Answer :

Let $A B C D$ be the given quadrilateral with vertices $ A($ -4, 5 $), B(0,7), C ( 5, - $5 $)$, and $D$ (4, 2).

Then, by plotting $A, B, C$, and $D$ on the Cartesian plane and joining $A B, B C, C D$, and $D A$, the given quadrilateral can be drawn as

To find the area of quadrilateral $A B C D$, we draw one diagonal, say $A C$.

Accordingly, area $(A B C D)=area(\triangle A B C)+area(\triangle A C D)$

We know that the area of a triangle whose vertices are $(x_1, y_1),(x_2, y_2)$, and $(x_3, y_3)$ is

$\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Therefore, area of $\triangle ABC$ $=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|$ unit $^{2}$

$=\frac{1}{2}|-4(12)+5(-2)|$ unit $^{2}$

$=\frac{1}{2}|-48-10|$ unit $^{2}$

$=\frac{1}{2}|-58|$ unit $^{2}$

$=\frac{1}{2} \times 58$ unit $^{2}$

$=29$ unit $^{2}$

Area of $\triangle A C D$

$=\frac{1}{2}|-4(-5+2)+5(-2-5)+(-4)(5+5)|$ unit $^{2}$

$=\frac{1}{2}|-4(-3)+5(-7)-4(10)|$ unit $^{2}$

$=\frac{1}{2}|12-35-40|$ unit $^{2}$

$=\frac{1}{2}|-63|$ unit $^{2}$

$=\frac{63}{2}$ unit $^{2}$

Thus, area (ABCD)

$ =(29+\frac{63}{2}) \text{ unit }^{2}=\frac{58+63}{2} \text{ unit }^{2}=\frac{121}{2} \text{ unit }^{2} $

Answer :

Let $A B C$ be the given equilateral triangle with side $2 a$.

Accordingly, $AB=BC=CA=2 a$

Assume that base $BC$ lies along the $y$-axis such that the mid-point of $BC$ is at the origin.

i.e., $BO=OC=a$, where $O$ is the origin.

Now, it is clear that the coordinates of point $C$ are $(0, a)$, while the coordinates of point $B$ are $(0, - a)$.

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex A lies on the $y$-axis.

On applying Pythagoras theorem to $\triangle A O C$, we obtain

$ \begin{aligned} & (AC)^{2}=(OA)^{2}+(OC)^{2} \\ & \Rightarrow(2 a)^{2}=(OA)^{2}+a^{2} \\ & \Rightarrow 4 a^{2} - a^{2}=(OA)^{2} \\ & \Rightarrow(OA)^{2}=3 a^{2} \\ & \Rightarrow OA=\sqrt{3} a \end{aligned} $

$\therefore$ Coordinates of point $A=( \pm \sqrt{3} a, 0)$

Thus, the vertices of the given equilateral triangle are $(0, a),(0, -{ } a)$, and $(\sqrt{3} a, 0)$ or $(0, a),(0, - a)$, and $(-\sqrt{3} a, 0)$.

Answer :

The given points are $P(x_1, y_1)$ and $Q(x_2, y_2)$.

(i) When PQ is parallel to the $y$-axis, $x_1=x_2$.

In this case, distance between $P$ and $Q=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

$ \begin{aligned} & =\sqrt{(y_2-y_1)^{2}} \\ & =|y_2-y_1| \end{aligned} $

(ii) When PQ is parallel to the $x$-axis, $y_1=y_2$.

In this case, distance between $P$ and $Q=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

$ \begin{aligned} & =\sqrt{(x_2-x_1)^{2}} \\ & =|x_2-x_1| \end{aligned} $

Answer :

Let $(a, 0)$ be the point on the $x$ axis that is equidistant from the points $(7,6)$ and $(3,4)$.

Accordingly, $\sqrt{(7-a)^{2}+(6-0)^{2}}=\sqrt{(3-a)^{2}+(4-0)^{2}}$

$\Rightarrow \sqrt{49+a^{2}-14 a+36}=\sqrt{9+a^{2}-6 a+16}$

$\Rightarrow \sqrt{a^{2}-14 a+85}=\sqrt{a^{2}-6 a+25}$

On squaring both sides, we obtain

$a^{2}$ - $14 a+85=a^{2}- 6 a+25$

$\Rightarrow a - 14 a+6 a=25 - 85$

$\Rightarrow $ - $8 a=- 60$

$\Rightarrow a=\frac{60}{8}=\frac{15}{2}$

Thus, the required point on the $x$-axis is

$ (\frac{15}{2}, 0) $

Answer :

The coordinates of the mid-point of the line segment joining the points

$P(0, - 4)$ and $B(8,0)$ are

$ (\frac{0+8}{2}, \frac{-4+0}{2})=(4,-2) $

It is known that the slope $(m)$ of a non-vertical line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}, x_2 \neq x_1$

Therefore, the slope of the line passing through $(0,0)$ and $ ( 4, \hat{a} 2 ) $ is

$\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}$.

Hence, the required slope of the line is $-\frac{1}{2}$.

Answer :

The vertices of the given triangle are A $(4,4), B(3,5)$, and C (-1, -1).

It is known that the slope $(m)$ of a non-vertical line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}, x_2 \neq x_1$

$\therefore$ Slope of $AB(m_1)=\frac{5-4}{3-4}=-1$

Slope of $BC(m_2)=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$

Slope of CA $(m_3)=\frac{4+1}{4+1}=\frac{5}{5}=1$

It is observed that $m_1 m_3=\hat{a} $ 1

This shows that line segments $A B$ and $C A$ are perpendicular to each other i.e., the given triangle is right-angled at $A(4,4)$.

Thus, the points $(4,4),(3,5)$, and (1, 1 ) are the vertices of a right-angled triangle.

Answer :

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 $\tan 120^{\circ}=\tan (180^{\circ} - 60^{\circ})=- \tan 60^{\circ}=-\sqrt{3}$

Answer :

Let points ( $$ - $2, $ -1), $(4,0),(3,3)$, and ( $-$3,2 ) be respectively denoted by A, B, C, and $D$.

Slope of $A B=\frac{0+1}{4+2}=\frac{1}{6}$

Slope of $CD=\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}$

$\Rightarrow$ Slope of $A B=$ Slope of $C D$

$\Rightarrow A B$ and $C D$ are parallel to each other.

Now, slope of $BC=\frac{3-0}{3-4}=\frac{3}{-1}=-3$

Slope of $AD=\frac{2+1}{-3+2}=\frac{3}{-1}=-3$

$\Rightarrow$ Slope of $BC=$ Slope of $AD$

$\Rightarrow B C$ and $A D$ are parallel to each other.

