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

(D) The given line is $3x + 4y = 12$.
Dividing by $12$,we get $\frac{x}{4} + \frac{y}{3} = 1$.
Thus,the $x-$intercept is $4$ and the $y-$intercept is $3$.
Let the line $L$ have $x-$intercept $a$ and $y-$intercept $b$.
According to the problem,$a = 2 \times 4 = 8$ and $b = \frac{3}{2}$.
The equation of line $L$ in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,which becomes $\frac{x}{8} + \frac{y}{3/2} = 1$.
This simplifies to $\frac{x}{8} + \frac{2y}{3} = 1$.
Multiplying by $24$,we get $3x + 16y = 24$.
Rewriting in slope-intercept form $y = mx + c$,we have $16y = -3x + 24$,or $y = -\frac{3}{16}x + \frac{24}{16}$.
Therefore,the slope of $L$ is $-\frac{3}{16}$.

(A) Let the two poles be $OA$ and $BC$ standing on a horizontal plane $OB$. Let $OA = 20\,m$ and $BC = 80\,m$. Let the distance between them be $OB = a$.
We set the coordinates as $O(0, 0)$,$A(0, 20)$,$B(a, 0)$,and $C(a, 80)$.
The line joining the top of the first pole $A(0, 20)$ to the foot of the second pole $B(a, 0)$ has the equation:
$y - 0 = \frac{20 - 0}{0 - a}(x - a) \Rightarrow y = -\frac{20}{a}(x - a) \quad \dots(1)$
The line joining the top of the second pole $C(a, 80)$ to the foot of the first pole $O(0, 0)$ has the equation:
$y - 0 = \frac{80 - 0}{a - 0}(x - 0) \Rightarrow y = \frac{80}{a}x \quad \dots(2)$
To find the intersection point $M$,we equate $(1)$ and $(2)$:
$\frac{80}{a}x = -\frac{20}{a}(x - a)$
$80x = -20x + 20a$
$100x = 20a \Rightarrow x = \frac{a}{5}$
Substituting $x = \frac{a}{5}$ into $(2)$:
$y = \frac{80}{a} \times \frac{a}{5} = 16\,m$.
Thus,the height of the point of intersection is $16\,m$.

(B) Let the line be $\frac{x}{a} + \frac{y}{b} = 1.$
Since the line passes through $P(-3, 4)$ and this point bisects the intercepted portion between the axes,the coordinates of the intercepts are $A(a, 0)$ and $B(0, b).$
By the midpoint formula,$P = \left( \frac{a+0}{2}, \frac{0+b}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right).$
Given $P(-3, 4),$ we have $\frac{a}{2} = -3 \implies a = -6$ and $\frac{b}{2} = 4 \implies b = 8.$
Substituting these values into the intercept form equation: $\frac{x}{-6} + \frac{y}{8} = 1.$
Multiplying by $24,$ we get $-4x + 3y = 24,$ or $4x - 3y + 24 = 0.$

(D) Let the slope of the line be $m = \tan \theta$. The equation of the line passing through $P(2, 3)$ is $x = 2 + r \cos \theta$ and $y = 3 + r \sin \theta$,where $r = 4$.
Substituting these into $x + y = 7$:
$(2 + 4 \cos \theta) + (3 + 4 \sin \theta) = 7$
$4(\cos \theta + \sin \theta) = 2$
$\cos \theta + \sin \theta = \frac{1}{2}$
Squaring both sides:
$1 + 2 \sin \theta \cos \theta = \frac{1}{4}$
$\sin 2\theta = \frac{1}{4} - 1 = -\frac{3}{4}$
Using $\sin 2\theta = \frac{2m}{1 + m^2}$:
$\frac{2m}{1 + m^2} = -\frac{3}{4}$
$8m = -3 - 3m^2 \Rightarrow 3m^2 + 8m + 3 = 0$
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$m = \frac{-8 \pm \sqrt{64 - 36}}{6} = \frac{-8 \pm \sqrt{28}}{6} = \frac{-8 \pm 2\sqrt{7}}{6} = \frac{-4 \pm \sqrt{7}}{3}$
Rationalizing $\frac{1 - \sqrt{7}}{1 + \sqrt{7}} = \frac{(1 - \sqrt{7})^2}{1 - 7} = \frac{1 + 7 - 2\sqrt{7}}{-6} = \frac{8 - 2\sqrt{7}}{-6} = \frac{-4 + \sqrt{7}}{3}$.

