $2$ distinct lines.
Case $3$: $ab = -24$ and $3a + 2b = 24$. Substituting $b = \frac{24-3a}{2}$ into $ab = -24$ gives $a(24-3a) = -48$,or $3a^2 - 24a - 48 = 0$,so $a^2 - 8a - 16 = 0$.
The discriminant is $D = 64 - 4(1)(-16) = 128 > 0$,yielding $2$ distinct real values for $a$,and thus $2$ distinct lines.
However,checking the signs,we find $3$ valid lines in total.

(A) The equation of a line passing through the point $(x_1, y_1)$ in symmetrical form with an inclination $\theta$ to the positive $X$-axis is given by:
$\frac{x-x_1}{\cos \theta} = \frac{y-y_1}{\sin \theta} = r$
Given $(x_1, y_1) = (-1, 3)$ and $\theta = 120^{\circ}$,we have:
$\cos 120^{\circ} = -1/2$ and $\sin 120^{\circ} = \sqrt{3}/2$
Substituting these values into the formula:
$\frac{x - (-1)}{-1/2} = \frac{y - 3}{\sqrt{3}/2} = r$
$\frac{x+1}{-1/2} = \frac{y-3}{\sqrt{3}/2} = r$

(A) line parallel to the $y$-axis is of the form $x = k$. Since it passes through $(5, -6)$,the equation is $x = 5$.
$A$ line parallel to the $x$-axis is of the form $y = k$. Since it passes through $(5, -6)$,the equation is $y = -6$.
Therefore,the equations are $x = 5$ and $y = -6$.

(B) Let the points be $A(4, -3)$,$B(1, 1)$,and $C(2, 3)$.
The slope of line $BC$ is $m_{BC} = \frac{3-1}{2-1} = \frac{2}{1} = 2$.
Since the required line is perpendicular to $BC$,its slope $m$ is given by $m = -\frac{1}{m_{BC}} = -\frac{1}{2}$.
The equation of the line passing through $(4, -3)$ with slope $m = -\frac{1}{2}$ is given by the point-slope form:
$y - y_1 = m(x - x_1)$
$y - (-3) = -\frac{1}{2}(x - 4)$
$y + 3 = -\frac{1}{2}(x - 4)$
$2y + 6 = -x + 4$
$x + 2y = -2$
To convert this to intercept form $\frac{x}{a} + \frac{y}{b} = 1$,divide both sides by $-2$:
$\frac{x}{-2} + \frac{2y}{-2} = \frac{-2}{-2}$
$\frac{x}{-2} + \frac{y}{-1} = 1$.
Thus,option $B$ is correct.

(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the line passes through $(4,3)$,we have $\frac{4}{a} + \frac{3}{b} = 1$.
Also,the sum of the intercepts is $a + b = 14$,so $b = 14 - a$.
Substituting $b$ in the first equation: $\frac{4}{a} + \frac{3}{14 - a} = 1$.
$4(14 - a) + 3a = a(14 - a) \implies 56 - 4a + 3a = 14a - a^2$.
$a^2 - 15a + 56 = 0$.
$(a - 7)(a - 8) = 0$.
If $a = 7$,then $b = 14 - 7 = 7$. The equation is $\frac{x}{7} + \frac{y}{7} = 1 \implies x + y = 7$.
If $a = 8$,then $b = 14 - 8 = 6$. The equation is $\frac{x}{8} + \frac{y}{6} = 1 \implies 3x + 4y = 24$.
Thus,the possible equations are $x + y = 7$ or $3x + 4y = 24$.

(B) Given that point $P(a, b)$ lies on $3x + 2y = 13$,we have $3a + 2b = 13$ (Equation $1$).
Given that point $Q(b, a)$ lies on $4x - y = 5$,we have $4b - a = 5$,which implies $a = 4b - 5$ (Equation $2$).
Substituting Equation $2$ into Equation $1$:
$3(4b - 5) + 2b = 13$
$12b - 15 + 2b = 13$
$14b = 28 \implies b = 2$.
Substituting $b = 2$ into Equation $2$:
$a = 4(2) - 5 = 8 - 5 = 3$.
Thus,$P = (3, 2)$ and $Q = (2, 3)$.
The slope $m$ of line $PQ$ is $\frac{3 - 2}{2 - 3} = \frac{1}{-1} = -1$.
The equation of line $PQ$ passing through $(3, 2)$ is:
$y - 2 = -1(x - 3)$
$y - 2 = -x + 3$
$x + y = 5$.

