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

(B) Let $A = (-2, -5)$,$B = (2, -2)$,and $C = (8, a)$.
Since the points are collinear,the slope of $AB$ must be equal to the slope of $BC$.
Slope of $AB = \frac{-2 - (-5)}{2 - (-2)} = \frac{-2 + 5}{2 + 2} = \frac{3}{4}$.
Slope of $BC = \frac{a - (-2)}{8 - 2} = \frac{a + 2}{6}$.
Equating the slopes: $\frac{3}{4} = \frac{a + 2}{6}$.
$18 = 4(a + 2) \Rightarrow 18 = 4a + 8$.
$4a = 10 \Rightarrow a = \frac{10}{4} = 2.5$.

(A) The slope of the line $AB$ is $m = \frac{6 - 3}{7 - 3} = \frac{3}{4}$.
The equation of the line $AB$ is $y - 3 = \frac{3}{4}(x - 3)$.
$4y - 12 = 3x - 9 \Rightarrow 3x - 4y + 3 = 0$.
To find the intercepts,we rewrite the equation as $3x - 4y = -3$,or $\frac{x}{-1} + \frac{y}{3/4} = 1$.
The intercepts are $(-1, 0)$ and $(0, 3/4)$.
The length of the segment intercepted between the axes is $\sqrt{(-1 - 0)^2 + (0 - 3/4)^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(C) The given line is $2x + 3y = 5$.
Rewriting this in slope-intercept form $y = mx + c$,we get $3y = -2x + 5$,which simplifies to $y = -\frac{2}{3}x + \frac{5}{3}$.
Comparing this with $y = mx + c$,the slope $m = -\frac{2}{3}$.
Two lines are parallel if they have the same slope but different $y$-intercepts (or are identical).
Since the question only specifies that the lines are parallel,the slope $m$ must be $-2/3$,while $c$ can be any real number except $5/3$ (if they are distinct parallel lines). However,in the context of general parallel lines,$m = -2/3$ is the necessary condition,and $c$ remains an arbitrary constant.

(B) The given equation of the line is $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$.
Rearranging the terms to group $x$ and $y$,we get:
$(3 + 2\lambda)x + (-1 - 3\lambda)y + (5 - 4\lambda) = 0$.
$A$ line is parallel to the $y$-axis if the coefficient of $y$ is $0$ and the coefficient of $x$ is non-zero.
Setting the coefficient of $y$ to $0$:
$-1 - 3\lambda = 0$
$-3\lambda = 1$
$\lambda = -1/3$.
Thus,the correct option is $B$.

(C) Solving $y = m_r x$ and $x + y = 1$,we get $x = \frac{1}{1 + m_r}$ and $y = \frac{m_r}{1 + m_r}$.
Thus,the points of intersection of the three lines on the transversal are $P_r = \left( \frac{1}{1 + m_r}, \frac{m_r}{1 + m_r} \right)$ for $r = 1, 2, 3$.
Since the intercepts between consecutive points are equal,the distance between $P_1$ and $P_2$ equals the distance between $P_2$ and $P_3$.
Using the distance formula,the condition for equal intercepts simplifies to $\frac{1}{1 + m_1} - \frac{1}{1 + m_2} = \frac{1}{1 + m_2} - \frac{1}{1 + m_3}$ (considering the projection on the x-axis).
This implies $\frac{2}{1 + m_2} = \frac{1}{1 + m_1} + \frac{1}{1 + m_3}$.
Therefore,$1 + m_1, 1 + m_2, 1 + m_3$ are in $H.P.$

(A) Given the curve $y = x^2 + 2x$.
For $x_1 = 1$,$y_1 = (1)^2 + 2(1) = 1 + 2 = 3$. So,the first point is $(1, 3)$.
For $x_2 = 3$,$y_2 = (3)^2 + 2(3) = 9 + 6 = 15$. So,the second point is $(3, 15)$.
The gradient $m$ of the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Substituting the values,$m = \frac{15 - 3}{3 - 1} = \frac{12}{2} = 6$.