(A) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$,where $p = 4$ is the perpendicular distance from the origin and $\alpha$ is the angle the normal makes with the positive $x$-axis.
The line $x + y = 0$ has a slope of $-1$,which corresponds to an angle of $135^o$ with the positive $x$-axis.
The perpendicular from the origin to line $L$ makes an angle of $60^o$ with $x + y = 0$. Thus,$\alpha = 135^o \pm 60^o$.
Case $1$: $\alpha = 135^o - 60^o = 75^o$.
$\cos 75^o = \cos(45^o + 30^o) = \frac{\sqrt 3 - 1}{2\sqrt 2}$ and $\sin 75^o = \sin(45^o + 30^o) = \frac{\sqrt 3 + 1}{2\sqrt 2}$.
The equation is $x \left( \frac{\sqrt 3 - 1}{2\sqrt 2} \right) + y \left( \frac{\sqrt 3 + 1}{2\sqrt 2} \right) = 4$,which simplifies to $(\sqrt 3 - 1)x + (\sqrt 3 + 1)y = 8\sqrt 2$.
Case $2$: $\alpha = 135^o + 60^o = 195^o$. Since the line makes positive intercepts,$\alpha$ must be in the first quadrant $(0^o < \alpha < 90^o)$. Thus,we reject this case.
Comparing with the options,the correct equation is $(\sqrt 3 - 1)x + (\sqrt 3 + 1)y = 8\sqrt 2$.

(C) The slope $m$ of the line is given by $m = \tan(135^\circ) = -1$.
Given the $y$-intercept $c = -2$.
The slope-intercept form of a line is $y = mx + c$.
Substituting the values,we get $y = (-1)x + (-2)$.
Therefore,the equation of the line is $y = -x - 2$.

(B) The equation of a line in slope-intercept form is $y = mx + c$,where $m$ is the slope and $c$ is the $y$-intercept.
From the given graph,the angle made by the line with the positive $x$-axis is $\theta = 60^\circ$.
Therefore,the slope $m = \tan(60^\circ) = \sqrt{3}$.
The line intersects the $y$-axis at a distance of $3$ units above the origin,so the $y$-intercept $c = 3$.
Substituting these values into the slope-intercept form,we get $y = \sqrt{3}x + 3$.
Thus,the correct option is $B$.

(A) The slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Given points are $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (-1, 4)$.
Substituting these values into the formula:
$m = \frac{4 - (-2)}{-1 - 3}$
$m = \frac{4 + 2}{-4}$
$m = \frac{6}{-4}$
$m = -\frac{3}{2}$

(A) The slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Given points are $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (7, -2)$.
Substituting the values into the formula:
$m = \frac{-2 - (-2)}{7 - 3}$
$m = \frac{-2 + 2}{4}$
$m = \frac{0}{4} = 0$
Therefore,the slope of the line is $0$.

(C) The slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Substituting the given points $(3, -2)$ and $(3, 4)$:
$m = \frac{4 - (-2)}{3 - 3} = \frac{6}{0}$
Since division by zero is not defined,the slope of the line is undefined.

(C) The inclination of the line is given as $\alpha = 60^{\circ}$.
The slope $m$ of a line is defined as $m = \tan(\alpha)$.
Substituting the given value,we get $m = \tan(60^{\circ})$.
Therefore,the slope $m = \sqrt{3}$.