Therefore, both pairs of opposite sides of quadrilateral $A B C D$ are parallel. Hence, $A B C D$ is a parallelogram.

Thus, points (-2, -1), $(4,0),(3,3)$, and (-3,2) are the vertices of a parallelogram.

Answer :

The slope of the line joining the points ($ 3, \hat{a} $ 1 ) and ( $4, \hat{a} $ 2 ) is

$ m=\frac{-2-(-1)}{4-3}=-2+1=-1 $

Now, the inclination $(\theta)$ of the line joining the points $(3, -$1 $)$ and $(4, -{ } 2)$ is given by $\tan \theta=- 1$

$\Rightarrow \theta=(90^{\circ}+45^{\circ})=135^{\circ}$

Thus, the angle between the $x$-axis and the line joining the points $(3, $ - 1 $)$ and ( $ 4, - $ 2 ) is $ 135^{\circ} $.

Answer :

Let $m_1$ and $m$ be the slopes of the two given lines such that $m_1=2 m$.

We know that if $\theta$ isthe angle between the lines $l_1$ and $I_2$ with slopes $m_1$ and $m_2$, then

$ \tan \theta=|\frac{m_2-m_1}{1+m_1 m_2}| $

It is given that the tangent of the angle between the two lines is $\frac{1}{3}$.

$\therefore \frac{1}{3}=|\frac{m-2 m}{1+(2 m) \cdot m}|$

$\Rightarrow \frac{1}{3}=|\frac{-m}{1+2 m^{2}}|$

$\Rightarrow \frac{1}{3}=\frac{-m}{1+2 m^{2}}$ or $\frac{1}{3}=-(\frac{-m}{1+2 m^{2}})=\frac{m}{1+2 m^{2}}$

If $m=$ - 1 , then the slopes of the lines are - 1 and - 2 .

If $m=-\frac{1}{2}$, then the slopes of the lines are $-\frac{1}{2}$ and -1.

Case II

$\frac{1}{3}=\frac{m}{1+2 m^{2}}$

$\Rightarrow 2 m^{2}+1=3 m$

$\Rightarrow 2 m^{2}-3 m+1=0$

$\Rightarrow 2 m^{2}-2 m-m+1=0$

$\Rightarrow 2 m(m-1)-1(m-1)=0$

$\Rightarrow(m-1)(2 m-1)=0$

$\Rightarrow m=1$ or $m=\frac{1}{2}$

If $m=1$, then the slopes of the lines are 1 and 2 .

If $m=\frac{1}{2}$, then the slopes of the lines are $\frac{1}{2}$ and 1

Hence, the slopes of the lines are - 1 and - 2 or $\quad-\frac{1}{2}$ and - 1 or 1 and 2 or $\frac{1}{2}$ and 1 .

Answer :

The slope of the line passing through $(x_1, y_1)$ and $(h, k)$ is $\frac{k-y_1}{h-x_1}$.

It is given that the slope of the line is $m$.

$\therefore \frac{k-y_1}{h-x_1}=m$

$\Rightarrow k-y_1=m(h-x_1)$

Hence, $k-y_1=m(h-x_1)$

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 $\frac{1}{2}$, is

$ \begin{aligned} & (y-3)=\frac{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

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

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

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})=\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} \\ & (\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 $m=\tan 30^{\circ}$

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

Thus, the required equation of the given line is

$ \begin{aligned} & y=\frac{1}{\sqrt{3}} x+2 \\ & y=\frac{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 $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$. Therefore, the equation of the line passing through the points ( $.-{ }^{\text{" }} 1,1,1) $ and

(2, - 4 ) is

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

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

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

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

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

Answer :

It is given that the vertices of $\triangle P Q R$ are $P(2,1), Q(\hat{a}.$ “" $.^{2}, 3)$, 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 $(\frac{2-2}{2}, \frac{1+3}{2})=(0,2)$

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

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

Hence,

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

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

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

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

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

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=\frac{6-5}{-3-2}=\frac{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 )

$ =-\frac{1}{m}=-\frac{1}{(\frac{-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

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

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

$ m=\frac{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 $(2,3) \quad m \quad 3$

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

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

$ \begin{aligned} & (y-\frac{3}{n+1})=\frac{-1}{3}(x-\frac{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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer :

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

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 $m_1=\frac{9-0}{-2-0}=-\frac{9}{2}$

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

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

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

$ \begin{aligned} & (y-9)=\frac{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 $(110,125.134)$.

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

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

i.e., $L=\frac{0.192}{90}(C-20)+124.942$, 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=\frac{1220-980}{16-14}(x-14) \\ & y-980=\frac{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 $AB$,

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

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

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

$\therefore x=2 a$ 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

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

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

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

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

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

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

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

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

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,

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

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

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

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

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

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

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

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

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

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)=\frac{(-2-0)}{(-2-3)}(x-3) \\ & y=\frac{-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.

Answer :

(i) The given equation is $x+7 y=0$.

It can be written as

$y=-\frac{1}{7} x+0$

This equation is of the form $y=m x+c$, where

$ m=-\frac{1}{7} \text{ and } c=0 $

Therefore, equation (1) is in the slope-intercept form, where the slope and the $y$-intercept are $-\frac{1}{7}$ and 0 respectively.

(ii) The given equation is $6 x+3 y -5=0$.

It can be written as

$y=\frac{1}{3}(-6 x+5)$

$y=-2 x+\frac{5}{3}$

This equation is of the form $y=m x+c$, where $m=-2$ and $c=\frac{5}{3}$.

Therefore, equation (2) is in the slope-intercept form, where the slope and the $y$-intercept are- 2 and $\frac{5}{3}$ respectively.

(iii) The given equation is $y=0$.

It can be written as

$y=0 . x+0$

This equation is of the form $y=m x+c$, where $m=0$ and $c=0$.

Therefore, equation (3) is in the slope-intercept form, where the slope and the $y$-intercept are 0 and 0 respectively.

Answer :

(i) The given equation is $3 x+2 y$ - $12=0$.

It can be written as

$3 x+2 y=12$

$\frac{3 x}{12}+\frac{2 y}{12}=1$ i.e., $\frac{x}{4}+\frac{y}{6}=1$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$, where $a=4$ and $b=6$.

Therefore, equation (1) is in the intercept form, where the intercepts on the $x$ and $y$ axes are 4 and 6 respectively.

(ii) The given equation is $4 x - 3 y=6$.

It can be written as

$\frac{4 x}{6}-\frac{3 y}{6}=1$

$\frac{2 x}{3}-\frac{y}{2}=1$

i.e., $\frac{x}{(\frac{3}{2})}+\frac{y}{(-2)}=1$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$, where $a=\frac{3}{2}$ and $b=-$2 .

