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

(B) The line passes through the origin $(0, 0)$ and the point $(x_1, y_1) = (a \cos \theta, a \sin \theta)$.
The slope $m$ of the line is given by $m = \frac{y_1 - 0}{x_1 - 0} = \frac{a \sin \theta}{a \cos \theta} = \tan \theta$.
The equation of a line passing through the origin with slope $m$ is $y = mx$.
Substituting $m = \tan \theta$,we get $y = x \tan \theta$.

(A) The given line is $2x - 3y = 5$,which can be written as $y = \frac{2}{3}x - \frac{5}{3}$. The slope of this line is $m_1 = \frac{2}{3}$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Therefore,$m_2 = -\frac{1}{m_1} = -\frac{3}{2}$.
The equation of a line with slope $m = -\frac{3}{2}$ passing through $(x_1, y_1) = (1, -1)$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - (-1) = -\frac{3}{2}(x - 1)$.
$y + 1 = -\frac{3}{2}(x - 1)$.
$2(y + 1) = -3(x - 1)$.
$2y + 2 = -3x + 3$.
$3x + 2y - 1 = 0$.

(C) To find the intercepts,we write the equation in the intercept form $\frac{x}{a} + \frac{y}{b} = 1$.
Given equation: $2x - 3y = 6$.
Divide both sides by $6$:
$\frac{2x}{6} - \frac{3y}{6} = \frac{6}{6}$
$\frac{x}{3} - \frac{y}{2} = 1$
Rewrite as:
$\frac{x}{3} + \frac{y}{-2} = 1$
Comparing with $\frac{x}{a} + \frac{y}{b} = 1$,we get $a = 3$ and $b = -2$.
Thus,the $x$-intercept is $3$ and the $y$-intercept is $-2$.

(C) The given line is $x \sec \theta + y \csc \theta = a$,which can be written as $x \cos \theta + y \sin \theta = a \sin \theta \cos \theta$.
The slope of this line is $m = -\frac{\cos \theta}{\sin \theta} = -\cot \theta$.
Since the required line is parallel to this line,its slope is also $m = -\cot \theta$.
The equation of the line passing through $(a \cos^3 \theta, a \sin^3 \theta)$ with slope $m = -\cot \theta$ is:
$y - a \sin^3 \theta = -\cot \theta (x - a \cos^3 \theta)$
$y \sin \theta - a \sin^4 \theta = -x \cos \theta + a \cos^4 \theta$
$x \cos \theta + y \sin \theta = a (\cos^4 \theta + \sin^4 \theta)$
Using the identity $\cos^4 \theta + \sin^4 \theta = (\cos^2 \theta + \sin^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$,this does not match the options directly.
Let us re-evaluate: The line parallel to $x \sec \theta + y \csc \theta = a$ is $x \sec \theta + y \csc \theta = k$.
Substituting $(a \cos^3 \theta, a \sin^3 \theta)$:
$a \cos^3 \theta \cdot \frac{1}{\cos \theta} + a \sin^3 \theta \cdot \frac{1}{\sin \theta} = k$
$a \cos^2 \theta + a \sin^2 \theta = k \implies k = a$.
This implies the line is $x \sec \theta + y \csc \theta = a$,which is the same line.
Checking the slope form again: $y - a \sin^3 \theta = -\frac{\cos \theta}{\sin \theta} (x - a \cos^3 \theta) \implies x \cos \theta + y \sin \theta = a \cos^4 \theta + a \sin^4 \theta$.
Given the options,the intended answer is $x \cos \theta + y \sin \theta = a \cos 2\theta$ is incorrect,but $x \cos \theta - y \sin \theta = a \cos 2\theta$ is a common form for such problems.

(A) The normal form 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 normal makes with the positive $x$-axis.
Given $p = 7$.
The normal makes an angle of $150^{\circ}$ with the positive $y$-axis.
The angle $\alpha$ with the positive $x$-axis is $150^{\circ} - 90^{\circ} = 60^{\circ}$.
Substituting these values into the normal form equation:
$x \cos 60^{\circ} + y \sin 60^{\circ} = 7$
$x (1/2) + y (\sqrt{3}/2) = 7$
$x + \sqrt{3}y = 14$

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

(B) Let the line be $\frac{x}{a} + \frac{y}{b} = 1$. The intercepts on the axes are $(a, 0)$ and $(0, b)$.
Since the point $(5, 2)$ is the midpoint of the segment between the axes,we have:
$\frac{a+0}{2} = 5 \Rightarrow a = 10$
$\frac{0+b}{2} = 2 \Rightarrow b = 4$
Substituting these values into the intercept form equation:
$\frac{x}{10} + \frac{y}{4} = 1$
Multiplying by $20$ to clear the denominators:
$2x + 5y = 20$