(A) Let the slope of the first line be $m_{1}$ and the slope of the second line be $m_{2}$.
The slope of the line through $(-2, 6)$ and $(4, 8)$ is:
$m_{1} = \frac{8 - 6}{4 - (-2)} = \frac{2}{6} = \frac{1}{3}$.
The slope of the line through $(8, 12)$ and $(x, 24)$ is:
$m_{2} = \frac{24 - 12}{x - 8} = \frac{12}{x - 8}$.
Since the two lines are perpendicular,the product of their slopes must be $-1$:
$m_{1} \times m_{2} = -1$.
Substituting the values:
$\frac{1}{3} \times \frac{12}{x - 8} = -1$.
Simplify the equation:
$\frac{4}{x - 8} = -1$.
Multiply both sides by $(x - 8)$:
$4 = -(x - 8)$.
$4 = -x + 8$.
$x = 8 - 4$.
$x = 4$.

(N/A) Since points $P$,$Q$,and $R$ are collinear,the slope of line segment $PQ$ must be equal to the slope of line segment $QR$.
The slope of $PQ$ is given by $m_{PQ} = \frac{y_{1}-k}{x_{1}-h}$.
The slope of $QR$ is given by $m_{QR} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$.
Equating the slopes: $\frac{y_{1}-k}{x_{1}-h} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$.
Multiplying both sides by $-1$ in the numerator and denominator of the left side: $\frac{k-y_{1}}{h-x_{1}} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$.
Cross-multiplying gives: $(h-x_{1})(y_{2}-y_{1}) = (k-y_{1})(x_{2}-x_{1})$.
Hence,the points are collinear.

(N/A) Let $(T, D)$ be any point on the line,where $D$ denotes the distance at time $T$.
Since the points $(0, 2)$,$(3, 8)$,and $(T, D)$ are collinear,the slope between any two pairs of points must be equal.
Slope $= \frac{8 - 2}{3 - 0} = \frac{D - 8}{T - 3}$
$\frac{6}{3} = \frac{D - 8}{T - 3}$
$2 = \frac{D - 8}{T - 3}$
$2(T - 3) = D - 8$
$2T - 6 = D - 8$
$D = 2T + 2$
$D = 2(T + 1)$
This is the required relation.

(A) If a line makes an angle of $30^{\circ}$ with the positive direction of the $y$-axis measured anticlockwise,then the angle made by the line with the positive direction of the $x$-axis measured anticlockwise is $90^{\circ} + 30^{\circ} = 120^{\circ}$.
Thus,the slope of the given line is $m = \tan 120^{\circ} = \tan(180^{\circ} - 60^{\circ}) = -\tan 60^{\circ} = -\sqrt{3}$.

(A) If points $A(x, -1), B(2, 1),$ and $C(4, 5)$ are collinear,then the slope of $AB$ must be equal to the slope of $BC$.
Slope of $AB = \frac{1 - (-1)}{2 - x} = \frac{2}{2 - x}$
Slope of $BC = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2$
Equating the slopes: $\frac{2}{2 - x} = 2$
$\Rightarrow 2 = 2(2 - x)$
$\Rightarrow 2 = 4 - 2x$
$\Rightarrow 2x = 2$
$\Rightarrow x = 1$
Thus,the required value of $x$ is $1$.

(A) The slope $m$ of the line joining the points $(x_1, y_1) = (3, -1)$ and $(x_2, y_2) = (4, -2)$ is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Substituting the values,we get $m = \frac{-2 - (-1)}{4 - 3} = \frac{-2 + 1}{1} = -1$.
The inclination $\theta$ of the line with the $x$-axis is given by $\tan \theta = m$.
Therefore,$\tan \theta = -1$.
Since $\tan 135^{\circ} = -1$,the angle $\theta$ is $135^{\circ}$.

(N/A) The slope of a line passing through two points $(x_{1}, y_{1})$ and $(h, k)$ is given by the formula:
$m = \frac{k - y_{1}}{h - x_{1}}$
It is given that the slope of the line is $m$.
Therefore,we have:
$\frac{k - y_{1}}{h - x_{1}} = m$
Multiplying both sides by $(h - x_{1})$,we get:
$k - y_{1} = m(h - x_{1})$
Hence,the equation is proved.