Therefore, equation (2) is in the intercept form, where the intercepts on the $x$ and $y$ axes are $\frac{3}{2}$ and -2 respectively.

(iii) The given equation is $3 y+2=0$.

It can be written as

$3 y=-2$

i.e., $\frac{y}{(-\frac{2}{3})}=1$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$, where $a=0$ and $b=-\frac{2}{3}$.

Therefore, equation (3) is in the intercept form, where the intercept on the $y$-axis is $-\frac{2}{3}$ and it has no intercept on the $x$-axis.

Answer :

The given equation of the line is $12(x+6)=5(y - 2)$.

$\Rightarrow 12 x+72=5 y$ 10

$\Rightarrow 12 x - 5 y+82=0$..

On comparing equation (1) with general equation of line $A x+B y+C=0$, we obtain $A=12, B=$ - 5 , and $C=82$.

It is known that the perpendicular distance ( $d$ ) of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by

$d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}}$.

The given point is $(x_1, y_1)=(- 1,1)$.

Therefore, the distance of point $( - 1,1 )$ from the given line

$=\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^{2}+(-5)^{2}}}$ units $=\frac{|-12-5+82|}{\sqrt{169}}$ units $=\frac{|65|}{13}$ units $=5$ units

Answer :

The given equation of line is

$\frac{x}{3}+\frac{y}{4}=1$

or, $4 x+3 y-12=0$

On comparing equation (1) with general equation of line $A x+B y+C=0$, we obtain $A=4, B=3$, and $C=$ - 12 .

Let $(a, 0)$ be the point on the $x$-axis whose distance from the given line is 4 units.

It is known that the perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by

$d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}}$.

Therefore,

$4=\frac{|4 a+3 \times 0-12|}{\sqrt{4^{2}+3^{2}}}$

$\Rightarrow 4=\frac{|4 a-12|}{5}$

$\Rightarrow|4 a-12|=20$

$\Rightarrow \pm(4 a-12)=20$

$\Rightarrow(4 a-12)=20$ or $-(4 a-12)=20$

$\Rightarrow 4 a=20+12$ or $4 a=-20+12$

$\Rightarrow a=8$ or -2

Thus, the required points on the $x$-axis are (-2,0) and $(8,0)$.

Answer :

It is known that the distance (d) between parallel lines $A x+B y+C_1=0$ and $A x+B y+C_2=0$ is given by $d=\frac{|C_1-C_2|}{\sqrt{A^{2}+B^{2}}}$

(i) The given parallel lines are $15 x+8 y$ âe" $34=0$ and $15 x+8 y+31=0$.

Here, $A=15, B=8, C_1=$ - 34 , and $C_2=31$.

Therefore, the distance between the parallel lines is $d=\frac{|C_1-C_2|}{\sqrt{A^{2}+B^{2}}}=\frac{|-34-31|}{\sqrt{(15)^{2}+(8)^{2}}}$ units $=\frac{|-65|}{17}$ units $=\frac{65}{17}$ units

(ii) The given parallel lines are $I(x+y)+p=0$ and $I(x+y) - r=0$.

$l x+l y+p=0$ and $l x+l y - r=0$

Here, $A=I, B=I, C_1=p$, and $C_2=- r$.

Therefore, the distance between the parallel lines is

$d=\frac{|C_1-C_2|}{\sqrt{A^{2}+B^{2}}}=\frac{|p+r|}{\sqrt{l^{2}+l^{2}}}$ units $=\frac{|p+r|}{\sqrt{2 l^{2}}}$ units $=\frac{|p+r|}{l \sqrt{2}}$ units $=\frac{1}{\sqrt{2}}|\frac{p+r}{l}|$ units

Answer :

The equation of the given line is

$3 x-4 y+2=0$

or $y=\frac{3 x}{4}+\frac{2}{4}$

or $y=\frac{3}{4} x+\frac{1}{2}$, which is of the form $y=m x+c$

$\therefore$ Slope of the given line

$ =\frac{3}{4} $

It is known that parallel lines have the same slope.

$\therefore$ Slope of the other line $=$

$ m=\frac{3}{4} $

Now, the equation of the line that has a slope of $\frac{3}{4}$ and passes through the point $(-$ $2,3)$ is

$ \begin{aligned} & (y-3)=\frac{3}{4}{x-(-2)} \\ & 4 y-12=3 x+6 \\ & \text{ i.e., } 3 x-4 y+18=0 \end{aligned} $

Answer :

The given equation of line is $x-7 y+5=0$.

Or, $y=\frac{1}{7} x+\frac{5}{7}$, which is of the form $y=m x+c$

$\therefore$ Slope of the given line

$ =\frac{1}{7} $

The slope of the line perpendicular to the line having a slope of $\frac{1}{7}$ is $m=-\frac{1}{(\frac{1}{7})}=-7$

The equation of the line with slope $- 7$ and $x$-intercept 3 is given by

$y=m(x - d)$

$\Rightarrow y=- 7(x - 3)$

$\Rightarrow y=- 7 x+21$

$\Rightarrow 7 x+y=21$

Answer :

The given lines are $\sqrt{3} x+y=1$ and $x+\sqrt{3} y=1$.

$y=-\sqrt{3} x+1 \quad \ldots(1) \quad$ and $y=-\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}}$

The slope of line (1) is $m_1=-\sqrt{3}$, while the slope of line (2) is $m_2=-\frac{1}{\sqrt{3}}$.

The acute angle i.e., $\theta$ between the two lines is given by $\tan \theta=|\frac{m_1-m_2}{1+m_1 m_2}|$

$\tan \theta=|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})(-\frac{1}{\sqrt{3}})}|$

$\tan \theta=|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}|=|\frac{-2}{2 \times \sqrt{3}}|$

$\tan \theta=\frac{1}{\sqrt{3}}$

$\theta=30^{\circ}$

Thus, the angle between the given lines is either $30^{\circ}$ or $180^{\circ} - 30^{\circ}=150^{\circ}$.

Answer :

The slope of the line passing through points $(h, 3)$ and $(4,1)$ is

$m_1=\frac{1-3}{4-h}=\frac{-2}{4-h}$

The slope of line $7 x$ - $9 y$ - $19=0$ or $y=\frac{7}{9} x-\frac{19}{9} \quad m_2=\frac{7}{9}$.

It is given that the two lines are perpendicular.

$\therefore m_1 \times m_2=-1$

$\Rightarrow(\frac{-2}{4-h}) \times(\frac{7}{9})=-1$

$\Rightarrow \frac{-14}{36-9 h}=-1$

$\Rightarrow 14=36-9 h$

$\Rightarrow 9 h=36-14$

$\Rightarrow h=\frac{22}{9}$

Thus, the value of $h$ is $\frac{22}{9}$.