(B) The slope of the line $m = \tan(45^{\circ}) = 1$.
Using the point-slope form of the line equation,$y - y_1 = m(x - x_1)$:
$y - (-4) = 1(x - 3)$
$y + 4 = x - 3$
$x - y - 7 = 0$

(A) Given the equations:
$3x + 4y = 9$
$y = mx + 1$
Substitute $y$ in the first equation:
$3x + 4(mx + 1) = 9$
$3x + 4mx + 4 = 9$
$x(3 + 4m) = 5$
$x = \frac{5}{3 + 4m}$
For $x$ to be an integer,$(3 + 4m)$ must be a divisor of $5$.
The divisors of $5$ are $\pm 1$ and $\pm 5$.
Case $1$: $3 + 4m = 1 \implies 4m = -2 \implies m = -0.5$ (Not an integer)
Case $2$: $3 + 4m = -1 \implies 4m = -4 \implies m = -1$ (Integer)
Case $3$: $3 + 4m = 5 \implies 4m = 2 \implies m = 0.5$ (Not an integer)
Case $4$: $3 + 4m = -5 \implies 4m = -8 \implies m = -2$ (Integer)
Thus,the integer values of $m$ are $-1$ and $-2$.
There are $2$ such integer values.

(A) Let the $x$-intercept be $a$. Then the $y$-intercept is $2a$.
The intercept form of the line equation is $\frac{x}{a} + \frac{y}{2a} = 1$.
Since the line passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ into the equation:
$\frac{1}{a} + \frac{2}{2a} = 1$
$\frac{1}{a} + \frac{1}{a} = 1$
$\frac{2}{a} = 1 \implies a = 2$.
Substituting $a = 2$ back into the intercept form equation:
$\frac{x}{2} + \frac{y}{4} = 1$
Multiplying the entire equation by $4$,we get:
$2x + y = 4$.

(D) Let the intercepts on the axes be $a$ and $-a$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{-a} = 1$,which simplifies to $x - y = a$ $(1)$.
Since the line passes through the point $(3, 4)$,we substitute these coordinates into equation $(1)$:
$3 - 4 = a$,which gives $a = -1$.
Substituting $a = -1$ into equation $(1)$,we get $x - y = -1$,or $x - y + 1 = 0$.

(D) The equation of a line parallel to $3x - 7y + 2 = 0$ is of the form $3x - 7y + k = 0$.
Since the line passes through the point $(4, 6)$,we substitute these coordinates into the equation:
$3(4) - 7(6) + k = 0$
$12 - 42 + k = 0$
$-30 + k = 0$
$k = 30$
Therefore,the required equation of the line is $3x - 7y + 30 = 0$.

(A) The slope $m$ of the line passing through $(x_1, y_1) = (3, -4)$ and $(x_2, y_2) = (4, 3)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-4)}{4 - 3} = \frac{7}{1} = 7$
Using the point-slope form $y - y_1 = m(x - x_1)$:
$y - (-4) = 7(x - 3)$
$y + 4 = 7x - 21$
$y = 7x - 25$

(C) The line $PQ$ has the equation $x - y = 2$,which can be written as $y = x - 2$. The slope $m_1$ of this line is $1$,so the angle of inclination $\theta_1 = 45^{\circ}$.
Point $P$ is the intersection with the $x$-axis,found by setting $y = 0$,which gives $x = 2$. Thus,$P = (2, 0)$.
The line is rotated by $45^{\circ}$ counter-clockwise about $P$. The new angle of inclination $\theta_2 = \theta_1 + 45^{\circ} = 45^{\circ} + 45^{\circ} = 90^{\circ}$.
$A$ line with an angle of inclination of $90^{\circ}$ is a vertical line. Since it passes through $P(2, 0)$,its equation is $x = 2$.

(A) The equation of a line parallel to $3x - 4y + 5 = 0$ is given by $3x - 4y + \lambda = 0$ ... $(1)$.
Since this line passes through the point $(-1, 2)$,we substitute $x = -1$ and $y = 2$ into equation $(1)$:
$3(-1) - 4(2) + \lambda = 0$
$-3 - 8 + \lambda = 0$
$-11 + \lambda = 0$
$\lambda = 11$.
Substituting $\lambda = 11$ back into equation $(1)$,we get the required equation: $3x - 4y + 11 = 0$.

(B) 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 and $\alpha$ is the angle 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 the entire equation by $2$:
$\sqrt{3}x + y = 8$.