(N/A) If the points $A(h, 0), B(a, b),$ and $C(0, k)$ lie on a line,then the slope of $AB$ must be equal to the slope of $BC$.
Slope of $AB = \frac{b - 0}{a - h} = \frac{b}{a - h}$
Slope of $BC = \frac{k - b}{0 - a} = \frac{k - b}{-a}$
Equating the slopes: $\frac{b}{a - h} = \frac{k - b}{-a}$
Cross-multiplying: $-ab = (k - b)(a - h)$
Expanding the right side: $-ab = ka - kh - ab + bh$
Simplifying: $ka + bh = kh$
Dividing both sides by $kh$: $\frac{ka}{kh} + \frac{bh}{kh} = \frac{kh}{kh}$
Result: $\frac{a}{h} + \frac{b}{k} = 1$.

(N/A) Since line $AB$ passes through points $A(1985, 92)$ and $B(1995, 97)$,its slope is
$\frac{97-92}{1995-1985} = \frac{5}{10} = \frac{1}{2}$
Let $y$ be the population in the year $2010$. Then,according to the given graph,line $AB$ must pass through point $C(2010, y)$.
$\therefore$ Slope of $AB = \text{Slope of } BC$
$\Rightarrow \frac{1}{2} = \frac{y-97}{2010-1995}$
$\Rightarrow \frac{1}{2} = \frac{y-97}{15}$
$\Rightarrow \frac{15}{2} = y-97$
$\Rightarrow y-97 = 7.5$
$\Rightarrow y = 7.5 + 97 = 104.5$
Thus,the slope of line $AB$ is $\frac{1}{2}$,while in the year $2010$,the population will be $104.5$ crores.

(N/A) The position of the lines is shown in the figure.
For a line parallel to the $x$-axis,the $y$-coordinate of every point on the line remains constant. Since the line passes through $(-2, 3)$,the $y$-coordinate is $3$. Therefore,the equation of the line parallel to the $x$-axis is $y = 3$.
Similarly,for a line parallel to the $y$-axis,the $x$-coordinate of every point on the line remains constant. Since the line passes through $(-2, 3)$,the $x$-coordinate is $-2$. Therefore,the equation of the line parallel to the $y$-axis is $x = -2$.

(A) Given slope $m = -4$ and the point $(x_{0}, y_{0}) = (-2, 3)$.
Using the point-slope form of the equation of a line,which is $y - y_{0} = m(x - x_{0})$:
Substituting the values,we get $y - 3 = -4(x - (-2))$.
$y - 3 = -4(x + 2)$
$y - 3 = -4x - 8$
Rearranging the terms to the standard form $Ax + By + C = 0$:
$4x + y - 3 + 8 = 0$
$4x + y + 5 = 0$
Thus,the required equation of the line is $4x + y + 5 = 0$.

(A) Given points are $(x_{1}, y_{1}) = (1, -1)$ and $(x_{2}, y_{2}) = (3, 5)$.
Using the two-point form of the equation of a line: $y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}(x - x_{1})$.
Substituting the values: $y - (-1) = \frac{5 - (-1)}{3 - 1}(x - 1)$.
$y + 1 = \frac{6}{2}(x - 1)$.
$y + 1 = 3(x - 1)$.
$y + 1 = 3x - 3$.
Rearranging the terms,we get $3x - y - 4 = 0$.

(A) The slope of the line is given by $m = \tan \theta = \frac{1}{2}$.
The $y$-intercept is given by $c = -\frac{3}{2}$.
Using the slope-intercept form of a line,$y = mx + c$,we substitute the values:
$y = \frac{1}{2}x - \frac{3}{2}$.
Multiplying the entire equation by $2$,we get:
$2y = x - 3$.
Rearranging the terms,we get the equation of the line as:
$x - 2y - 3 = 0$.