Answer :

The slope of line $A x+B y+C=0$ or $y=(\frac{-A}{B}) x+(\frac{-C}{B})$ is $\quad m=-\frac{A}{B}$

It is known that parallel lines have the same slope.

$\therefore$ Slope of the other line $=m=-\frac{A}{B}$

The equation of the line passing through point $(x_1, y_1)$ and having a slope $m=-\frac{A}{B}$ is

$y-y_1=m(x-x_1)$

$y-y_1=-\frac{A}{B}(x-x_1)$

$B(y-y_1)=-A(x-x_1)$

$A(x-x_1)+B(y-y_1)=0$

Hence, the line through point $(x_1, y_1)$ and parallel to line $A x+B y+C=0$ is

$A(x - x_1)+B(y - y_1)=0$

Answer :

It is given that the slope of the first line, $m_1=2$.

Let the slope of the other line be $m_2$.

The angle between the two lines is $60^{\circ}$. $\therefore \tan 60^{\circ}=|\frac{m_1-m_2}{1+m_1 m_2}|$

$\Rightarrow \sqrt{3}=|\frac{2-m_2}{1+2 m_2}|$

$\Rightarrow \sqrt{3}= \pm(\frac{2-m_2}{1+2 m_2})$

$\Rightarrow \sqrt{3}=\frac{2-m_2}{1+2 m_2}$ or $\sqrt{3}=-(\frac{2-m_2}{1+2 m_2})$

$\Rightarrow \sqrt{3}(1+2 m_2)=2-m_2$ or $\sqrt{3}(1+2 m_2)=-(2-m_2)$

$\Rightarrow \sqrt{3}+2 \sqrt{3} m_2+m_2=2$ or $\sqrt{3}+2 \sqrt{3} m_2-m_2=-2$

$\Rightarrow \sqrt{3}+(2 \sqrt{3}+1) m_2=2$ or $\sqrt{3}+(2 \sqrt{3}-1) m_2=-2$

$\Rightarrow m_2=\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}$ or $m_2=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$

Case I: $\quad m_2=(\frac{2-\sqrt{3}}{2 \sqrt{3}+1})$

The equation of the line passing through point $(2,3)$ and having a slope of $\frac{(2-\sqrt{3})}{(2 \sqrt{3}+1)}$ is

$(y-3)=\frac{2-\sqrt{3}}{2 \sqrt{3}+1}(x-2)$

$(2 \sqrt{3}+1) y-3(2 \sqrt{3}+1)=(2-\sqrt{3}) x-2(2-\sqrt{3})$

$(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-4+2 \sqrt{3}+6 \sqrt{3}+3$

$(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$

In this case, the equation of the other line is $(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$.

Case II : $\quad m_2=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$

The equation of the line passing through point $(2,3)$ and having a slope of $\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$ is

$ \begin{aligned} & (y-3)=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}(x-2) \\ & (2 \sqrt{3}-1) y-3(2 \sqrt{3}-1)=-(2+\sqrt{3}) x+2(2+\sqrt{3}) \\ & (2 \sqrt{3}-1) y+(2+\sqrt{3}) x=4+2 \sqrt{3}+6 \sqrt{3}-3 \\ & (2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3} \end{aligned} $

In this case, the equation of the other line is $(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}$.

Thus, the required equation of the other line is $(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$

or $(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}$

Answer :

The right bisector of a line segment bisects the line segment at $90^{\circ}$.

The end-points of the line segment are given as A $(3,4)$ and $B(-{ } 1,2)$.

Accordingly, mid-point of $A B$

$ =(\frac{3-1}{2}, \frac{4+2}{2})=(1,3) $

Slope of $A B=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}$

$\therefore$ Slope of the line perpendicular to $A B=$

$ =-\frac{1}{(\frac{1}{2})}=-2 $

The equation of the line passing through $(1,3)$ and having a slope of - 2 is

(y - 3$)=$ -2 (x - 1)

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

$2 x+y=5$

Thus, the required equation of the line is $2 x+y=5$.

Answer :

Let $(a, b)$ be the coordinates of the foot of the perpendicular from the point $(- 1,3)$ to the line $3 x - 4 y - 16=0$.

Slope of the line joining $( - 1,3) \text{ and } (a, b), m_1=\frac{b-3}{a+1}$

Slope of the line $3 x$- $4 y$- $16=0$ or $y=\frac{3}{4} x-4, m_2=\frac{3}{4}$

Since these two lines are perpendicular, $m_1 m_2=$ - 1

$\therefore(\frac{b-3}{a+1}) \times(\frac{3}{4})=-1$

$\Rightarrow \frac{3 b-9}{4 a+4}=-1$

$\Rightarrow 3 b-9=-4 a-4$

$\Rightarrow 4 a+3 b=5$

Point $(a, b)$ lies on line $3 x$ - $4 y=16$.

$\therefore 3 a$ - $4 b=16$

On solving equations (1) and (2), we obtain

$a=\frac{68}{25}$ and $b=-\frac{49}{25}$

Thus, the required coordinates of the foot of the perpendicular are $(\frac{68}{25},-\frac{49}{25})$

Answer :

The given equation of line is $y=m x+c$.

It is given that the perpendicular from the origin meets the given line at (-1,2).

Therefore, the line joining the points $(0,0)$ and $(- 1,2)$ is perpendicular to the given line.

$\therefore$ Slope of the line joining $(0,0)$ and $( - 1,2)$

$ =\frac{2}{-1}=-2 $

The slope of the given line is $m$.

$\therefore m \times-2=-$

[The two lines are perpendicular]

$\Rightarrow m=\frac{1}{2}$

Since point $( - 1,2)$ lies on the given line, it satisfies the equation $y=m x+c$.

$\therefore 2=m(-1)+c$

$\Rightarrow 2=\frac{1}{2}(-1)+c$

$\Rightarrow c=2+\frac{1}{2}=\frac{5}{2}$

Thus, the respective values of $m$ and $c$ are $\frac{1}{2}$ and $\frac{5}{2}$.

Answer :

The equations of given lines are

$x \cos \theta - y \sin \theta=k \cos 2 \theta \ldots$ (1)

$x \sec \theta+y cosec \theta=k$.

The perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by

$$ \begin{equation*} d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}} . \tag{2} \end{equation*} $$

On comparing equation (1) to the general equation of line i.e., $A x+B y+C=0$, we obtain $A=\cos \theta, B=$ -sin $\theta$, and $C=- k \cos 2 \theta$.