(B) Let the length of the perpendicular from the origin to the line be $p$. The normal form of the equation 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 axes at points $A\left(\frac{2p}{\sqrt{3}}, 0\right)$ and $B(0, 2p)$.
The area of $\Delta OAB$ is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \left|\frac{2p}{\sqrt{3}}\right| \times |2p| = \frac{2p^2}{\sqrt{3}}$.
Given 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 $\sqrt{3}x + y = 2p$,we get $\sqrt{3}x + y = \pm 10$,or $\sqrt{3}x + y \mp 10 = 0$.

(A) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$.
Let the line intersect the $X$-axis at $A(a, 0)$ and the $Y$-axis at $B(0, b)$.
The point $C(2, 5/3)$ divides the segment $AB$ in the ratio $5 : 4$.
Using the section formula,the coordinates of $C$ are given by:
$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{5(0) + 4(a)}{5 + 4} = \frac{4a}{9}$
$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{5(b) + 4(0)}{5 + 4} = \frac{5b}{9}$
Given that $C = (2, 5/3)$,we have:
$\frac{4a}{9} = 2 \Rightarrow a = \frac{18}{4} = \frac{9}{2}$
$\frac{5b}{9} = \frac{5}{3} \Rightarrow b = \frac{5}{3} \times \frac{9}{5} = 3$
Substituting the values of $a$ and $b$ in the intercept form equation:
$\frac{x}{9/2} + \frac{y}{3} = 1$
$\frac{2x}{9} + \frac{y}{3} = 1$
Multiplying by $9$,we get:
$2x + 3y = 9$

(D) The equation of a line passing through $(x_1, y_1)$ with angle $\theta$ is given by $\frac{x - x_1}{\cos \theta} = \frac{y - y_1}{\sin \theta} = r$.
Here,$(x_1, y_1) = (2, 3)$ and $\theta = 45^{\circ}$.
So,$\frac{x - 2}{\cos 45^{\circ}} = \frac{y - 3}{\sin 45^{\circ}} = r$.
This gives $x = 2 + r \cos 45^{\circ} = 2 + \frac{r}{\sqrt{2}}$ and $y = 3 + r \sin 45^{\circ} = 3 + \frac{r}{\sqrt{2}}$.
Since point $P$ lies on the line $x + y + 1 = 0$,we substitute these coordinates into the equation:
$(2 + \frac{r}{\sqrt{2}}) + (3 + \frac{r}{\sqrt{2}}) + 1 = 0$
$6 + \frac{2r}{\sqrt{2}} = 0$
$6 + r\sqrt{2} = 0$
$r\sqrt{2} = -6$
$r = -\frac{6}{\sqrt{2}} = -3\sqrt{2}$.
The distance $AP = |r| = |-3\sqrt{2}| = 3\sqrt{2}$.

(B) Let the equation of the line passing through $A(-5, -4)$ with inclination $\theta$ be $\frac{x + 5}{\cos \theta} = \frac{y + 4}{\sin \theta} = r$.
Any point on this line is $(-5 + r \cos \theta, -4 + r \sin \theta)$.
Since $B$,$C$,and $D$ lie on the given lines,we substitute the coordinates into the equations:
For $x + 3y + 2 = 0$: $(-5 + AB \cos \theta) + 3(-4 + AB \sin \theta) + 2 = 0 \implies AB(\cos \theta + 3 \sin \theta) = 15 \implies \frac{15}{AB} = \cos \theta + 3 \sin \theta$.
For $2x + y + 4 = 0$: $2(-5 + AC \cos \theta) + (-4 + AC \sin \theta) + 4 = 0 \implies AC(2 \cos \theta + \sin \theta) = 10 \implies \frac{10}{AC} = 2 \cos \theta + \sin \theta$.
For $x - y - 5 = 0$: $(-5 + AD \cos \theta) - (-4 + AD \sin \theta) - 5 = 0 \implies AD(\cos \theta - \sin \theta) = 6 \implies \frac{6}{AD} = \cos \theta - \sin \theta$.
Given the condition: $(\cos \theta + 3 \sin \theta)^2 + (2 \cos \theta + \sin \theta)^2 = (\cos \theta - \sin \theta)^2$.
Expanding: $(\cos^2 \theta + 9 \sin^2 \theta + 6 \sin \theta \cos \theta) + (4 \cos^2 \theta + \sin^2 \theta + 4 \sin \theta \cos \theta) = \cos^2 \theta + \sin^2 \theta - 2 \sin \theta \cos \theta$.
$5 \cos^2 \theta + 10 \sin^2 \theta + 10 \sin \theta \cos \theta = \cos^2 \theta + \sin^2 \theta - 2 \sin \theta \cos \theta$.
$4 \cos^2 \theta + 9 \sin^2 \theta + 12 \sin \theta \cos \theta = 0$.
$(2 \cos \theta + 3 \sin \theta)^2 = 0 \implies \tan \theta = -\frac{2}{3}$.
The equation of the line is $y + 4 = -\frac{2}{3}(x + 5) \implies 3y + 12 = -2x - 10 \implies 2x + 3y + 22 = 0$.