(C) The first line is given by $ax + by + c = 0$. The slope of this line is $m_1 = -\frac{a}{b}$.
The second line is given parametrically by $x = \alpha t + \beta$ and $y = \gamma t + \delta$. The slope of this line is $m_2 = \frac{\gamma}{\alpha}$.
For two lines to be parallel,their slopes must be equal,so $m_1 = m_2$.
Substituting the slopes,we get $-\frac{a}{b} = \frac{\gamma}{\alpha}$.
Cross-multiplying,we get $-a\alpha = b\gamma$,which simplifies to $a\alpha + b\gamma = 0$.

(C) The intercept form of a line cutting off equal intercepts $a$ from the axes is $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$.
Since the line passes through the point $(1, -2)$,we substitute these coordinates into the equation:
$1 + (-2) = a$
$a = -1$.
Substituting $a = -1$ back into the equation $x + y = a$,we get $x + y = -1$,or $x + y + 1 = 0$.

(C) The equation of a line with slope $m$ and $y$-intercept $c$ is given by $y = mx + c$.
Given that the $y$-intercept $c = -1$.
Since the lines are equally inclined to the axes,the angle of inclination $\theta$ is $45^{\circ}$ or $135^{\circ}$.
Thus,the slope $m = \tan(45^{\circ}) = 1$ or $m = \tan(135^{\circ}) = -1$.
Case $1$: For $m = 1$ and $c = -1$,the equation is $y = 1x - 1$,which simplifies to $x - y - 1 = 0$.
Case $2$: For $m = -1$ and $c = -1$,the equation is $y = -1x - 1$,which simplifies to $x + y + 1 = 0$.
Therefore,the required equations are $x - y - 1 = 0$ and $x + y + 1 = 0$.

(B) The given line is $5x - y = 1$.
Any line perpendicular to $5x - y = 1$ is of the form $x + 5y = k$.
Writing this in intercept form: $\frac{x}{k} + \frac{y}{k/5} = 1$.
The intercepts on the coordinate axes are $a = k$ and $b = k/5$.
The area of the triangle formed by the line and the coordinate axes is given by $\frac{1}{2} |ab| = 5$.
Substituting the values: $\frac{1}{2} |k \cdot \frac{k}{5}| = 5$.
$|k^2| = 50$,which implies $k = \pm \sqrt{50} = \pm 5\sqrt{2}$.
Therefore,the equation of the line $L$ is $x + 5y = \pm 5\sqrt{2}$.

(A) The line has a slope $m = 3$ and cuts an intercept of $3$ on the positive $x$-axis. This means the line passes through the point $(3, 0)$.
Using the point-slope form of the equation of a line,$y - y_1 = m(x - x_1)$:
$y - 0 = 3(x - 3)$
$y = 3x - 9$
Thus,the required equation is $y = 3x - 9$.

(B) The midpoint $E$ of $AB$ is $\left( \frac{a + a'}{2}, \frac{b + b'}{2} \right)$.
The midpoint $F$ of $CD$ is $\left( \frac{-a + a'}{2}, \frac{b - b'}{2} \right)$.
The slope $m$ of the line $EF$ is given by:
$m = \frac{\frac{b - b'}{2} - \frac{b + b'}{2}}{\frac{a' - a}{2} - \frac{a' + a}{2}} = \frac{\frac{-2b'}{2}}{\frac{-2a}{2}} = \frac{b'}{a}$.
The equation of the line passing through $E$ with slope $m$ is:
$y - \frac{b + b'}{2} = \frac{b'}{a} \left( x - \frac{a + a'}{2} \right)$.
Multiplying by $2a$:
$2ay - a(b + b') = 2b'x - b'(a + a')$.
$2ay - ab - ab' = 2b'x - ab' - a'b'$.
$2ay - 2b'x = ab - a'b'$.

(B) The slope of the given line $y = x$ is $m_1 = 1$.
Since the required line is perpendicular to $y = x$,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 $m = -1$ passing through $(x_1, y_1) = (3, 2)$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values,we get $y - 2 = -1(x - 3)$.
$y - 2 = -x + 3$.
Rearranging the terms,we get $x + y = 5$.