It is given that $p$ is the length of the perpendicular from $(0,0)$ to line (1). $\therefore p=\frac{|A(0)+B(0)+C|}{\sqrt{A^{2}+B^{2}}}=\frac{|C|}{\sqrt{A^{2}+B^{2}}}=\frac{|-k \cos 2 \theta|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=|-k \cos 2 \theta|$

On comparing equation (2) to the general equation of line i.e., $A x+B y+C=0$, we obtain $A=\sec \theta, B=cosec \theta$, and $C=$ -k.

It is given that $q$ is the length of the perpendicular from $(0,0)$ to line (2).

$\therefore q=\frac{|A(0)+B(0)+C|}{\sqrt{A^{2}+B^{2}}}=\frac{|C|}{\sqrt{A^{2}+B^{2}}}=\frac{|-k|}{\sqrt{\sec ^{2} \theta+cosec^{2} \theta}}$

From (3) and (4), we have

$ \begin{aligned} & p^{2}+4 q^{2}=(\mid-k \cos 2 \theta)^{2}+4(\frac{|-k|}{\sqrt{\sec ^{2} \theta+cosec^{2} \theta}})^{2} \\ & =k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{(\sec ^{2} \theta+cosec^{2} \theta)} \\ & =k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{(\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta})} \\ & =k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta})} \\ & =k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{(\frac{1}{\sin ^{2} \theta \cos ^{2} \theta})} \\ & =k^{2} \cos ^{2} 2 \theta+4 k^{2} \sin ^{2} \theta \cos ^{2} \theta \\ & =k^{2} \cos ^{2} 2 \theta+k^{2}(2 \sin ^{2} \theta \cos ^{2} \theta. \\ & =k^{2} \cos ^{2} 2 \theta+k^{2} \sin ^{2} 2 \theta \\ & =k^{2}(\cos ^{2} 2 \theta+\sin ^{2} 2 \theta) \\ & =k^{2} \end{aligned} $

Hence, we proved that $p^{2}+4 q^{2}=k^{2}$.

Answer :

Let $A D$ be the altitude of triangle $A B C$ from vertex $A$.

Accordingly, $A D \perp B C$

The equation of the line passing through point $(2,3)$ and having a slope of 1 is

$(y - 3)=1(x -2 )$

$\Rightarrow x - y+1=0$

$\Rightarrow y $ - $x=1$

Therefore, equation of the altitude from vertex $A=y $ - $x=1$.

Length of $A D=$ Length of the perpendicular from $A(2,3)$ to $B C$

The equation of $B C$ is

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

$\Rightarrow(y+1)=-1(x-4)$

$\Rightarrow y+1=-x+4$

$\Rightarrow x+y-3=0$

The perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by $\sqrt{A^{2}+B^{2}}$.

On comparing equation (1) to the general equation of line $A x+B y+C=0$, we obtain $A=1, B=1$, and $C=$ - 3 .

$\therefore$ Length of $A D$

$ =\frac{|1 \times 2+1 \times 3-3|}{\sqrt{1^{2}+1^{2}}} \text{ units }=\frac{|2|}{\sqrt{2}} \text{ units }=\frac{2}{\sqrt{2}} \text{ units }=\sqrt{2} \text{ units } $

Thus, the equation and the length of the altitude from vertex $A$ are $y -$ $x=1$ and $\sqrt{2}$ units respectively.

Answer :

It is known that the equation of a line whose intercepts on the axes are $a$ and $b$ is

$$ \begin{align*} & \frac{x}{a}+\frac{y}{b}=1 \\ & \text{ or } b x+a y=a b \\ & \text{ or } b x+a y-a b=0 \tag{1} \end{align*} $$

The perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by

$ d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}} $

On comparing equation (1) to the general equation of line $A x+B y+C=0$, we obtain $A=b, B=a$, and $C=a -a b$.

Therefore, if $p$ is the length of the perpendicular from point $(x_1, y_1)=(0,0)$ to line (1), we obtain

$p=\frac{|A(0)+B(0)-a b|}{\sqrt{b^{2}+a^{2}}}$

$\Rightarrow p=\frac{|-a b|}{\sqrt{a^{2}+b^{2}}}$

On squaring both sides, we obtain

$p^{2}=\frac{(-a b)^{2}}{a^{2}+b^{2}}$

$\Rightarrow p^{2}(a^{2}+b^{2})=a^{2} b^{2}$

$\Rightarrow \frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{1}{p^{2}}$

$\Rightarrow \frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

Hence, we showed that $\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$.

Answer :

The given equation of line is

(k - 3) x - $(4 - k^{2}) y+k^{2} $ -$7 k+6=0$

Answer :

Let the intercepts cut by the given lines on the axes be $a$ and $b$.

It is given that

$a+b=1 \ldots(1)$

$a b=- 6$

On solving equations (1) and (2), we obtain

$a=3$ and $b=$ - 2 or $a=-2$ and $b=3$

It is known that the equation of the line whose intercepts on the axes are $a$ and $b$ is

$\frac{x}{a}+\frac{y}{b}=1$ or $b x+a y-a b=0$

Case I: $a=3$ and $b=$ - 2

In this case, the equation of the line is - $2 x+3 y+6=0$, i.e., $2 x$ - $3 y=6$.

Case II: $a=$ - 2 and $b=3$

In this case, the equation of the line is $3 x $ - $2 y+6=0$, i.e., - $3 x+2 y=6$.

Thus, the required equation of the lines are $2 x $ - $3 y=6$ and $-3 x+2 y=6$.

Answer :

Let $(0, b)$ be the point on the $y$-axis whose distance from line $\frac{x}{3}+\frac{y}{4}=1$ is 4 units.

The given line can be written as $4 x+3 y$ - $12=0$

On comparing equation (1) to the general equation of line $A x+B y+C=0$, we obtain $A=4, B=3$, and $C=\hat{a} \ell^{\text{" }} 12$.

It is known that the perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by

$d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}}$.

Therefore, if $(0, b)$ is the point on the $y$-axis whose distance from line $\frac{x}{3}+\frac{y}{4}=1$ is 4 units, then: $4=\frac{|4(0)+3(b)-12|}{\sqrt{4^{2}+3^{2}}}$

$\Rightarrow 4=\frac{|3 b-12|}{5}$

$\Rightarrow 20=|3 b-12|$

$\Rightarrow 20= \pm(3 b-12)$

$\Rightarrow 20=(3 b-12)$ or $20=-(3 b-12)$

$\Rightarrow 3 b=20+12$ or $3 b=-20+12$

$\Rightarrow b=\frac{32}{3}$ or $b=-\frac{8}{3}$

Thus, the required points are $(0, \frac{32}{3})$ and $(0,-\frac{8}{3})$.