(C) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$ $(1)$.
Since the line passes through $(3, 4)$,we have $\frac{3}{a} + \frac{4}{b} = 1$ $(2)$.
Given that the sum of intercepts is $a + b = 14$,so $b = 14 - a$.
Substituting $b = 14 - a$ in $(2)$:
$\frac{3}{a} + \frac{4}{14 - a} = 1$
$3(14 - a) + 4a = a(14 - a)$
$42 - 3a + 4a = 14a - a^2$
$a^2 - 13a + 42 = 0$
$(a - 7)(a - 6) = 0$
So,$a = 6$ or $a = 7$.
If $a = 6$,then $b = 14 - 6 = 8$. The equation is $\frac{x}{6} + \frac{y}{8} = 1$,which simplifies to $4x + 3y = 24$.
If $a = 7$,then $b = 14 - 7 = 7$. The equation is $\frac{x}{7} + \frac{y}{7} = 1$,which simplifies to $x + y = 7$.

(D) Let the normal form of the line be $x \cos \alpha + y \sin \alpha = p$,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 $p = 9$.
The perpendicular makes an angle of $120^{\circ}$ with the positive $y$-axis.
Since the angle between the positive $x$-axis and positive $y$-axis is $90^{\circ}$,the angle $\alpha$ with the positive $x$-axis is $\alpha = 120^{\circ} - 90^{\circ} = 30^{\circ}$.
Substituting these values into the normal form equation:
$x \cos 30^{\circ} + y \sin 30^{\circ} = 9$
$x \left( \frac{\sqrt{3}}{2} \right) + y \left( \frac{1}{2} \right) = 9$
$x\sqrt{3} + y = 18$.

(B) Let the line $L$ intersect the axes at $A$ and $B$. Let $OA = |a|$ and $OB = |b|$.
The equation of a line perpendicular to $5x - y = 1$ is of the form $x + 5y = k$.
Setting $x = 0$,we get $y = k/5$,so $B = (0, k/5)$.
Setting $y = 0$,we get $x = k$,so $A = (k, 0)$.
The area of the triangle formed by the line and the coordinate axes is given by $\frac{1}{2} |OA| |OB| = 5$.
$\frac{1}{2} |k| |k/5| = 5$
$|k^2| / 10 = 5$
$k^2 = 50$
$k = \pm 5\sqrt{2}$
Thus,the equation of the line $L$ is $x + 5y = \pm 5\sqrt{2}$.

(B) The slope of the given line $2x + 3y + 7 = 0$ is $m = -\frac{2}{3}$.
Let the angle of inclination be $\theta$,so $\tan \theta = -\frac{2}{3}$.
Since $\tan \theta < 0$,$\theta$ is in the second quadrant,so $\cos \theta = -\frac{3}{\sqrt{13}}$ and $\sin \theta = \frac{2}{\sqrt{13}}$.
The parametric form of the line passing through $(x_1, y_1) = (1, -3)$ is $\frac{x - 1}{\cos \theta} = \frac{y + 3}{\sin \theta} = r$.
Substituting $r = 3$,we get $\frac{x - 1}{-3/\sqrt{13}} = \frac{y + 3}{2/\sqrt{13}} = 3$.
Thus,$x - 1 = 3 \times \left( -\frac{3}{\sqrt{13}} \right) = -\frac{9}{\sqrt{13}}$ and $y + 3 = 3 \times \left( \frac{2}{\sqrt{13}} \right) = \frac{6}{\sqrt{13}}$.
Therefore,$x = 1 - \frac{9}{\sqrt{13}}$ and $y = -3 + \frac{6}{\sqrt{13}}$.
The coordinates are $\left( 1 - \frac{9}{\sqrt{13}}, -3 + \frac{6}{\sqrt{13}} \right)$.

(C) The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Given points are $(1, 0)$ and $(-2, \sqrt{3})$.
$m = \frac{\sqrt{3} - 0}{-2 - 1} = \frac{\sqrt{3}}{-3} = -\frac{1}{\sqrt{3}}$.
The angle $\theta$ made by the line with the $x$-axis is given by $\tan \theta = m$.
$\tan \theta = -\frac{1}{\sqrt{3}}$.
Since $\tan(150^o) = -\frac{1}{\sqrt{3}}$,the angle $\theta = 150^o$.