(B) Let the midpoints be $E(1, 3)$ on $BC$,$D(5, 7)$ on $CA$,and $F(-5, 7)$ on $AB$.
By the midpoint theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side.
Therefore,the side $AB$ is parallel to the line segment $ED$.
The slope of $ED = \frac{7 - 3}{5 - 1} = \frac{4}{4} = 1$.
Since $AB$ is parallel to $ED$,the slope of $AB$ is also $1$.
The side $AB$ passes through the midpoint $F(-5, 7)$.
The equation of line $AB$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values,$y - 7 = 1(x - (-5))$.
$y - 7 = x + 5$.
$x - y + 12 = 0$.

(D) The given line is $\frac{x}{a} - \frac{y}{b} = 1$,which can be written as $bx - ay = ab$.
The line cuts the $x$-axis where $y = 0$. Substituting $y = 0$ in the equation,we get $bx = ab$,so $x = a$. Thus,the point of intersection is $(a, 0)$.
The slope of the given line $bx - ay = ab$ is $m_1 = \frac{b}{a}$.
The slope of the line perpendicular to it is $m_2 = -\frac{1}{m_1} = -\frac{a}{b}$.
The equation of the line with slope $m_2 = -\frac{a}{b}$ passing through $(a, 0)$ is $y - 0 = -\frac{a}{b}(x - a)$.
Multiplying by $b$,we get $by = -ax + a^2$,which simplifies to $ax + by = a^2$.
Dividing both sides by $ab$,we get $\frac{ax}{ab} + \frac{by}{ab} = \frac{a^2}{ab}$,which results in $\frac{x}{b} + \frac{y}{a} = \frac{a}{b}$.

(B) The given line is $x + y + 1 = 0$,which can be written as $y = -x - 1$. The slope of this line is $m_1 = -1$.
The slope of a line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{-1} = 1$.
The equation of a line with slope $m = 1$ passing through the point $(x_1, y_1) = (1, 2)$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values,we get $y - 2 = 1(x - 1)$.
Simplifying this,we get $y - 2 = x - 1$,which results in $y - x - 1 = 0$.

(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively.
Given that $a + b = 14$,so $a = 14 - b$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting $a = 14 - b$,we get $\frac{x}{14 - b} + \frac{y}{b} = 1$.
Since the line passes through $(3, 4)$,we have $\frac{3}{14 - b} + \frac{4}{b} = 1$.
Multiplying by $b(14 - b)$,we get $3b + 4(14 - b) = b(14 - b)$.
$3b + 56 - 4b = 14b - b^2$.
$b^2 - 15b + 56 = 0$.
$(b - 7)(b - 8) = 0$.
So,$b = 7$ or $b = 8$.
If $b = 7$,then $a = 14 - 7 = 7$. The equation is $\frac{x}{7} + \frac{y}{7} = 1 \Rightarrow x + y = 7$.
If $b = 8$,then $a = 14 - 8 = 6$. The equation is $\frac{x}{6} + \frac{y}{8} = 1 \Rightarrow 4x + 3y = 24$.
Comparing with the given options,$4x + 3y = 24$ is the correct choice.

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.
Since $(x_1, y_1)$ is the midpoint of $AB$,we have $x_1 = \frac{a+0}{2} = \frac{a}{2}$ and $y_1 = \frac{0+b}{2} = \frac{b}{2}$.
This implies $a = 2x_1$ and $b = 2y_1$.
The intercept form of the equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values of $a$ and $b$,we get $\frac{x}{2x_1} + \frac{y}{2y_1} = 1$.
Multiplying both sides by $2x_1y_1$,we obtain $x y_1 + y x_1 = 2x_1y_1$.