Answer :

The equation of the line joining the points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ is given by

$ \begin{aligned} & y-\sin \theta=\frac{\sin \phi-\sin \theta}{\cos \phi-\cos \theta}(x-\cos \theta) \\ & y(\cos \phi-\cos \theta)-\sin \theta(\cos \phi-\cos \theta)=x(\sin \phi-\sin \theta)-\cos \theta(\sin \phi-\sin \theta) \\ & x(\sin \theta-\sin \phi)+y(\cos \phi-\cos \theta)+\cos \theta \sin \phi-\cos \theta \sin \theta-\sin \theta \cos \phi+\sin \theta \cos \theta=0 \\ & x(\sin \theta-\sin \phi)+y(\cos \phi-\cos \theta)+\sin (\phi-\theta)=0 \\ & A x+B y+C=0, \text{ where } A=\sin \theta-\sin \phi, B=\cos \phi-\cos \theta, \text{ and } C=\sin (\phi-\theta) \end{aligned} $

It is known that the perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $(x_1, y_1)$ is given by $d=\frac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}}$.

Therefore, the perpendicular distance $(d)$ of the given line from point $(x_1, y_1)=(0,0)$ is

$ \begin{aligned} & d=\frac{|(\sin \theta-\sin \phi)(0)+(\cos \phi-\cos \theta)(0)+\sin (\phi-\theta)|}{\sqrt{(\sin \theta-\sin \phi)^{2}+(\cos \phi-\cos \theta)^{2}}} \\ & =\frac{|\sin (\phi-\theta)|}{\sqrt{\sin ^{2} \theta+\sin ^{2} \phi-2 \sin \theta \sin \phi+\cos ^{2} \phi+\cos ^{2} \theta-2 \cos \phi \cos \theta}} \\ & =\frac{|\sin (\phi-\theta)|}{\sqrt{(\sin ^{2} \theta+\cos ^{2} \theta)+(\sin ^{2} \phi+\cos ^{2} \phi)-2(\sin \theta \sin \phi+\cos \theta \cos \phi)}} \\ & =\frac{|\sin (\phi-\theta)|}{\sqrt{1+1-2(\cos (\phi-\theta))}} \\ & =\frac{|\sin (\phi-\theta)|}{\sqrt{2(1-\cos (\phi-\theta))}} \\ & =\frac{|\sin (\phi-\theta)|}{\sqrt{2(2 \sin ^{2}(\frac{\phi-\theta}{2}))}} \\ & =\frac{|\sin (\phi-\theta)|}{|2 \sin (\frac{\phi-\theta}{2})|} \end{aligned} $

Answer :

The equation of any line parallel to the $y$-axis is of the form

$x=a$

The two given lines are

$x$ - $7 y+5=0$

$3 x+y=0$

On solving equations (2) and (3), we obtain $x=-\frac{5}{22}$ and $y=\frac{15}{22}$.

Therefore, $(-\frac{5}{22}, \frac{15}{22})$

is the point of intersection of lines (2) and (3).

Since line $x=a$ passes through point $(-\frac{5}{22}, \frac{15}{22}), a=-\frac{5}{22}$.

Thus, the required equation of the line is $x=-\frac{5}{22}$.

Answer :

The equation of the given line is $\frac{x}{4}+\frac{y}{6}=1$.

This equation can also be written as $3 x+2 y$ âє“ $12=0$

$y=\frac{-3}{2} x+6$

, which is of the form $y=m x+c$

$\therefore$ Slope of the given line $=-\frac{3}{2}$

$\therefore$ Slope of line perpendicular to the given line

$ =-\frac{1}{(-\frac{3}{2})}=\frac{2}{3} $

Let the given line intersect the $y$-axis at $(0, y)$.

On substituting $x$ with 0 in the equation of the given line, we obtain $\frac{y}{6}=1 \Rightarrow y=6$

$\therefore$ The given line intersects the $y$-axis at $(0,6)$.

The equation of the line that has a slope of $\frac{2}{3}$ and passes through point $(0,6)$ is

$(y-6)=\frac{2}{3}(x-0)$

$3 y-18=2 x$

$2 x-3 y+18=0$

Thus, the required equation of the line is $2 x-3 y+18=0$.

Answer :

The equations of the given lines are

$y$ - $x=0$

$x+y=0$.

$x$ - $k=0$

The point of intersection of lines (1) and (2) is given by

$x=0$ and $y=0$

The point of intersection of lines (2) and (3) is given by

$x=k$ and $y=$ -k

The point of intersection of lines (3) and (1) is given by

$x=k$ and $y=k$

Thus, the vertices of the triangle formed by the three given lines are $(0,0),(k, -k) $, and $(k, k)$.

We know that the area of a triangle whose vertices are $(x_1, y_1),(x_2, y_2)$, and $(x_3, y_3)$ is

$\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Therefore, area of the triangle formed by the three given lines

$=\frac{1}{2}|0(-k-k)+k(k-0)+k(0+k)|$ square units

$=\frac{1}{2}|k^{2}+k^{2}|$ square units

$=\frac{1}{2}|2 k^{2}|$ square units

$=k^{2}$ square units

Answer :

The equations of the given lines are

$3 x+y-2=0$

$p x+2 y-3=0$ $2 x-y-3=0$…

On solving equations (1) and (3), we obtain

$x=1$ and $y=-1$

Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).

$p(1)+2(-1)-3=0$

$p-2-3=0$

$p=5$

Thus, the required value of $p$ is 5 .

Answer :

The equations of the given lines are

$y=m_1 x+c_1 \ldots(1)$

$y=m_2 x+c_2 \ldots$

$y=m_3 x+c_3$.

On subtracting equation (1) from (2), we obtain

$0=(m_2-m_1) x+(c_2-c_1)$

$\Rightarrow(m_1-m_2) x=c_2-c_1$

$\Rightarrow x=\frac{c_2-c_1}{m_1-m_2}$

On substituting this value of $x$ in (1), we obtain

$ \begin{aligned} & y=m_1(\frac{c_2-c_1}{m_1-m_2})+c_1 \\ & y=\frac{m_1 c_2-m_1 c_1}{m_1-m_2}+c_1 \\ & y=\frac{m_1 c_2-m_1 c_1+m_1 c_1-m_2 c_1}{m_1-m_2} \\ & y=\frac{m_1 c_2-m_2 c_1}{m_1-m_2} \end{aligned} $

$ \therefore(\frac{c_2-c_1}{m_1-m_2}, \frac{m_1 c_2-m_2 c_1}{m_1-m_2}) \text{ is the point of intersection of lines (1) and (2). } $

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).