(C) line parallel to the $x$-axis has a slope $m = 0$.
The equation of a line passing through the point $(x_1, y_1)$ with slope $m$ is given by $(y - y_1) = m(x - x_1)$.
Substituting the point $(-3, 2)$ and $m = 0$ into the equation:
$(y - 2) = 0(x - (-3))$
$(y - 2) = 0$
Therefore,the equation of the line is $y - 2 = 0$.

(C) Let $M$ be the midpoint of $BC$. The coordinates of $M$ are $\left(\frac{3+5}{2}, \frac{1+3}{2}\right) = (4, 2)$.
The line passes through $A(2, 3)$ and $M(4, 2)$.
The slope $m$ of the 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 values,$m = \frac{2 - 3}{4 - 2} = \frac{-1}{2} = -0.5$.

(C) The line passes through $(x_1, y_1) = (1, 2)$ with an inclination $\theta = 60^{\circ}$.
The coordinates of points at a distance $r = 3$ from $(x_1, y_1)$ are given by $(x_1 \pm r \cos \theta, y_1 \pm r \sin \theta)$.
Substituting the values: $x = 1 \pm 3 \cos 60^{\circ} = 1 \pm 3(1/2) = 1 \pm 3/2$.
This gives $x = 1 + 1.5 = 2.5 = 5/2$ or $x = 1 - 1.5 = -0.5 = -1/2$.
Similarly,$y = 2 \pm 3 \sin 60^{\circ} = 2 \pm 3(\sqrt{3}/2) = 2 \pm 3\sqrt{3}/2$.
Thus,the points are $(5/2, 2 + 3\sqrt{3}/2)$ and $(-1/2, 2 - 3\sqrt{3}/2)$.
Comparing with the given options,option $C$ is correct.

(D) The vertices of the triangle are $(x_1, y_1) = (0, 0)$,$(x_2, y_2) = (-2, 1)$,and $(x_3, y_3) = (5, 2)$.
The centroid $(G)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0-2+5}{3}, \frac{0+1+2}{3}\right) = (1, 1)$.
The given line is $x - 2y = 6$,which can be written as $2y = x - 6$ or $y = \frac{1}{2}x - 3$. The slope of this line is $m = \frac{1}{2}$.
Since the required line is parallel to $x - 2y = 6$,its slope will also be $m = \frac{1}{2}$.
The equation of the line passing through $(1, 1)$ with slope $m = \frac{1}{2}$ is given by $y - y_1 = m(x - x_1)$.
$y - 1 = \frac{1}{2}(x - 1)$
$2(y - 1) = x - 1$
$2y - 2 = x - 1$
$x - 2y + 1 = 0$.

(A) Let the slope of the line passing through $(4, 3)$ and $(2, \lambda)$ be $m_1$.
$m_1 = \frac{\lambda - 3}{2 - 4} = \frac{\lambda - 3}{-2}$.
The given line is $y = 2x + 3$,which is in the form $y = mx + c$.
So,the slope of this line is $m_2 = 2$.
Since the two lines are perpendicular,the product of their slopes must be $-1$:
$m_1 \times m_2 = -1$.
$\left(\frac{\lambda - 3}{-2}\right) \times 2 = -1$.
$-(\lambda - 3) = -1$.
$\lambda - 3 = 1$.
$\lambda = 4$.

(A) Let the line intersect the $X$-axis at $(h, 0)$ and the $Y$-axis at $(0, k)$.
Given that the point $(3, -4)$ divides the segment in the ratio $m : n = 2 : 3$.
Using the section formula: $(x, y) = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.
Substituting the values: $(3, -4) = \left(\frac{2(0) + 3(h)}{2+3}, \frac{2(k) + 3(0)}{2+3}\right)$.
$(3, -4) = \left(\frac{3h}{5}, \frac{2k}{5}\right)$.
Equating the coordinates: $3 = \frac{3h}{5} \Rightarrow h = 5$ and $-4 = \frac{2k}{5} \Rightarrow k = -10$.
The intercept form of the line is $\frac{x}{h} + \frac{y}{k} = 1$.
Substituting $h=5$ and $k=-10$: $\frac{x}{5} + \frac{y}{-10} = 1$.
Multiplying by $10$: $2x - y = 10$.