(B) Given that the slope $m = \tan \theta = \frac{1}{2}$ and the $x$-intercept $d = 4$.
The equation of a line with slope $m$ and $x$-intercept $d$ is given by $y = m(x - d)$.
Substituting the values $m = \frac{1}{2}$ and $d = 4$ into the formula:
$y = \frac{1}{2}(x - 4)$
Multiplying both sides by $2$:
$2y = x - 4$
Rearranging the terms to the standard form:
$x - 2y - 4 = 0$
Thus,the required equation is $x - 2y - 4 = 0$.

(A) The intercept form of a line is given by $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the $x$-intercept and $y$-intercept respectively.
Given $a = -3$ and $b = 2$.
Substituting these values into the formula:
$\frac{x}{-3} + \frac{y}{2} = 1$
Multiplying the entire equation by $6$ to clear the denominators:
$-2x + 3y = 6$
Rearranging the terms to the standard form $Ax + By + C = 0$:
$2x - 3y + 6 = 0$

(N/A) The normal form of the equation of a line is given by $x \cos \omega + y \sin \omega = p$,where $p$ is the perpendicular distance from the origin and $\omega$ is the angle the normal makes with the positive $x$-axis.
Given $p = 4$ and $\omega = 15^{\circ}$.
We know that $\cos 15^{\circ} = \cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Similarly,$\sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.
Substituting these values into the normal form equation:
$x \left( \frac{\sqrt{3} + 1}{2\sqrt{2}} \right) + y \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) = 4$.
Multiplying both sides by $2\sqrt{2}$,we get:
$(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 8\sqrt{2}$.

(A) Assuming $F$ along the $x$-axis and $K$ along the $y$-axis,we have two points $(32, 273)$ and $(212, 373)$ in the $XY$-plane.
By the two-point form,the point $(F, K)$ satisfies the equation:
$K - 273 = \frac{373 - 273}{212 - 32}(F - 32)$
$K - 273 = \frac{100}{180}(F - 32)$
$K - 273 = \frac{5}{9}(F - 32)$
$K = \frac{5}{9}(F - 32) + 273$ ... $(1)$
When $K = 0$,Equation $(1)$ gives:
$0 = \frac{5}{9}(F - 32) + 273$
$-\frac{5}{9}(F - 32) = 273$
$F - 32 = -\frac{273 \times 9}{5}$
$F - 32 = -491.4$
$F = -491.4 + 32 = -459.4$

(A) The equation of a line passing through a point $(x_0, y_0)$ with slope $m$ is given by the point-slope form: $(y-y_0) = m(x-x_0)$.
Given point $(x_0, y_0) = (-4, 3)$ and slope $m = \frac{1}{2}$.
Substituting these values into the formula:
$(y-3) = \frac{1}{2}(x - (-4))$
$(y-3) = \frac{1}{2}(x+4)$
Multiply both sides by $2$:
$2(y-3) = x+4$
$2y-6 = x+4$
Rearranging the terms to the form $Ax+By+C=0$:
$x - 2y + 4 + 6 = 0$
$x - 2y + 10 = 0$.

(B) We know that the equation of a line passing through a point $(x_{0}, y_{0})$ with slope $m$ is given by the point-slope form: $(y - y_{0}) = m(x - x_{0})$.
Given that the line passes through the origin $(0, 0)$,we substitute $x_{0} = 0$ and $y_{0} = 0$ into the equation:
$(y - 0) = m(x - 0)$.
Simplifying this,we get:
$y = mx$.