(B) The midpoint of the line segment joining $(1, 3)$ and $(1, -7)$ is given by $\left( \frac{1+1}{2}, \frac{3-7}{2} \right) = (1, -2)$.
Since the required line is parallel to $2x - 3y = 1$,its equation is of the form $2x - 3y = k$.
Substituting the point $(1, -2)$ into the equation: $2(1) - 3(-2) = k$.
$2 + 6 = k \Rightarrow k = 8$.
Thus,the equation of the line is $2x - 3y = 8$.

(A) Let the intercepts be $a$ and $-a$. The intercept form of the line is $\frac{x}{a} + \frac{y}{-a} = 1$.
This simplifies to $x - y = a$.
Since the line passes through $(-3, 2)$,we substitute these coordinates into the equation:
$-3 - 2 = a \Rightarrow a = -5$.
Substituting $a = -5$ back into the equation $x - y = a$,we get $x - y = -5$,which is $x - y + 5 = 0$.

(A) The intercept form of the equation 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.
Given that the $x$-intercept $a = 3$ and the $y$-intercept $b = -2$.
Substituting these values into the intercept form equation,we get:
$\frac{x}{3} + \frac{y}{-2} = 1$
This simplifies to:
$\frac{x}{3} - \frac{y}{2} = 1$

(D) The given line is $3x + 4y = 5$,which can be written as $4y = -3x + 5$ or $y = -\frac{3}{4}x + \frac{5}{4}$.
The slope of this line is $m_1 = -\frac{3}{4}$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -\frac{1}{m_1} = -\frac{1}{-3/4} = \frac{4}{3}$.
The equation of a line passing through $(x_1, y_1) = (3, -4)$ with slope $m = \frac{4}{3}$ is given by the point-slope form: $y - y_1 = m(x - x_1)$.
Substituting the values,we get $y - (-4) = \frac{4}{3}(x - 3)$,which simplifies to $y + 4 = \frac{4}{3}(x - 3)$.

(C) The given line is $y = 3x - 1$,which is in the slope-intercept form $y = mx + c$.
Comparing this,the slope of the given line is $m = 3$.
Since the required line is parallel to the given line,its slope will also be $m = 3$.
The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is given by $(y - y_1) = m(x - x_1)$.
Substituting the point $(1, 2)$ and slope $m = 3$,we get $(y - 2) = 3(x - 1)$.

(A) The given line is $2x + 3y + 4 = 0$,which can be written as $3y = -2x - 4$,or $y = -\frac{2}{3}x - \frac{4}{3}$.
The slope of this line is $m_1 = -\frac{2}{3}$.
Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Therefore,$m_2 = -\frac{1}{m_1} = -\frac{1}{-2/3} = \frac{3}{2}$.
The equation of a line passing through $(x_1, y_1) = (-1, 1)$ with slope $m_2 = \frac{3}{2}$ is given by $y - y_1 = m_2(x - x_1)$.
Substituting the values,we get $y - 1 = \frac{3}{2}(x - (-1))$,which simplifies to $y - 1 = \frac{3}{2}(x + 1)$.
Multiplying both sides by $2$,we get $2(y - 1) = 3(x + 1)$.

(C) The intersection of the lines $x = 0$ (the $y$-axis) and $y = 0$ (the $x$-axis) is the origin $(0, 0)$.
We need to find the equation of a line passing through the points $(0, 0)$ and $(2, 2)$.
The slope $m$ of the line is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{2 - 0} = 1$.
Using the point-slope form $y - y_1 = m(x - x_1)$,we get $y - 0 = 1(x - 0)$,which simplifies to $y = x$.

(B) The line joining $(a, 0)$ and $(-a, 0)$ is the $x$-axis,which has a slope of $0$.
Since the required line is perpendicular to the $x$-axis,it must be a vertical line.
$A$ vertical line passing through the origin $(0, 0)$ has the equation $x = 0$.

(B) The general equation of a straight line is given by $ax + by + c = 0$.
However,this can be rewritten as $y = mx + c$ (slope-intercept form) or $\frac{x}{a} + \frac{y}{b} = 1$ (intercept form).
In all these standard forms,there are exactly $2$ independent constants (parameters) that define the line.
Therefore,$2$ geometrical parameters are required to specify a straight line.