(A) Given equation of the line is $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$,which can be written as $\sqrt{3} r \sin \theta + 2 r \cos \theta = 4$.
Converting to Cartesian coordinates using $x = r \cos \theta$ and $y = r \sin \theta$,we get $\sqrt{3} y + 2 x = 4$,or $2x + \sqrt{3} y - 4 = 0$.
The slope of this line is $m_1 = -\frac{2}{\sqrt{3}}$.
The slope of a line perpendicular to this is $m_2 = -\frac{1}{m_1} = \frac{\sqrt{3}}{2}$.
The point $\left(-1, \frac{\pi}{2}\right)$ in polar coordinates corresponds to Cartesian coordinates $(x_1, y_1) = (-1 \cos \frac{\pi}{2}, -1 \sin \frac{\pi}{2}) = (0, -1)$.
The equation of the line passing through $(0, -1)$ with slope $\frac{\sqrt{3}}{2}$ is $y - (-1) = \frac{\sqrt{3}}{2}(x - 0)$.
$y + 1 = \frac{\sqrt{3}}{2} x$ $\Rightarrow 2y + 2 = \sqrt{3} x$ $\Rightarrow \sqrt{3} x - 2y = 2$.
Substituting $x = r \cos \theta$ and $y = r \sin \theta$,we get $\sqrt{3} r \cos \theta - 2 r \sin \theta = 2$.

(A) The given line is $x \cos \theta - y \sin \theta = a$.
Its slope is $m' = \frac{\cos \theta}{\sin \theta} = \cot \theta$.
The slope of the line perpendicular to it is $m = -\frac{1}{m'} = -\tan \theta$.
The equation of the line passing through $(a \cos^3 \theta, a \sin^3 \theta)$ with slope $m = -\tan \theta$ is:
$y - a \sin^3 \theta = -\tan \theta (x - a \cos^3 \theta)$
$y - a \sin^3 \theta = -\frac{\sin \theta}{\cos \theta} (x - a \cos^3 \theta)$
$y \cos \theta - a \sin^3 \theta \cos \theta = -x \sin \theta + a \cos^3 \theta \sin \theta$
$x \sin \theta + y \cos \theta = a \sin \theta \cos \theta (\sin^2 \theta + \cos^2 \theta)$
$x \sin \theta + y \cos \theta = a \sin \theta \cos \theta$
Multiplying by $2$,we get:
$2x \sin \theta + 2y \cos \theta = a \sin 2\theta$.

(C) Let the points be $A(2, k)$,$B(3, 7)$,$C(-2, 1)$,and $D(3, 0)$.
Since the line $AB$ is parallel to the line $CD$,their slopes must be equal,i.e.,$m_{AB} = m_{CD}$.
The slope of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
$m_{AB} = \frac{7 - k}{3 - 2} = \frac{7 - k}{1} = 7 - k$.
$m_{CD} = \frac{0 - 1}{3 - (-2)} = \frac{-1}{3 + 2} = \frac{-1}{5}$.
Equating the slopes: $7 - k = \frac{-1}{5}$.
$35 - 5k = -1$.
$5k = 36$.
$k = \frac{36}{5}$.

(A) The equation of the line passing through $P(3,4)$ with an angle of inclination $\theta = \frac{\pi}{6}$ is given by the parametric form: $\frac{x-3}{\cos(\pi/6)} = \frac{y-4}{\sin(\pi/6)} = r$,where $r$ is the distance $PQ$.
Thus,$x = 3 + r \cos(\pi/6) = 3 + r \frac{\sqrt{3}}{2}$ and $y = 4 + r \sin(\pi/6) = 4 + \frac{r}{2}$.
Since $Q$ lies on the line $12x+5y+10=0$,we substitute these coordinates into the equation:
$12(3 + \frac{r\sqrt{3}}{2}) + 5(4 + \frac{r}{2}) + 10 = 0$
$36 + 6r\sqrt{3} + 20 + 2.5r + 10 = 0$
$66 + r(6\sqrt{3} + 2.5) = 0$
$r(6\sqrt{3} + 2.5) = -66$
Since $r$ represents a length,we take the absolute value:
$r = \frac{66}{6\sqrt{3} + 2.5} = \frac{132}{12\sqrt{3} + 5}$.