(A) The slope of the line that inclines with the $x-$axis at an angle of $75^{\circ}$ is $m = \tan 75^{\circ}$.
$m = \tan(45^{\circ} + 30^{\circ}) = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \cdot \tan 30^{\circ}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.
The equation of a line passing through $(x_0, y_0)$ with slope $m$ is $(y - y_0) = m(x - x_0)$.
Substituting $(x_0, y_0) = (2, 2\sqrt{3})$ and $m = \frac{\sqrt{3}+1}{\sqrt{3}-1}$:
$(y - 2\sqrt{3}) = \frac{\sqrt{3}+1}{\sqrt{3}-1}(x - 2)$.
$(y - 2\sqrt{3})(\sqrt{3}-1) = (\sqrt{3}+1)(x - 2)$.
$y(\sqrt{3}-1) - 2\sqrt{3}(\sqrt{3}-1) = x(\sqrt{3}+1) - 2(\sqrt{3}+1)$.
$(\sqrt{3}+1)x - (\sqrt{3}-1)y = 2\sqrt{3} + 2 - (6 - 2\sqrt{3}) = 4\sqrt{3} - 4$.
$(\sqrt{3}+1)x - (\sqrt{3}-1)y = 4(\sqrt{3}-1)$.

(A) The equation of a line with slope $m$ and $x$-intercept $d$ is given by $y = m(x - d)$.
Given that the line intersects the $x$-axis at a distance of $3$ units to the left of the origin,the $x$-intercept is $d = -3$.
The slope of the line is given as $m = -2$.
Substituting these values into the formula:
$y = -2(x - (-3))$
$y = -2(x + 3)$
$y = -2x - 6$
Rearranging the terms to the standard form $Ax + By + C = 0$:
$2x + y + 6 = 0$.

(A) The equation of a line with slope $m$ and $y$-intercept $c$ is given by $y=mx+c$.
Here,the $y$-intercept is $c=2$ (since it intersects $2$ units above the origin).
The slope $m$ is given by $m=\tan(30^{\circ})=\frac{1}{\sqrt{3}}$.
Substituting these values into the equation,we get:
$y=\frac{1}{\sqrt{3}}x+2$
Multiplying both sides by $\sqrt{3}$:
$\sqrt{3}y=x+2\sqrt{3}$
Rearranging the terms,we get:
$x-\sqrt{3}y+2\sqrt{3}=0$.

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

(A) The slope $m_1$ of the line joining the points $(2, 5)$ and $(-3, 6)$ is given by $m_1 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5}$.
Two lines are perpendicular if the product of their slopes is $-1$. Let the slope of the required line be $m_2$.
Then $m_1 \times m_2 = -1 \implies -\frac{1}{5} \times m_2 = -1 \implies m_2 = 5$.
The equation of the line passing through $(-3, 5)$ with slope $m_2 = 5$ is given by the point-slope form: $y - y_1 = m_2(x - x_1)$.
Substituting the values: $y - 5 = 5(x - (-3))$.
$y - 5 = 5(x + 3)$.
$y - 5 = 5x + 15$.
$5x - y + 20 = 0$.

(A) The equation of a line in the intercept form is $\frac{x}{a} + \frac{y}{b} = 1$ $(i)$.
Given that the line cuts off equal intercepts on both axes,we have $a = b$.
Substituting $b = a$ in $(i)$,we get $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$ $(ii)$.
Since the line passes through the point $(2, 3)$,we substitute these coordinates into $(ii)$:
$2 + 3 = a \Rightarrow a = 5$.
Substituting $a = 5$ back into $(ii)$,the required equation of the line is $x + y = 5$.