(B) Let the rectangle be $ABCD$. The points $A (1, 3)$ and $C (5, 1)$ are opposite vertices.
The midpoint of the diagonal $AC$ is $M = (\frac{1+5}{2}, \frac{3+1}{2}) = (3, 2)$.
Since $M$ is the center of the rectangle,it must also lie on the diagonal $BD$ connecting the other two vertices.
The line passing through the other two vertices ($B$ and $D$) has a gradient $m = 2$.
Using the point-slope form $y - y_1 = m(x - x_1)$ for the line passing through $M (3, 2)$ with slope $m = 2$:
$y - 2 = 2(x - 3)$
$y - 2 = 2x - 6$
$2x - y - 4 = 0$.

(A) Let the $x$-intercept be $a$ and the $y$-intercept be $2a$.
The intercept form of the line equation is $\frac{x}{a} + \frac{y}{2a} = 1$.
Since the line passes through $(1, 2)$,we substitute these coordinates 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 by $4$,we get $2x + y = 4$.

(A) Step $1$: Find the midpoint of the line segment joining $(2, -19)$ and $(6, 1)$.
Midpoint $= (\frac{2+6}{2}, \frac{-19+1}{2}) = (4, -9)$.
Step $2$: Find the slope of the line joining $(-1, 3)$ and $(5, -1)$.
Slope $(m_1) = \frac{-1-3}{5-(-1)} = \frac{-4}{6} = -\frac{2}{3}$.
Step $3$: The required line is perpendicular to this line,so its slope $(m_2)$ is $-\frac{1}{m_1} = \frac{3}{2}$.
Step $4$: Use the point-slope form $(y - y_1) = m(x - x_1)$ with point $(4, -9)$ and slope $\frac{3}{2}$.
$y - (-9) = \frac{3}{2}(x - 4)$
$2(y + 9) = 3(x - 4)$
$2y + 18 = 3x - 12$
$3x - 2y = 30$.

(A) Let the line intersect the $x$-axis at $(a, 0)$ and the $y$-axis at $(0, b)$.
Since the midpoint of the segment between the axes is $(x_1, y_1)$,we have:
$x_1 = \frac{a+0}{2} \implies a = 2x_1$
$y_1 = \frac{0+b}{2} \implies b = 2y_1$
The intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values of $a$ and $b$,we get $\frac{x}{2x_1} + \frac{y}{2y_1} = 1$.
Multiplying by $2$,we obtain $\frac{x}{x_1} + \frac{y}{y_1} = 2$.

(C) The given line is $ax + by + c = 0$.
Its slope is $m = -\frac{a}{b}$.
Since the required line is parallel to the given line,its slope will also be $m = -\frac{a}{b}$.
The equation of a line passing through $(c, d)$ with slope $m$ is given by $y - d = m(x - c)$.
Substituting $m = -\frac{a}{b}$,we get $y - d = -\frac{a}{b}(x - c)$.
Multiplying both sides by $b$,we get $b(y - d) = -a(x - c)$.
Rearranging the terms,we get $a(x - c) + b(y - d) = 0$.

(C) The given line is $ax + by + c = 0$. The slope of this line is $m_1 = -\frac{a}{b}$.
Since the required line is perpendicular to the given line,its slope $m_2$ satisfies $m_1 \times m_2 = -1$.
Therefore,$m_2 = \frac{b}{a}$.
The equation of a line passing through $(a, b)$ with slope $m_2 = \frac{b}{a}$ is given by $y - b = \frac{b}{a}(x - a)$.
Multiplying by $a$,we get $a(y - b) = b(x - a)$.
$ay - ab = bx - ab$.
$bx - ay = 0$.

(A) The slope of the line $m = \tan(45^\circ) = 1$.
Using the point-slope form of the line equation,$y - y_1 = m(x - x_1)$,where $(x_1, y_1) = (4, -6)$:
$y - (-6) = 1(x - 4)$
$y + 6 = x - 4$
$x - y - 10 = 0$.