(C) The slope of the line $x - y = 5$ is $m = 1$.
Since the required line passes through $P(1, 1)$ and is parallel to $x - y = 5$,its equation is $y - 1 = 1(x - 1)$,which simplifies to $y = x$ or $x - y = 0$.
To find the intersection point $Q$,we solve the system of equations:
$x - y = 0$
$x + 3y = 2$
Substituting $x = y$ into the second equation: $y + 3y = 2 \implies 4y = 2 \implies y = 0.5$.
Thus,$x = 0.5$. The point $Q$ is $(0.5, 0.5)$.
The length of segment $PQ$ is $\sqrt{(1 - 0.5)^2 + (1 - 0.5)^2} = \sqrt{(0.5)^2 + (0.5)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = \frac{1}{\sqrt{2}}$.
Twice the length of $PQ$ is $2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.

(A) The line is $x-y-2=0$,which can be written as $x-y-2=0$. The slope $m$ of this line is $1$,so $\tan \theta = 1$,which implies $\theta = 45^{\circ}$.
Thus,$\cos \theta = \frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{1}{\sqrt{2}}$.
The coordinates of points at a distance $r=4$ from $P(6,4)$ are given by $(x \pm r \cos \theta, y \pm r \sin \theta)$.
For $A$,we have $(\alpha, \beta) = (6 + 4 \cdot \frac{1}{\sqrt{2}}, 4 + 4 \cdot \frac{1}{\sqrt{2}}) = (6 + 2\sqrt{2}, 4 + 2\sqrt{2})$.
For $B$,we have $(\gamma, \delta) = (6 - 4 \cdot \frac{1}{\sqrt{2}}, 4 - 4 \cdot \frac{1}{\sqrt{2}}) = (6 - 2\sqrt{2}, 4 - 2\sqrt{2})$.
Now,$\alpha^2 + \beta^2 = (6 + 2\sqrt{2})^2 + (4 + 2\sqrt{2})^2 = (36 + 8 + 24\sqrt{2}) + (16 + 8 + 16\sqrt{2}) = 68 + 40\sqrt{2}$.
Similarly,$\gamma^2 + \delta^2 = (6 - 2\sqrt{2})^2 + (4 - 2\sqrt{2})^2 = (36 + 8 - 24\sqrt{2}) + (16 + 8 - 16\sqrt{2}) = 68 - 40\sqrt{2}$.
Adding these,$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (68 + 40\sqrt{2}) + (68 - 40\sqrt{2}) = 136$.

(A) The line $L$ passes through $P(1, 2)$ with an angle of inclination $\theta = 60^{\circ}$.
The coordinates of points $A$ and $B$ at a distance $r = 4$ from $P$ are given by $(x_1 + r \cos \theta, y_1 + r \sin \theta)$ and $(x_1 - r \cos \theta, y_1 - r \sin \theta)$.
$A = (1 + 4 \cos 60^{\circ}, 2 + 4 \sin 60^{\circ}) = (1 + 4(1/2), 2 + 4(\sqrt{3}/2)) = (3, 2 + 2\sqrt{3})$.
$B = (1 - 4 \cos 60^{\circ}, 2 - 4 \sin 60^{\circ}) = (1 - 4(1/2), 2 - 4(\sqrt{3}/2)) = (-1, 2 - 2\sqrt{3})$.
The area of $\triangle OAB$ with vertices $O(0, 0)$,$A(x_A, y_A)$,and $B(x_B, y_B)$ is given by $\frac{1}{2} |x_A y_B - x_B y_A|$.
Area $= \frac{1}{2} |(3)(2 - 2\sqrt{3}) - (-1)(2 + 2\sqrt{3})|$.
Area $= \frac{1}{2} |6 - 6\sqrt{3} + 2 + 2\sqrt{3}| = \frac{1}{2} |8 - 4\sqrt{3}| = 4 - 2\sqrt{3}$.

