Answer :

The $y$-coordinate of every point on the $x$-axis is 0 .

Therefore, the equation of the $x$-axis is $y=0$.

The $x$-coordinate of every point on the $y$-axis is 0 .

Therefore, the equation of the $y$-axis is $x=0$.

Answer :

We know that the equation of the line passing through point $(x_0, y_0)$, whose slope is $m$, is $(y-y_0)=m(x-x_0)$.

Thus, the equation of the line passing through point ( $-4,3$), whose slope is $\dfrac{1}{2}$, is

$ \begin{aligned} & (y-3)=\dfrac{1}{2}(x+4) \\ & 2(y-3)=x+4 \\ & 2 y-6=x+4 \\ & \text{ i.e., } x-2 y+10=0 \end{aligned} $

Answer :

We know that the equation of the line passing through point $(x_0, y_0)$, whose slope is $m$, is $(y-y_0)=m(x-x_0)$.

Thus, the equation of the line passing through point $(0,0)$, whose slope is $m$, is

( $y$ - 0$)=m(x$ - 0$)$

i.e., $y=m x$

Answer :

The slope of the line that inclines with the $x$-axis at an angle of $75^{\circ}$ is

$m=\tan 75^{\circ}$

$\Rightarrow m=\tan (45^{\circ}+30^{\circ})=\dfrac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \cdot \tan 30^{\circ}}=\dfrac{1+\dfrac{1}{\sqrt{3}}}{1-1 \cdot \dfrac{1}{\sqrt{3}}}=\dfrac{\dfrac{\sqrt{3}+1}{\sqrt{3}}}{\dfrac{\sqrt{3}-1}{\sqrt{3}}}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$

We know that the equation of the line passing through point $(x_0, y_0)$, whose slope is $m$, is $(y-y_0)=m(x-x_0)$.

Thus, if a line passes though $(2,2 \sqrt{3})$ and inclines with the $x$-axis at an angle of $75^{\circ}$, then the equation of the line is given as

$ \begin{aligned} & (y-2 \sqrt{3})=\dfrac{\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} \\ & (\sqrt{3}+1) x-(\sqrt{3}-1) y=4 \sqrt{3}-4 \\ & \text{ i.e., }(\sqrt{3}+1) x-(\sqrt{3}-1) y=4(\sqrt{3}-1) \end{aligned} $

Answer :

It is known that if a line with slope $m$ makes $x$-intercept $d$, then the equation of the line is given as

$y=m(x-d)$

For the line intersecting the $x$-axis at a distance of 3 units to the left of the origin, $d=-3$.

The slope of the line is given as $m=-2$

Thus, the required equation of the given line is

$y=-2[x-(-3)]$

$y=-2 x-6$

i.e., $2 x+y+6=0$

Answer :

It is known that if a line with slope $m$ makes $y$-intercept $c$, then the equation of the line is given as

$y=m x+c$

Here, $c=2$ and $m=\tan 30^{\circ}$

$ =\dfrac{1}{\sqrt{3}} \text{. } $

Thus, the required equation of the given line is

$ \begin{aligned} & y=\dfrac{1}{\sqrt{3}} x+2 \\ & y=\dfrac{x+2 \sqrt{3}}{\sqrt{3}} \\ & \sqrt{3} y=x+2 \sqrt{3} \\ & \text{ i.e., } x-\sqrt{3} y+2 \sqrt{3}=0 \end{aligned} $

Answer :

It is known that the equation of the line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is $y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$. Therefore, the equation of the line passing through the points $(-1,1)$ and

(2, - 4 ) is

$(y-1)=\dfrac{-4-1}{2+1}(x+1)$

$(y-1)=\dfrac{-5}{3}(x+1)$

$3(y-1)=-5(x+1)$

$3 y-3=-5 x-5$

i.e., $5 x+3 y+2=0$

Answer :

It is given that the vertices of $\triangle P Q R$ are $P(2,1), Q (-2,3),$ and $R(4,5)$.

Let $RL$ be the median through vertex $R$.

Accordingly, $L$ is the mid-point of $PQ$.