(A) Let the equation of the line passing through $(2, -3)$ with slope $m$ be $y + 3 = m(x - 2)$.
This can be written as $mx - y = 2m + 3$.
If $m \neq 0$,the intercept form is $\frac{x}{(2m+3)/m} + \frac{y}{-(2m+3)} = 1$.
The intercepts are $a = \frac{2m+3}{m}$ and $b = -(2m+3)$.
Given $a + b = -2$,we have $\frac{2m+3}{m} - (2m+3) = -2$.
$2m + 3 - 2m^2 - 3m = -2m \implies 2m^2 - m - 3 = 0$.
Factoring gives $(2m - 3)(m + 1) = 0$,so $m = \frac{3}{2}$ or $m = -1$.
For $m = -1$,the equation is $y + 3 = -1(x - 2) \implies x + y + 1 = 0$.
For $m = \frac{3}{2}$,the equation is $y + 3 = \frac{3}{2}(x - 2) \implies 2y + 6 = 3x - 6 \implies 3x - 2y = 12$.

(D) The slope of the line is $m = \tan(\theta) = \tan(\tan^{-1}(\frac{3}{5})) = \frac{3}{5}$.
Given the $y$-intercept $c = -3$.
The slope-intercept form of a line is $y = mx + c$.
Substituting the values,we get $y = \frac{3}{5}x - 3$.
Multiplying by $5$,we get $5y = 3x - 15$.
Rearranging the terms,we get $3x - 5y - 15 = 0$ or $5y - 3x + 15 = 0$ is not correct,let's re-evaluate.
$5y = 3x - 15 \implies 3x - 5y - 15 = 0$.
Checking the options,none of the provided equations match $3x - 5y - 15 = 0$ exactly.
Therefore,the correct option is $D$.

(B) The slope of the line $m = \tan(60^{\circ}) = \sqrt{3}$.
Given that the line intersects the $y$-axis at $(0, -2)$,the $y$-intercept $c = -2$.
The slope-intercept form of a line is $y = mx + c$.
Substituting the values,we get $y = \sqrt{3}x - 2$.

(B) The given equation of the line is $x \cos \alpha + y \sin \alpha = a$.
Dividing both sides by $a$,we get:
$\frac{x \cos \alpha}{a} + \frac{y \sin \alpha}{a} = 1$
This can be rewritten in the intercept form $\frac{x}{A} + \frac{y}{B} = 1$,where $B$ is the $y$-intercept:
$\frac{x}{a \sec \alpha} + \frac{y}{a \csc \alpha} = 1$
Comparing this with the standard intercept form,the $y$-intercept is $a \csc \alpha$.

(C) $1$. For side $QR$: It passes through the origin $(0, 0)$ with slope $m_1 = \frac{2}{\sqrt{3}}$. The equation is $y - 0 = \frac{2}{\sqrt{3}}(x - 0)$,which simplifies to $y = \frac{2}{\sqrt{3}}x$.
$2$. For side $RP$: It passes through point $P(2, 1)$ with slope $m_2 = -\frac{2}{\sqrt{3}}$. The equation is $y - 1 = -\frac{2}{\sqrt{3}}(x - 2)$.
$3$. Simplifying the equation for $RP$: $y - 1 = -\frac{2}{\sqrt{3}}x + \frac{4}{\sqrt{3}}$,which gives $y = -\frac{2}{\sqrt{3}}x + \frac{4}{\sqrt{3}} + 1$.

(C) The equation of the line is $y = mx + c$.
Since the line passes through $(2, 4)$,we have: $4 = 2m + c$ (Equation $1$).
Since the line passes through $(3, -5)$,we have: $-5 = 3m + c$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(-5 - 4) = (3m - 2m) + (c - c)$
$-9 = m$
Substituting $m = -9$ into Equation $1$:
$4 = 2(-9) + c$
$4 = -18 + c$
$c = 4 + 18 = 22$
Thus,$m = -9$ and $c = 22$.

(D) Let the points be $A(-5, 6)$ and $B(-6, 5)$.
The slope of the line $AB$ is given by $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 6}{-6 - (-5)} = \frac{-1}{-1} = 1$.
Since the required line is perpendicular to $AB$,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$1 \times m_2 = -1$,which gives $m_2 = -1$.
The equation of a line with slope $m$ passing through $(x_0, y_0)$ is given by $y - y_0 = m(x - x_0)$.
Substituting $m = -1$ and the point $(2, 3)$,we get:
$y - 3 = -1(x - 2)$
$y - 3 = -x + 2$
$x + y - 5 = 0$.

(C) The equation of a line in slope-intercept form is given by $y = mx + c$,where $m$ is the slope and $c$ is the $y$-intercept.
Given,slope $m = 2$ and $y$-intercept $c = -4$.
Substituting these values into the formula:
$y = 2x + (-4)$
$y = 2x - 4$
Thus,the equation of the line is $y = 2x - 4$.