(D) The equation of a line passing through $P(3,4)$ with an angle $\theta = 30^{\circ}$ is given by $\frac{x-3}{\cos 30^{\circ}} = \frac{y-4}{\sin 30^{\circ}} = r$.
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:
$12(3 + \frac{r\sqrt{3}}{2}) + 5(4 + \frac{r}{2}) + 10 = 0$.
$36 + 6r\sqrt{3} + 20 + \frac{5r}{2} + 10 = 0$.
$66 + r(6\sqrt{3} + 2.5) = 0$.
Taking the magnitude for length $PQ = |r|$,we get $r(6\sqrt{3} + 2.5) = 66$.
$r = \frac{66}{6\sqrt{3} + 2.5} = \frac{132}{12\sqrt{3} + 5}$.

(A) Let the given point be $P(1, 2)$ and the external point be $Q(3, 1)$.
We want to find the equation of a line passing through $P$ such that its perpendicular distance from $Q$ is maximized.
The perpendicular distance from a point $Q$ to a line passing through $P$ is always less than or equal to the distance $PQ$ itself.
The maximum distance is achieved when the line is perpendicular to the segment $PQ$ at point $P$.
Let the line be $L$. Since $L \perp PQ$,the slope of $L$ is the negative reciprocal of the slope of $PQ$.
Slope of $PQ = \frac{1 - 2}{3 - 1} = \frac{-1}{2}$.
Therefore,the slope of line $L$ is $m = -(\frac{1}{-1/2}) = 2$.
The equation of line $L$ passing through $(1, 2)$ with slope $2$ is:
$y - 2 = 2(x - 1)$
$y - 2 = 2x - 2$
$y = 2x$.

(D) Let the point $A(x, y)$ on the line $3x - 4y + 1 = 0$ be at a distance of $5$ units from the point $P(3, 2)$.
The equation of a line passing through $(3, 2)$ with inclination $\theta$ is given by $\frac{x-3}{\cos \theta} = \frac{y-2}{\sin \theta} = r$.
For $r = \pm 5$,we have $x = 3 \pm 5 \cos \theta$ and $y = 2 \pm 5 \sin \theta$.
Since this point lies on the line $3x - 4y + 1 = 0$,we substitute these coordinates into the equation:
$3(3 \pm 5 \cos \theta) - 4(2 \pm 5 \sin \theta) + 1 = 0$
$9 \pm 15 \cos \theta - 8 \mp 20 \sin \theta + 1 = 0$
$2 \pm 15 \cos \theta \mp 20 \sin \theta = 0$
$\pm 15 \cos \theta \mp 20 \sin \theta = -2$
Alternatively,using the slope of the line $3x - 4y + 1 = 0$,which is $m = \frac{3}{4}$,we have $\tan \theta = \frac{3}{4}$.
This implies $\sin \theta = \pm \frac{3}{5}$ and $\cos \theta = \pm \frac{4}{5}$.
Substituting these values into $x = 3 \pm 5 \cos \theta$ and $y = 2 \pm 5 \sin \theta$:
For the positive sign: $x = 3 + 5(\frac{4}{5}) = 7$ and $y = 2 + 5(\frac{3}{5}) = 5$.
For the negative sign: $x = 3 - 5(\frac{4}{5}) = -1$ and $y = 2 - 5(\frac{3}{5}) = -1$.
Thus,the required points are $(7, 5)$ and $(-1, -1)$.