$ \begin{aligned} & \frac{m_1 c_2-m_2 c_1}{m_1-m_2}=m_3(\frac{c_2-c_1}{m_1-m_2})+c_3 \\ & \frac{m_1 c_2-m_2 c_1}{m_1-m_2}=\frac{m_3 c_2-m_3 c_1+c_3 m_1-c_3 m_2}{m_1-m_2} \\ & m_1 c_2-m_2 c_1-m_3 c_2+m_3 c_1-c_3 m_1+c_3 m_2=0 \\ & m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0 \\ & \text{ Hence, } m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0 . \end{aligned} $

Answer :

Let the slope of the required line be $m_1$.

The given line can be written as $y=\frac{1}{2} x-\frac{3}{2}$, which is of the form $y=m x+c$

$\therefore$ Slope of the given line $=m_2=\frac{1}{2}$

It is given that the angle between the required line and line $x - 2 y=3$ is $45^{\circ}$.

We know that if $\theta$ isthe acute angle between lines $I_1$ and $I_2$ with slopes $m_1$ and $m_2$ respectively, then

$\tan \theta=|\frac{m_2-m_1}{1+m_1 m_2}|$ $\therefore \tan 45^{\circ}=\frac{|m_1-m_2|}{1+m_1 m_2}$

$\Rightarrow I=|\frac{\frac{1}{2}-m_1}{1+\frac{m_1}{2}}|$

$\Rightarrow 1=|\frac{(\frac{1-2 m_1}{2})}{\frac{2+m_1}{2}}|$

$\Rightarrow 1=|\frac{1-2 m_1}{2+m_1}|$

$\Rightarrow 1= \pm(\frac{1-2 m_1}{2+m_1})$

$\Rightarrow 1=\frac{1-2 m_1}{2+m_1}$ or $1=-(\frac{1-2 m_1}{2+m_1})$

$\Rightarrow 2+m_1=1-2 m_1$ or $2+m_1=-1+2 m_1$

$\Rightarrow m_1=-\frac{1}{3}$ or $m_1=3$

Case I: $m_1=3$

The equation of the line passing through $(3,2)$ and having a slope of 3 is:

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

$y$ - $2=3 x$ - 9

$3 x -y=7$

Case II: $m_1=-\frac{1}{3}$

The equation of the line passing through $(3,2)$ and having a slope of $-\frac{1}{3}$ is:

$y-2=-\frac{1}{3}(x-3)$

$3 y-6=-x+3$

$x+3 y=9$

Thus, the equations of the lines are $3 x$ - $y=7$ and $x+3 y=9$.

Answer :

Let the equation of the line having equal intercepts on the axes be

$\frac{x}{a}+\frac{y}{a}=1$

Or $x+y=a$

On solving equations $4 x+7 y - 3=0$ and $2 x $ - $3 y+1=0$, we obtain $x=\frac{1}{13}$ and $y=\frac{5}{13}$.

$\therefore(\frac{1}{13}, \frac{5}{13})$ is the point of intersection of the two given lines.

Since equation (1) passes through point $(\frac{1}{13}, \frac{5}{13})$,

$\frac{1}{13}+\frac{5}{13}=a$

$\Rightarrow a=\frac{6}{13}$

$\therefore$ Equation (1) becomes

$ x+y=\frac{6}{13} \text{, i.e., } 13 x+13 y=6 $

Thus, the required equation of the line is $13 x+13 y=6$

Answer :

Let the equation of the line passing through the origin be $y=m_1 x$.

If this line makes an angle of $\theta$ with line $y=m x+c$, then angle $\theta$ is given by $\therefore \tan \theta=|\frac{m_1-m}{1+m_1 m}|$

$\Rightarrow \tan \theta=|\frac{\frac{y}{x}-m}{1+\frac{y}{x} m}|$

$\Rightarrow \tan \theta= \pm(\frac{\frac{y}{x}-m}{1+\frac{y}{x} m})$

$\Rightarrow \tan \theta=\frac{\frac{y}{x}-m}{1+\frac{y}{x} m}$ or $\tan \theta=-(\frac{\frac{y}{x}-m}{1+\frac{y}{x} m})$

Case I: $\quad \tan \theta=\frac{\frac{y}{x}-m}{1+\frac{y}{x} m}$

$ \tan \theta=\frac{\frac{y}{x}-m}{1+\frac{y}{x} m} $

$\Rightarrow \tan \theta+\frac{y}{x} m \tan \theta=\frac{y}{x}-m$

$\Rightarrow m+\tan \theta=\frac{y}{x}(1-m \tan \theta)$

$\Rightarrow \frac{y}{x}=\frac{m+\tan \theta}{1-m \tan \theta}$

Case II:

$ \tan \theta=-(\frac{\frac{y}{x}-m}{1+\frac{y}{x} m}) $

$\tan \theta=-(\frac{\frac{y}{x}-m}{1+\frac{y}{x} m})$

$\Rightarrow \tan \theta+\frac{y}{x} m \tan \theta=-\frac{y}{x}+m$

$\Rightarrow \frac{y}{x}(1+m \tan \theta)=m-\tan \theta$

$\Rightarrow \frac{y}{x}=\frac{m-\tan \theta}{1+m \tan \theta}$

Therefore, the required line is given by $\frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}$

Answer :

The equation of the line joining the points ( $.-{ }^{*} 1,1)$ and $(5,7)$ is given by

$y-1=\frac{7-1}{5+1}(x+1)$

$y-1=\frac{6}{6}(x+1)$

$x-y+2=0$

The equation of the given line is

$x+y-=0$…

The point of intersection of lines (1) and (2) is given by

$x=1$ and $y=3$

Let point $(1,3)$ divide the line segment joining $(- 1,1)$ and $(5,7)$ in the ratio $1: k$. Accordingly, by section formula, $(1,3)=(\frac{k(-1)+1(5)}{1+k}, \frac{k(1)+1(7)}{1+k})$

$\Rightarrow(1,3)=(\frac{-k+5}{1+k}, \frac{k+7}{1+k})$

$\Rightarrow \frac{-k+5}{1+k}=1, \frac{k+7}{1+k}=3$

$\therefore \frac{-k+5}{1+k}=1$

$\Rightarrow-k+5=1+k$

$\Rightarrow 2 k=4$

$\Rightarrow k=2$

Thus, the line joining the points ( $-$ “ 1,1$)$ and $(5,7)$ is divided by line $x+y=4$ in the ratio $1: 2$.

Answer :

The given lines are

$2 x - y=0$…

$4 x+7 y+5=0 \ldots$

A $(1,2)$ is a point on line (1).