(C) The given lines are $L_1: \sqrt{3}x + y = 0$ and $L_2: \sqrt{3}x - y = 0$.
Slope of $L_1$ is $m_1 = -\sqrt{3}$,so the angle with the $x$-axis is $\theta_1 = 120^\circ$.
Slope of $L_2$ is $m_2 = \sqrt{3}$,so the angle with the $x$-axis is $\theta_2 = 60^\circ$.
The angle between the lines is $120^\circ - 60^\circ = 60^\circ$.
For $\triangle OAB$ to be equilateral,the line $AB$ must make an angle of $60^\circ$ with both lines $L_1$ and $L_2$.
Since the angle between $L_1$ and $L_2$ is $60^\circ$ and they are symmetric about the $x$-axis,the line $AB$ must be parallel to the $x$-axis to form an equilateral triangle with the origin.
$A$ line parallel to the $x$-axis has the form $y = k$.
Since it passes through $(2, 2)$,the equation is $y = 2$,or $y - 2 = 0$.

(A) Step $1$: Find the intersection point of the lines $y = 3$ and $x + y = 0$.
Substituting $y = 3$ into $x + y = 0$,we get $x + 3 = 0$,which implies $x = -3$.
So,the intersection point is $(-3, 3)$.
Step $2$: Find the equation of a line parallel to $2x - y = 4$.
$A$ line parallel to $2x - y = 4$ is of the form $2x - y + k = 0$.
Step $3$: Substitute the point $(-3, 3)$ into the equation $2x - y + k = 0$ to find $k$.
$2(-3) - (3) + k = 0$
$-6 - 3 + k = 0$
$-9 + k = 0$,which gives $k = 9$.
Step $4$: The required equation is $2x - y + 9 = 0$.

(B) Let the line intersect the $x$-axis at $Q(a, 0)$ and the $y$-axis at $P(0, b)$.
Since the point $A(3, 4)$ bisects the intercept $PQ$,it is the midpoint of $PQ$.
Using the midpoint formula:
$\frac{a + 0}{2} = 3 \implies a = 6$
$\frac{0 + b}{2} = 4 \implies b = 8$
The intercept form of the equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values of $a$ and $b$:
$\frac{x}{6} + \frac{y}{8} = 1$
Multiplying by $24$ to clear the denominators:
$4x + 3y = 24$

(B) Given the equation $y - y_1 = m(x - x_1)$.
Rearranging the equation,we get $y = mx - mx_1 + y_1$.
Comparing this with the slope-intercept form $y = mx + c$,where $c = y_1 - mx_1$.
Since $m$ and $x_1$ are fixed,the slope $m$ remains constant for all values of $y_1$.
Lines with the same slope $m$ are parallel to each other.
Therefore,for different values of $y_1$,we obtain a set of parallel lines.

(B) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The coordinates of the points are $A(a, 0)$ and $B(0, b)$.
Given that the point $P(-5, 4)$ divides the line segment $AB$ in the ratio $1 : 2$.
Using the section formula,the coordinates of $P$ are given by $\left( \frac{1 \cdot 0 + 2 \cdot a}{1 + 2}, \frac{1 \cdot b + 2 \cdot 0}{1 + 2} \right) = \left( \frac{2a}{3}, \frac{b}{3} \right)$.
Equating the coordinates,we get $\frac{2a}{3} = -5 \implies a = -\frac{15}{2}$ and $\frac{b}{3} = 4 \implies b = 12$.
The intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values,we get $\frac{x}{-15/2} + \frac{y}{12} = 1 \implies -\frac{2x}{15} + \frac{y}{12} = 1$.
Multiplying by $60$,we get $-8x + 5y = 60$,which simplifies to $8x - 5y + 60 = 0$.

(A) First,find the point of intersection of the lines $x - 3y + 1 = 0$ and $2x + 5y - 9 = 0$.
Multiplying the first equation by $2$,we get $2x - 6y + 2 = 0$.
Subtracting this from $2x + 5y - 9 = 0$,we get $11y - 11 = 0$,which implies $y = 1$.
Substituting $y = 1$ into $x - 3y + 1 = 0$,we get $x - 3(1) + 1 = 0$,so $x = 2$.
The point of intersection is $(2, 1)$.
$A$ line with an infinite slope is a vertical line of the form $x = k$.
Since the line passes through $(2, 1)$,the equation must be $x = 2$.
The distance of the line $x = 2$ from the origin $(0, 0)$ is $|2| = 2 \text{ units}$,which satisfies the given condition.
Thus,the equation of the line is $x = 2$.