(A) The equation of a line in the intercept form is $\frac{x}{a} + \frac{y}{b} = 1$ ..... $(i)$
Here,$a$ and $b$ are the intercepts on $x$ and $y$ axes respectively.
It is given that $a + b = 9 \Rightarrow b = 9 - a$ ..... $(ii)$
From equation $(i)$ and $(ii)$,we obtain $\frac{x}{a} + \frac{y}{9 - a} = 1$ ..... $(iii)$
It is given that the line passes through point $(2,2)$. Therefore,equation $(iii)$ becomes $\frac{2}{a} + \frac{2}{9 - a} = 1$
$\Rightarrow 2 \left( \frac{1}{a} + \frac{1}{9 - a} \right) = 1$
$\Rightarrow 2 \left( \frac{9 - a + a}{a(9 - a)} \right) = 1$
$\Rightarrow \frac{18}{9a - a^2} = 1$
$\Rightarrow 18 = 9a - a^2$
$\Rightarrow a^2 - 9a + 18 = 0$
$\Rightarrow a^2 - 6a - 3a + 18 = 0$
$\Rightarrow a(a - 6) - 3(a - 6) = 0$
$\Rightarrow (a - 6)(a - 3) = 0$
$\Rightarrow a = 6$ or $a = 3$
If $a = 6$,then $b = 9 - 6 = 3$. The equation is $\frac{x}{6} + \frac{y}{3} = 1 \Rightarrow x + 2y - 6 = 0$.
If $a = 3$,then $b = 9 - 3 = 6$. The equation is $\frac{x}{3} + \frac{y}{6} = 1 \Rightarrow 2x + y - 6 = 0$.

(A) The slope $m$ of the line making an angle $\theta = \frac{2 \pi}{3}$ with the positive $x$-axis is $m = \tan\left(\frac{2 \pi}{3}\right) = -\sqrt{3}$.
The equation of the line passing through $(0, 2)$ with slope $m = -\sqrt{3}$ is given by the point-slope form $(y - y_1) = m(x - x_1)$:
$(y - 2) = -\sqrt{3}(x - 0)$
$y - 2 = -\sqrt{3}x$
$\sqrt{3}x + y - 2 = 0$.
For the second line,it is parallel to the first line,so its slope is also $m = -\sqrt{3}$.
It crosses the $y$-axis $2$ units below the origin,meaning it passes through the point $(0, -2)$.
Using the point-slope form $(y - (-2)) = -\sqrt{3}(x - 0)$:
$y + 2 = -\sqrt{3}x$
$\sqrt{3}x + y + 2 = 0$.

It is given that when $C = 20,$ the value of $L$ is $124.942,$ and when $C = 110,$ the value of $L$ is $125.134.$
Accordingly,the points $(20, 124.942)$ and $(110, 125.134)$ satisfy the linear relation between $L$ and $C.$
Using the two-point form of a line,$(L - L_1) = \frac{L_2 - L_1}{C_2 - C_1}(C - C_1),$ we substitute the given values:
$(L - 124.942) = \frac{125.134 - 124.942}{110 - 20}(C - 20)$
$(L - 124.942) = \frac{0.192}{90}(C - 20)$
$L = \frac{0.192}{90}(C - 20) + 124.942.$

(A) The relationship between selling price and demand is linear.
Assuming selling price per liter along the $x$-axis and demand along the $y$-axis,we have two points,$(14, 980)$ and $(16, 1220)$,in the $XY$ plane that satisfy the linear relationship.
The equation of the line passing through $(14, 980)$ and $(16, 1220)$ is given by:
$y - 980 = \frac{1220 - 980}{16 - 14}(x - 14)$
$y - 980 = \frac{240}{2}(x - 14)$
$y - 980 = 120(x - 14)$
$y = 120(x - 14) + 980$
When $x = 17$ (price per liter):
$y = 120(17 - 14) + 980$
$y = 120(3) + 980$
$y = 360 + 980 = 1340$
Thus,the owner could sell $1340$ liters of milk weekly at Rs $17$ per liter.

(A) The given equation is $3x - 4y + 10 = 0$.
Rearranging the equation into the slope-intercept form $y = mx + c$:
$-4y = -3x - 10$
$y = \frac{-3}{-4}x + \frac{-10}{-4}$
$y = \frac{3}{4}x + \frac{5}{2}$
Comparing this with $y = mx + c$,the slope $m$ is $\frac{3}{4}$.

(A) The given equation is $3x - 4y + 10 = 0$.
To find the $x$-intercept,set $y = 0$:
$3x - 4(0) + 10 = 0 \implies 3x = -10 \implies x = -\frac{10}{3}$.
To find the $y$-intercept,set $x = 0$:
$3(0) - 4y + 10 = 0 \implies -4y = -10 \implies y = \frac{10}{4} = \frac{5}{2}$.
Thus,the $x$-intercept is $-\frac{10}{3}$ and the $y$-intercept is $\frac{5}{2}$.