By mid-point formula, the coordinates of point $L$ are given by $(\dfrac{2-2}{2}, \dfrac{1+3}{2})=(0,2)$

It is known that the equation of the line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is $y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$.

Therefore, the equation of $RL$ can be determined by substituting $(x_1, y_1)=(4,5)$ and $(x_2, y_2)=(0,2)$.

Hence,

$y-5=\dfrac{2-5}{0-4}(x-4)$

$\Rightarrow y-5=\dfrac{-3}{-4}(x-4)$

$\Rightarrow 4(y-5)=3(x-4)$

$\Rightarrow 4 y-20=3 x-12$

$\Rightarrow 3 x-4 y+8=0$

Thus, the required equation of the median through vertex $R$ is $3 x-4 y+8=0$.

Answer :

The slope of the line joining the points $(2,5)$ and $( - 3,6)$ is

$ m=\dfrac{6-5}{-3-2}=\dfrac{1}{-5} $

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points $(2,5)$ and (-3,6 )

$ =-\dfrac{1}{m}=-\dfrac{1}{(\dfrac{-1}{5})}=5 $

Now, the equation of the line passing through point ( $- 3,5$ ), whose slope is 5 , is

$ \begin{aligned} & (y-5)=5(x+3) \\ & y-5=5 x+15 \\ & \text{ i.e., } 5 x-y+20=0 \end{aligned} $

Answer :

According to the section formula, the coordinates of the point that divides the line segment joining the points $(1,0)$ and $(2,3)$ in the ratio $1: n$ is given by

$ (\dfrac{n(1)+1(2)}{1+n}, \dfrac{n(0)+1(3)}{1+n})=(\dfrac{n+2}{n+1}, \dfrac{3}{n+1}) $

The slope of the line joining the points $(1,0)$ and $(2,3)$ is

$ m=\dfrac{3-0}{2-1}=3 $

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points $(1,0)$ and $(2,3) \quad m \quad 3$

$ =-\dfrac{1}{m}=-\dfrac{1}{3} $

Now, the equation of the line passing through $(\dfrac{n+2}{n+1}, \dfrac{3}{n+1})$ and whose slope is $-\dfrac{1}{3}$ is given by

$ \begin{aligned} & (y-\dfrac{3}{n+1})=\dfrac{-1}{3}(x-\dfrac{n+2}{n+1}) \\ & \Rightarrow 3[(n+1) y-3]=-[x(n+1)-(n+2)] \\ & \Rightarrow 3(n+1) y-9=-(n+1) x+n+2 \\ & \Rightarrow(1+n) x+3(1+n) y=n+11 \end{aligned} $

Answer :

The equation of a line in the intercept form is

$$ \begin{equation*} \dfrac{x}{a} + \dfrac{y}{b}=1 \tag{i} \end{equation*} $$

Here, $a$ and $b$ are the intercepts on $x$ and $y$ axes respectively.

It is given that the line cuts off equal intercepts on both the axes. This means that $a=b$.

Accordingly, equation (i) reduces to

$$ \begin{align*} & \dfrac{x}{a} + \dfrac{y}{a}=1 \\ & \Rightarrow x+y=a \end{align*} $$

Since the given line passes through point $(2,3)$, equation (ii) reduces to

$2+3=a \Rightarrow a=5$

On substituting the value of $a$ in equation (ii), we obtain

$x+y=5$, which is the required equation of the line

Answer :

The equation of a line in the intercept form is

$$ \begin{equation*} \dfrac{x}{a} + \dfrac{y}{b}=1 \tag{i} \end{equation*} $$

Here, $a$ and $b$ are the intercepts on $x$ and $y$ axes respectively.