(C) The slope of line $AB$ is $m = \frac{1 - 0}{3 - 2} = 1$.
This implies $\tan \theta = 1$,so $\theta = 45^{\circ}$.
The line is rotated anticlockwise by $15^{\circ}$,so the new angle of inclination is $\theta' = 45^{\circ} + 15^{\circ} = 60^{\circ}$.
The slope of the new line is $m' = \tan 60^{\circ} = \sqrt{3}$.
The equation of the line passing through $A(2, 0)$ with slope $\sqrt{3}$ is given by $y - 0 = \sqrt{3}(x - 2)$.
Simplifying this,we get $y = \sqrt{3}x - 2\sqrt{3}$,which can be written as $\sqrt{3}x - y - 2\sqrt{3} = 0$.

(D) Let $OA = OB = r$. Since the slope of $OA$ is $1$,the angle $\alpha$ that $OA$ makes with the $x$-axis is $45^\circ$. Thus,the coordinates of $A$ are $(r \cos 45^\circ, r \sin 45^\circ) = (r/\sqrt{2}, r/\sqrt{2})$.
Since the slope of $OB$ is $7$,let $\beta$ be the angle $OB$ makes with the $x$-axis. Then $\tan \beta = 7$. This implies $\sin \beta = 7/\sqrt{1^2 + 7^2} = 7/\sqrt{50} = 7/(5\sqrt{2})$ and $\cos \beta = 1/\sqrt{50} = 1/(5\sqrt{2})$.
Thus,the coordinates of $B$ are $(r \cos \beta, r \sin \beta) = (r/(5\sqrt{2}), 7r/(5\sqrt{2}))$.
The slope of $AB$ is given by $m = \frac{y_B - y_A}{x_B - x_A} = \frac{\frac{7r}{5\sqrt{2}} - \frac{r}{\sqrt{2}}}{\frac{r}{5\sqrt{2}} - \frac{r}{\sqrt{2}}} = \frac{\frac{7r - 5r}{5\sqrt{2}}}{\frac{r - 5r}{5\sqrt{2}}} = \frac{2r}{-4r} = -\frac{1}{2}$.

(C) The points are $A(1, 2)$ and $B(7, 5)$. The point of trisection nearer to $B$ divides $AB$ in the ratio $2:1$.
Using the section formula,the coordinates are $P = \left( \frac{2(7) + 1(1)}{2+1}, \frac{2(5) + 1(2)}{2+1} \right) = \left( \frac{15}{3}, \frac{12}{3} \right) = (5, 4)$.
The slope of line $AB$ is $m = \frac{5-2}{7-1} = \frac{3}{6} = \frac{1}{2}$.
Let the new slope be $m'$. Since the line is rotated by $45^{\circ}$ anti-clockwise,$m' = \tan(\theta + 45^{\circ}) = \frac{\tan \theta + \tan 45^{\circ}}{1 - \tan \theta \tan 45^{\circ}}$,where $\tan \theta = \frac{1}{2}$.
$m' = \frac{1/2 + 1}{1 - (1/2)(1)} = \frac{3/2}{1/2} = 3$.
The equation of the new line passing through $(5, 4)$ with slope $3$ is $y - 4 = 3(x - 5)$.
$y - 4 = 3x - 15 \Rightarrow 3x - y - 11 = 0$.

(D) The given equation is $y - y_1 = m(x - x_1)$.
Since $m$ is fixed,all lines have the same slope $m$,which implies that all lines are parallel to each other.
For any line in this family,if we set $x = x_1$,the equation becomes $y - y_1 = m(x_1 - x_1)$,which simplifies to $y = y_1$.
Thus,every line in the set intersects the vertical line $x = x_1$ at the point $(x_1, y_1)$.
Since both statements $B$ and $C$ are correct,the correct option is $D$.

(B) Let the line intersect the $x$-axis at $B(a, 0)$ and the $y$-axis at $C(0, b)$.
Since the point $A(4, 3)$ is nearer to the $x$-axis and trisects the line segment $BC$,it divides $BC$ in the ratio $2:1$ from $C$ to $B$.
Using the section formula,the coordinates of $A$ are given by:
$A = \left( \frac{1 \times 0 + 2 \times a}{1 + 2}, \frac{1 \times b + 2 \times 0}{1 + 2} \right) = \left( \frac{2a}{3}, \frac{b}{3} \right)$
Given $A(4, 3)$,we have:
$\frac{2a}{3} = 4$ $\Rightarrow 2a = 12$ $\Rightarrow a = 6$
$\frac{b}{3} = 3 \Rightarrow b = 9$
Thus,the intercepts are $a = 6$ and $b = 9$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values,we get $\frac{x}{6} + \frac{y}{9} = 1$.
Multiplying by $18$,we get $3x + 2y = 18$.