(B) The given line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $y = -\frac{b}{a}x + b$.
The slope of this line is $m = -\frac{b}{a}$.
Since the required line is parallel to this line,its slope will also be $m = -\frac{b}{a}$.
The equation of the line passing through $(a, b)$ with slope $m = -\frac{b}{a}$ is given by $y - b = m(x - a)$.
Substituting the values: $y - b = -\frac{b}{a}(x - a)$.
Dividing both sides by $b$: $\frac{y}{b} - 1 = -\frac{1}{a}(x - a)$.
$\frac{y}{b} - 1 = -\frac{x}{a} + 1$.
Rearranging the terms: $\frac{x}{a} + \frac{y}{b} = 2$.

(C) The hour hand,minute hand,and second hand of a clock always pass through the origin $(0,0)$ because one end of these hands is fixed at the center of the clock.
At $4$ o'clock,the hour hand makes an angle of $30^{\circ}$ with the positive $x$-axis in the fourth quadrant (as shown in the figure).
The angle made by the line with the positive $x$-axis is $\theta = -30^{\circ}$.
The slope of the line is $m = \tan(-30^{\circ}) = -\frac{1}{\sqrt{3}}$.
The equation of a line passing through the origin with slope $m$ is $y = mx$.
Substituting the value of $m$,we get $y = -\frac{1}{\sqrt{3}}x$.
Multiplying both sides by $\sqrt{3}$,we get $\sqrt{3}y = -x$,which simplifies to $x + \sqrt{3}y = 0$.

(C) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle the normal makes with the positive $x$-axis.
Given the line makes an angle of $120^{\circ}$ with the $x$-axis,its slope is $m = \tan(120^{\circ}) = -\sqrt{3}$.
The normal to this line makes an angle $\alpha = 120^{\circ} - 90^{\circ} = 30^{\circ}$ or $120^{\circ} + 90^{\circ} = 210^{\circ}$ with the $x$-axis.
Using $p = 4$,the equations are:
$x \cos(30^{\circ}) + y \sin(30^{\circ}) = 4$ $\Rightarrow x(\frac{\sqrt{3}}{2}) + y(\frac{1}{2}) = 4$ $\Rightarrow x\sqrt{3} + y = 8$.
$x \cos(210^{\circ}) + y \sin(210^{\circ}) = 4$ $\Rightarrow x(-\frac{\sqrt{3}}{2}) + y(-\frac{1}{2}) = 4$ $\Rightarrow x\sqrt{3} + y = -8$.
Comparing with the given options,$x\sqrt{3} + y = 8$ is option $C$.

(B) The point of intersection of the lines $4x - 3y - 1 = 0$ and $2x - 5y + 3 = 0$ is found by solving them simultaneously.
Multiplying the second equation by $2$,we get $4x - 10y + 6 = 0$.
Subtracting this from the first equation: $(4x - 3y - 1) - (4x - 10y + 6) = 0$,which gives $7y - 7 = 0$,so $y = 1$.
Substituting $y = 1$ into $2x - 5(1) + 3 = 0$,we get $2x - 2 = 0$,so $x = 1$.
The point of intersection is $(1, 1)$.
Lines equally inclined to the axes have slopes $m = \tan(45^{\circ}) = 1$ or $m = \tan(135^{\circ}) = -1$.
Using the point-slope form $y - y_1 = m(x - x_1)$,the equations are $y - 1 = 1(x - 1)$ and $y - 1 = -1(x - 1)$.
These can be written as $y - 1 = \pm 1(x - 1)$.