It is given thata $+b=9 \Rightarrow b=9$ - $a \ldots$ (ii)

From equations (i) and (ii), we obtain

$$ \begin{equation*} \dfrac{x}{a} + \dfrac{y}{9-a}=1 \tag{iii} \end{equation*} $$

It is given that the line passes through point $(2,2)$. Therefore, equation (iii) reduces to

$$ \begin{aligned} & \dfrac{2}{a} + \dfrac{2}{9-a}=1 \\ & \Rightarrow 2(\dfrac{1}{a} + \dfrac{1}{9-a})=1 \\ & \Rightarrow 2(\dfrac{9-a+a}{a(9-a)})=1 \\ & \Rightarrow \dfrac{18}{9 a-a^{2}}=1 \\ & \Rightarrow 18=9 a-a^{2} \\ & \Rightarrow a^{2}-9 a+18=0 \\ & \Rightarrow a^{2}-6 a-3 a+18=0 \\ & \Rightarrow a(a-6)-3(a-6)=0 \\ & \Rightarrow(a-6)(a-3)=0 \\ & \Rightarrow a=6 \text{ or } a=3 \end{aligned} $$

If $a=6$ and $b=9$ - $6=3$, then the equation of the line is

$\dfrac{x}{6}+\dfrac{y}{3}=1 \Rightarrow x+2 y-6=0$

If $a=3$ and $b=9 - 3=6$, then the equation of the line is

$\dfrac{x}{3}+\dfrac{y}{6}=1 \Rightarrow 2 x+y-6=0$

Answer :

The slope of the line making an angle $\dfrac{2 \pi}{3}$ with the positive $x$-axis is $\quad m=\tan (\dfrac{2 \pi}{3})=-\sqrt{3}$

Now, the equation of the line passing through point $(0,2)$ and having a slope $-\sqrt{3}$ is $(y-2)=-\sqrt{3}(x-0)$. $y-2=-\sqrt{3} x$

i.e., $\sqrt{3} x+y-2=0$

The slope of line parallel to line $\sqrt{3} x+y-2=0$ is $-\sqrt{3}$.

It is given that the line parallel to line $\sqrt{3} x+y-2=0$ crosses the $y$-axis 2 units below the origin i.e., it passes through point ( 0, - 2 ).

Hence, the equation of the line passing through point ( $0, -$ “ 2 ) and having a slope $-\sqrt{3}$ is

$y-(-2)=-\sqrt{3}(x-0)$

$y+2=-\sqrt{3} x$

$\sqrt{3} x+y+2=0$

Answer :

The slope of the line joining the origin $(0,0)$ and point (-2,9) is $m_1=\dfrac{9-0}{-2-0}=-\dfrac{9}{2}$

Accordingly, the slope of the line perpendicular to the line joining the origin and point (- 2,9 ) is

$m_2=-\dfrac{1}{m_1}=-\dfrac{1}{(-\dfrac{9}{2})}=\dfrac{2}{9}$

Now, the equation of the line passing through point $($ - 2,9$)$ and having a slope $m_2$ is

$ \begin{aligned} & (y-9)=\dfrac{2}{9}(x+2) \\ & 9 y-81=2 x+4 \\ & \text{ i.e., } 2 x-9 y+85=0 \end{aligned} $

Answer :

It is given that when $C=20$, the value of $L$ is 124.942 , whereas when $C=110$, the value of $L$ is 125.134 .

Accordingly, points $(20,124.942)$ and $(110,125.134)$ satisfy the linear relation between $L$ and $C$.

Now, assuming $C$ along the $x$-axis and $L$ along the $y$-axis, we have two points i.e., $(20,124.942)$ and $(110,125.134)$ in the $X Y$ plane.

Therefore, the linear relation between $L$ and $C$ is the equation of the line passing through points $(20,124.942)$ and $(110,125.134)$.

$(L - 124.942)=\dfrac{125.134-124.942}{110-20}(C-20)$

$L-124.942=\dfrac{0.192}{90}(C-20)$

i.e., $L=\dfrac{0.192}{90}(C-20)+124.942$, which is the required linear relation

Answer :

The relationship between selling price and demand is linear.

Assuming selling price per litre along the $x$-axis and demand along the $y$-axis, we have two points i.e., $(14,980)$ and $(16,1220)$ in the XY plane that satisfy the linear relationship between selling price and demand.

Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through points $(14,980)$ and $(16,1220)$.