(A) The given equation is $\sqrt{3} x + y = 8$.
Dividing the equation by $\sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2$,we get:
$\frac{\sqrt{3}}{2} x + \frac{1}{2} y = \frac{8}{2}$
$\frac{\sqrt{3}}{2} x + \frac{1}{2} y = 4$.
Comparing this with the normal form $x \cos \omega + y \sin \omega = p$,we have:
$\cos \omega = \frac{\sqrt{3}}{2}$ and $\sin \omega = \frac{1}{2}$.
Since both $\cos \omega$ and $\sin \omega$ are positive,$\omega$ lies in the first quadrant.
Thus,$\omega = 30^{\circ}$ and $p = 4$.

(N/A) The given lines can be written in slope-intercept form $(y = mx + c)$ as follows:
$y = -\frac{a_{1}}{b_{1}}x - \frac{c_{1}}{b_{1}}$ $(1)$
$y = -\frac{a_{2}}{b_{2}}x - \frac{c_{2}}{b_{2}}$ $(2)$
The slopes of lines $(1)$ and $(2)$ are $m_{1} = -\frac{a_{1}}{b_{1}}$ and $m_{2} = -\frac{a_{2}}{b_{2}}$,respectively.
Two lines are parallel if and only if their slopes are equal,i.e.,$m_{1} = m_{2}$.
Substituting the values of the slopes,we get:
$-\frac{a_{1}}{b_{1}} = -\frac{a_{2}}{b_{2}}$
Multiplying both sides by $-1$,we obtain:
$\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}}$
Thus,the lines are parallel if $\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}}$.

(A) The given line is $x-2y+3=0$,which can be rewritten as $2y = x+3$,or $y = \frac{1}{2}x + \frac{3}{2}$.
The slope of this line is $m_1 = \frac{1}{2}$.
The slope of a line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{1/2} = -2$.
The equation of the line passing through the point $(1, -2)$ with slope $m_2 = -2$ is given by the point-slope form: $y - y_1 = m(x - x_1)$.
Substituting the values: $y - (-2) = -2(x - 1)$.
$y + 2 = -2x + 2$.
$2x + y = 0$.

(A) The given equation is $x+7y=0$.
To reduce it to the slope-intercept form $y=mx+c$,we isolate $y$:
$7y = -x$
$y = -\frac{1}{7}x + 0$
Comparing this with the standard slope-intercept form $y=mx+c$,we get:
Slope $(m)$ = $-\frac{1}{7}$
$y$-intercept $(c)$ = $0$

(A) The given equation is $6x + 3y - 5 = 0$.
To reduce it to the slope-intercept form $(y = mx + c)$,we isolate $y$:
$3y = -6x + 5$
Divide the entire equation by $3$:
$y = -\frac{6}{3}x + \frac{5}{3}$
$y = -2x + \frac{5}{3}$
Comparing this with $y = mx + c$,we get the slope $m = -2$ and the $y$-intercept $c = \frac{5}{3}$.

(A) The given equation is $y=0$.
This can be written in the slope-intercept form $y=mx+c$ as:
$y = 0 \cdot x + 0$
Comparing this with $y=mx+c$,we get:
Slope $(m)$ = $0$
$y$-intercept $(c)$ = $0$
Thus,the slope is $0$ and the $y$-intercept is $0$.

(A) The given equation is $3x + 2y - 12 = 0$.
Adding $12$ to both sides,we get:
$3x + 2y = 12$.
Dividing both sides by $12$,we get:
$\frac{3x}{12} + \frac{2y}{12} = \frac{12}{12}$.
This simplifies to:
$\frac{x}{4} + \frac{y}{6} = 1$.
Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} = 1$,we find the $x$-intercept $a = 4$ and the $y$-intercept $b = 6$.