(C) Let the line intersect the $x$-axis at $A(a, 0)$ and the $y$-axis at $B(0, b)$.
Given that the point $P(-4, 3)$ divides the segment $AB$ in the ratio $5 : 3$ internally.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{5(0) + 3(a)}{5 + 3}, \frac{5(b) + 3(0)}{5 + 3} \right) = \left( \frac{3a}{8}, \frac{5b}{8} \right)$.
Equating this to $(-4, 3)$,we get:
$\frac{3a}{8} = -4$ $\Rightarrow 3a = -32$ $\Rightarrow a = -\frac{32}{3}$.
$\frac{5b}{8} = 3$ $\Rightarrow 5b = 24$ $\Rightarrow b = \frac{24}{5}$.
The intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values of $a$ and $b$:
$\frac{x}{(-32/3)} + \frac{y}{(24/5)} = 1$
$-\frac{3x}{32} + \frac{5y}{24} = 1$.
Multiplying by $96$ (the $LCM$ of $32$ and $24$):
$-9x + 20y = 96$
$9x - 20y + 96 = 0$.

(B) The slope $m$ of the line passing through points $(x_1, y_1) = (-5, -6)$ and $(x_2, y_2) = (3, 10)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - (-6)}{3 - (-5)} = \frac{16}{8} = 2$.
Using the point-slope form $(y - y_1) = m(x - x_1)$ with point $(-5, -6)$:
$y - (-6) = 2(x - (-5))$
$y + 6 = 2(x + 5)$
$y + 6 = 2x + 10$
$2x - y + 4 = 0$.
Thus,the correct option is $B$.

(B) The intercept form of a line is given by $\frac{x}{a'} + \frac{y}{b'} = 1$,where $a'$ and $b'$ are the intercepts on the $x$ and $y$ axes respectively.
Given intercepts are $a' = 2a \sec \theta$ and $b' = 2a \csc \theta$.
Substituting these into the intercept form:
$\frac{x}{2a \sec \theta} + \frac{y}{2a \csc \theta} = 1$
Since $\frac{1}{\sec \theta} = \cos \theta$ and $\frac{1}{\csc \theta} = \sin \theta$,we get:
$\frac{x \cos \theta}{2a} + \frac{y \sin \theta}{2a} = 1$
Multiplying both sides by $2a$:
$x \cos \theta + y \sin \theta = 2a$
Rearranging the terms,we get:
$x \cos \theta + y \sin \theta - 2a = 0$.

(B) The given equations are $y = mx + c$ (slope-intercept form) and $x \cos \alpha + y \sin \alpha = p$ (normal form).
Rewriting the first equation as $mx - y + c = 0$,we can compare it with $x \cos \alpha + y \sin \alpha - p = 0$.
Since both represent the same line,the ratio of their coefficients must be equal:
$\frac{\cos \alpha}{m} = \frac{\sin \alpha}{-1} = \frac{-p}{-c} = k$
From this,$\cos \alpha = mk$ and $\sin \alpha = -k$.
Using the identity $\cos^2 \alpha + \sin^2 \alpha = 1$,we get $(mk)^2 + (-k)^2 = 1$,which simplifies to $k^2(m^2 + 1) = 1$,so $k = \frac{1}{\sqrt{1 + m^2}}$.
Also,from $\frac{-p}{-c} = k$,we have $\frac{p}{c} = k = \frac{1}{\sqrt{1 + m^2}}$.
Therefore,$c = p \sqrt{1 + m^2}$.

(C) The slope $m$ of the line passing through $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (2, 5)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 2}{2 - 1} = \frac{3}{1} = 3$.
Using the point-slope form $y - y_1 = m(x - x_1)$,we get $y - 2 = 3(x - 1)$.
Expanding this,$y - 2 = 3x - 3$.
Rearranging the terms,$y - 3x + 1 = 0$ or $3x - y - 1 = 0$.

(A) The given line is $3x + 4y + 5 = 0$.
The slope of this line is $m_1 = -\frac{3}{4}$.
The slope of a line perpendicular to this line is $m_2 = -\frac{1}{m_1} = \frac{4}{3}$.
The equation of a line with slope $m = \frac{4}{3}$ passing through $(1, 2)$ is given by $y - y_1 = m(x - x_1)$.
$y - 2 = \frac{4}{3}(x - 1)$
$3(y - 2) = 4(x - 1)$
$3y - 6 = 4x - 4$
$4x - 3y + 2 = 0$
Thus,the correct option is $A$.