$ \begin{aligned} & y-980=\dfrac{1220-980}{16-14}(x-14) \\ & y-980=\dfrac{240}{2}(x-14) \\ & y-980=120(x-14) \\ & \text{ i.e., } y=120(x-14)+980 \end{aligned} $

When $x=$ Rs $17 /$ litre,

$ \begin{aligned} & y=120(17-14)+980 \\ & \Rightarrow y=120 \times 3+980=360+980=1340 \end{aligned} $

Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.

Answer :

Let $A B$ be the line segment between the axes and let $P(a, b)$ be its mid-point.

Let the coordinates of $A$ and $B$ be $(0, y)$ and $(x, 0)$ respectively.

Since $P(a, b)$ is the mid-point of $AB$,

$(\dfrac{0+x}{2}, \dfrac{y+0}{2})=(a, b)$

$\Rightarrow(\dfrac{x}{2}, \dfrac{y}{2})=(a, b)$

$\Rightarrow \dfrac{x}{2}=a$ and $\dfrac{y}{2}=b$

$\therefore x=2 a$ and $y=2 b$

Thus, the respective coordinates of A and B are $(0,2 b)$ and $(2 a, 0)$.

The equation of the line passing through points $(0,2 b)$ and $(2 a, 0)$ is

$(y-2 b)=\dfrac{(0-2 b)}{(2 a-0)}(x-0)$

$y-2 b=\dfrac{-2 b}{2 a}(x)$

$a(y-2 b)=-b x$

$a y-2 a b=-b x$

i.e., $b x+a y=2 a b$

On dividing both sides by $a b$, we obtain $\dfrac{b x}{a b}+\dfrac{a y}{a b}=\dfrac{2 a b}{a b}$

$\Rightarrow \dfrac{x}{a}+\dfrac{y}{b}=2$

Thus, the equation of the line is $\dfrac{x}{a}+\dfrac{y}{b}=2$.

Answer :

Let $A B$ be the line segment between the axes such that point $R(h, k)$ divides $A B$ in the ratio 1: 2 .

Let the respective coordinates of $A$ and $B$ be $(x, 0)$ and $(0, y)$.

Since point $R(h, k)$ divides $AB$ in the ratio $1: 2$, according to the section formula,

$(h, k)=(\dfrac{1 \times 0+2 \times x}{1+2}, \dfrac{1 \times y+2 \times 0}{1+2})$

$\Rightarrow(h, k)=(\dfrac{2 x}{3}, \dfrac{y}{3})$

$\Rightarrow h=\dfrac{2 x}{3}$ and $k=\dfrac{y}{3}$

$\Rightarrow x=\dfrac{3 h}{2}$ and $y=3 k$

Therefore, the respective coordinates of $A$ and $B$ are $(\dfrac{3 h}{2}, 0)$ and $(0,3 k)$.

Now, the equation of line $AB$ passing through points $(\dfrac{3 h}{2}, 0)$ and

$(0,3 k)$ is $(y-0)=\dfrac{3 k-0}{0-\dfrac{3 h}{2}}(x-\dfrac{3 h}{2})$

$y=-\dfrac{2 k}{h}(x-\dfrac{3 h}{2})$

$h y=-2 k x+3 h k$

i.e., $2 k x+h y=3 h k$

Thus, the required equation of the line is $2 k x+h y=3 h k$.

Answer :

In order to show that points $(3,0)$, $ (-2,- 2 )$ , and $(8,2)$ are collinear, it suffices to show that the line passing through points $(3,0)$ and $ (-2 , - 2 )$ also passes through point $(8,2)$.

The equation of the line passing through points $(3,0)$ and $( -2, - 2 )$ is

$ \begin{aligned} & (y-0)=\dfrac{(-2-0)}{(-2-3)}(x-3) \\ & y=\dfrac{-2}{-5}(x-3) \\ & 5 y=2 x-6 \\ & \text{ i.e., } 2 x-5 y=6 \end{aligned} $

It is observed that at $x=8$ and $y=2$,

L.H.S. $=2 \times 8 $ $ - 5 \times 2=16$ - $10=6=$ R.H.S.

Therefore, the line passing through points $(3,0)$ and ( $.- 2, - 2)$ also passes through point $(8,2)$. Hence, points $(3$, $0)$, (-2, -2), and (8, 2) are collinear.