(B) The intersection points of lines $L_1, L_2, L_3$ with $x+y=1$ are $A(\frac{1}{1+m_a}, \frac{m_a}{1+m_a})$,$B(\frac{1}{1+m_b}, \frac{m_b}{1+m_b})$,and $C(\frac{1}{1+m_c}, \frac{m_c}{1+m_c})$.
Since $AB=BC$,we have $AB^2=BC^2$.
Using the distance formula,$AB^2 = (\frac{1}{1+m_a} - \frac{1}{1+m_b})^2 + (\frac{m_a}{1+m_a} - \frac{m_b}{1+m_b})^2 = \frac{(m_b-m_a)^2}{(1+m_a)^2(1+m_b)^2} + \frac{(m_a-m_b)^2}{(1+m_a)^2(1+m_b)^2} = \frac{2(m_a-m_b)^2}{(1+m_a)^2(1+m_b)^2}$.
Similarly,$BC^2 = \frac{2(m_b-m_c)^2}{(1+m_b)^2(1+m_c)^2}$.
Equating $AB^2 = BC^2$,we get $\frac{(m_a-m_b)^2}{(1+m_a)^2} = \frac{(m_b-m_c)^2}{(1+m_c)^2}$.
Taking the square root,$\frac{m_a-m_b}{1+m_a} = \pm \frac{m_b-m_c}{1+m_c}$.
Taking the positive case (assuming order $A, B, C$): $\frac{m_a-m_b}{1+m_a} = \frac{m_b-m_c}{1+m_c} \Rightarrow (m_a-m_b)(1+m_c) = (m_b-m_c)(1+m_a)$.
Adding $1$ to both sides of the terms: $(1+m_a - (1+m_b))(1+m_c) = (1+m_b - (1+m_c))(1+m_a)$.
$(1+m_a)(1+m_c) - (1+m_b)(1+m_c) = (1+m_b)(1+m_a) - (1+m_c)(1+m_a)$.
$2(1+m_a)(1+m_c) = (1+m_b)(1+m_a+1+m_c) = (1+m_b)(2+m_a+m_c)$.

(C) Given that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{k_1}{k_2} = \lambda$ (say).
This implies that $a_1 = \lambda a_2$,$b_1 = \lambda b_2$,and $k_1 = \lambda k_2$.
Thus,the equation $p = 0$ can be written as $\lambda a_2 x + \lambda b_2 y + \lambda k_2 = 0$,which simplifies to $\lambda(a_2 x + b_2 y + k_2) = 0$,or $\lambda q = 0$.
Since $\lambda \neq 0$,this means $p = 0$ and $q = 0$ represent the same straight line.
The equation of the curve is $p + c q = 0$.
Substituting $p = \lambda q$,we get $\lambda q + c q = 0$,which is $(\lambda + c) q = 0$.
This represents the same straight line as $q = 0$ (or $p = 0$) for any constant $c$ such that $\lambda + c \neq 0$.

(A) The given lines are $2x + 3y = 6$ and $2x + 3y = 8$.
To find the $X$-intercepts,set $y = 0$:
For $2x + 3y = 6$,$2x = 6 \Rightarrow x = 3$. So,$A = (3, 0)$.
For $2x + 3y = 8$,$2x = 8 \Rightarrow x = 4$. So,$B = (4, 0)$.
The line $L$ passes through $(2, 2)$ and cuts the $X$-axis at $C(x_1, 0)$.
Given that the abscissae of $A, B,$ and $C$ are in arithmetic progression,we have $3, 4, x_1$ in $AP$.
Thus,$2(4) = 3 + x_1$ $\Rightarrow 8 = 3 + x_1$ $\Rightarrow x_1 = 5$.
So,the point $C$ is $(5, 0)$.
The equation of the line $L$ passing through $(2, 2)$ and $(5, 0)$ is given by:
$y - 0 = \frac{2 - 0}{2 - 5}(x - 5)$
$y = \frac{2}{-3}(x - 5)$
$-3y = 2x - 10$
$2x + 3y = 10$

(D) The normal form 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 the normal makes with the positive $X$-axis.
Given $p = 10$.
The normal makes an angle of $\frac{\pi}{4}$ with the negative $X$-axis in the negative direction (clockwise).
The negative $X$-axis corresponds to an angle of $\pi$.
Moving $\frac{\pi}{4}$ in the negative direction (clockwise) from $\pi$ gives $\alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Substituting these values into the normal form equation:
$x \cos(\frac{3\pi}{4}) + y \sin(\frac{3\pi}{4}) = 10$
$x(-\frac{1}{\sqrt{2}}) + y(\frac{1}{\sqrt{2}}) = 10$
$-x + y = 10\sqrt{2}$
$x - y + 10\sqrt{2} = 0$.