(B) The equation of any line parallel to $2x + 6y + 7 = 0$ is of the form $2x + 6y + k = 0$.
This line intersects the $x$-axis at $A\left(-\frac{k}{2}, 0\right)$ and the $y$-axis at $B\left(0, -\frac{k}{6}\right)$.
The length of the intercept between the axes is given as $AB = 10$.
Using the distance formula,$AB = \sqrt{\left(-\frac{k}{2} - 0\right)^2 + \left(0 - \left(-\frac{k}{6}\right)\right)^2} = 10$.
$\sqrt{\frac{k^2}{4} + \frac{k^2}{36}} = 10$.
$\sqrt{\frac{9k^2 + k^2}{36}} = 10 \Rightarrow \sqrt{\frac{10k^2}{36}} = 10$.
$\frac{\sqrt{10}|k|}{6} = 10 \Rightarrow |k| = \frac{60}{\sqrt{10}} = 6\sqrt{10}$.
Thus,$k = \pm 6\sqrt{10}$.
Therefore,there are $2$ such lines: $2x + 6y + 6\sqrt{10} = 0$ and $2x + 6y - 6\sqrt{10} = 0$.

(D) The given line is $3x + y = 3$,which can be written as $y = -3x + 3$. The slope of this line is $m_1 = -3$.
The slope of a line perpendicular to this line is $m_2 = -1/m_1 = -1/(-3) = 1/3$.
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,we get $y - 2 = \frac{1}{3}(x - 2)$.
Multiplying by $3$,we get $3y - 6 = x - 2$,which simplifies to $x - 3y + 4 = 0$.
To find the $y$-intercept,we set $x = 0$ in the equation: $0 - 3y + 4 = 0$,which gives $3y = 4$ or $y = 4/3$.
Thus,the $y$-intercept is $4/3$.

(D) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.
The intercepts on the $x$-axis and $y$-axis are $A(a, 0)$ and $B(0, b)$ respectively.
The point $P(1, 2)$ is the midpoint of the segment $AB$.
Using the midpoint formula,we have $\frac{a + 0}{2} = 1$ and $\frac{0 + b}{2} = 2$.
This gives $a = 2$ and $b = 4$.
Substituting these values into the intercept form equation,we get $\frac{x}{2} + \frac{y}{4} = 1$.
Multiplying by $4$,we get $2x + y = 4$,or $2x + y - 4 = 0$.

(B) Given that point $P(a, b)$ lies on $3x + 2y = 13$,we have:
$3a + 2b = 13$ ..... $(i)$
Given that point $Q(b, a)$ lies on $4x - y = 5$,we have:
$4b - a = 5$ or $-a + 4b = 5$ ..... $(ii)$
Multiplying equation $(ii)$ by $3$,we get $-3a + 12b = 15$ ..... $(iii)$
Adding $(i)$ and $(iii)$:
$(3a + 2b) + (-3a + 12b) = 13 + 15$
$14b = 28 \Rightarrow b = 2$
Substituting $b = 2$ in $(i)$:
$3a + 2(2) = 13$ $\Rightarrow 3a = 9$ $\Rightarrow a = 3$
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) The equation of a line parallel to $2x + 3y - 7 = 0$ is of the form $2x + 3y + \lambda = 0$.
Since the line passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:
$2(1) + 3(1) + \lambda = 0$
$2 + 3 + \lambda = 0$
$5 + \lambda = 0$
$\lambda = -5$
Substituting the value of $\lambda$ back into the equation,we get $2x + 3y - 5 = 0$.

(A) Let the line intersect the $x$-axis at $(a, 0)$ and the $y$-axis at $(0, b)$.
Since the point $(5, 2)$ bisects the segment between these axes,we have:
$\frac{a + 0}{2} = 5 \Rightarrow a = 10$
$\frac{0 + b}{2} = 2 \Rightarrow b = 4$
Using the intercept form of the line equation,$\frac{x}{a} + \frac{y}{b} = 1$,we get:
$\frac{x}{10} + \frac{y}{4} = 1$
Multiplying by $20$,we obtain $2x + 5y = 20$.