Let $B$ be the point of intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain

$ x=\frac{-5}{18} \text{ and } y=\frac{-5}{9} $

$\therefore$ Coordinates of point $B$ are $(\frac{-5}{18}, \frac{-5}{9})$.

By using distance formula, the distance between points $A$ and $B$ can be obtained as

$AB=\sqrt{(1+\frac{5}{18})^{2}+(2+\frac{5}{9})^{2}}$ units

$=\sqrt{(\frac{23}{18})^{2}+(\frac{23}{9})^{2}}$ units

$=\sqrt{(\frac{23}{2 \times 9})^{2}+(\frac{23}{9})^{2}}$ units

$=\sqrt{(\frac{23}{9})^{2}(\frac{1}{2})^{2}+(\frac{23}{9})^{2}}$ units

$=\sqrt{(\frac{23}{9})^{2}(\frac{1}{4}+1)}$ units

$=\frac{23}{9} \sqrt{\frac{5}{4}}$ units

$=\frac{23}{9} \times \frac{\sqrt{5}}{2}$ units

$=\frac{23 \sqrt{5}}{18}$ units

Thus, the required distance is 18

units

Answer :

Let $y=m x+c$ be the line through point $($ - 1,2$)$.

Accordingly, $2=m(- 1)+c$.

$\Rightarrow 2=- m+c$

$\Rightarrow c=m+2$

$\therefore y=m x+m+2$

The given line is

$x+y=4$.

On solving equations (1) and (2), we obtain

$ \begin{aligned} & x=\frac{2-m}{m+1} \text{ and } y=\frac{5 m+2}{m+1} \\ & \therefore(\frac{2-m}{m+1}, \frac{5 m+2}{m+1}) \text{ is the point of intersection of lines (1) and (2). } \end{aligned} $

Since this point is at a distance of 3 units from point (- 1,2 ), according to distance formula,

$ \begin{aligned} & \sqrt{(\frac{2-m}{m+1}+1)^{2}+(\frac{5 m+2}{m+1}-2)^{2}}=3 \\ \Rightarrow & (\frac{2-m+m+1}{m+1})^{2}+(\frac{5 m+2-2 m-2}{m+1})^{2}=3^{2} \\ \Rightarrow & \frac{9}{(m+1)^{2}}+\frac{9 m^{2}}{(m+1)^{2}}=9 \\ \Rightarrow & \frac{1+m^{2}}{(m+1)^{2}}=1 \\ \Rightarrow & 1+m^{2}=m^{2}+1+2 m \\ \Rightarrow & 2 m=0 \\ \Rightarrow & m=0 \end{aligned} $

Thus, the slope of the required line must be zero i.e., the line must be parallel to the $x$-axis.

Answer :

Let $A(1,3)$ and $B(-4,1)$ be the coordinates of the end points of the hypotenuse.

Now, plotting the line segment joining the points $A(1,3)$ and $B(-4,1)$ on the coordinate plane, we will get two right triangles with $A B$ as the hypotenuse. Now from the diagram, it is clear that the point of intersection of the other two legs of the right triangle having $A B$ as the hypotenuse can be either $P$ or $Q$.

CASE 1: When $\angle $ APB is taken.

The perpendicular sides in $\angle $ APB are AP and PB.

Now, side $P B$ is parallel to $x$-axis and at a distance of 1 units above $x$-axis.

So, equation of $P B$ is, $y=1$ or $y-1=0$.

The side AP is parallel to $y$-axis and at a distance of 1 units on the right of $y$-axis.

So, equation of $A P$ is $x=1$ or $x-1=0$.

CASE 2: When $\angle $ AQB is taken.

The perpendicular sides in $\angle $ AQB are AQ and QB.

Now, side $A Q$ is parallel to $x$-axis and at a distance of 3 units above $x$-axis.

So, equation of $A Q$ is, $y=3$ or $y-3=0$.

The side QB is parallel to $y$-axis and at a distance of 4 units on the left of $y$-axis.

So, equation of $QB$ is $x=-4$ or $x+4=0$.

Hence, the equation of the legs are :

$x=1, y=1$ or $x=-4, y=3$

Answer :

The equation of the given line is

$x+3 y=7$

Let point $B(a, b)$ be the image of point $A(3,8)$.

Accordingly, line (1) is the perpendicular bisector of $A B$.

Slope of $AB=\frac{b-8}{a-3}$, while the slope of line $(1)=-\frac{1}{3}$

Since line (1) is perpendicular to $A B$,

$$ \begin{align*} & (\frac{b-8}{a-3}) \times(-\frac{1}{3})=-1 \\ & \Rightarrow \frac{b-8}{3 a-9}=1 \\ & \Rightarrow b-8=3 a-9 \\ & \Rightarrow 3 a-b=1 \tag{2} \end{align*} $$

Mid-point of $AB=(\frac{a+3}{2}, \frac{b+8}{2})$

The mid-point of line segment $A B$ will also satisfy line (1).

Hence, from equation (1), we have

$(\frac{a+3}{2})+3(\frac{b+8}{2})=7$

$\Rightarrow a+3+3 b+24=14$

$\Rightarrow a+3 b=-13$

On solving equations (2) and (3), we obtain $a=$ - 1 and $b=$ -4.

Thus, the image of the given point with respect to the given line is (-1, -4).

Answer :

The equations of the given lines are

$y=3 x+1 \ldots(1)$

$2 y=x+3$ $y=m x+4 \ldots(3)$

Slope of line (1), $m_1=3$

Slope of line (2), $m_2=\frac{1}{2}$

Slope of line (3), $m_3=m$

It is given that lines (1) and (2) are equally inclined to line (3). This means that the angle between lines (1) and (3) equals the angle between lines (2) and (3).

$\therefore|\frac{m_1-m_3}{1+m_1 m_3}|=|\frac{m_2-m_3}{1+m_2 m_3}|$

$\Rightarrow|\frac{3-m}{1+3 m}|=|\frac{\frac{1}{2}-m}{1+\frac{1}{2} m}|$

$\Rightarrow|\frac{3-m}{1+3 m}|=|\frac{1-2 m}{m+2}|$

$\Rightarrow \frac{3-m}{1+3 m}= \pm(\frac{1-2 m}{m+2})$

$\Rightarrow \frac{3-m}{1+3 m}=\frac{1-2 m}{m+2}$ or $\frac{3-m}{1+3 m}=-(\frac{1-2 m}{m+2})$

If $\frac{3-m}{1+3 m}=\frac{1-2 m}{m+2}$, then

$(3-m)(m+2)=(1-2 m)(1+3 m)$

$\Rightarrow-m^{2}+m+6=1+m-6 m^{2}$

$\Rightarrow 5 m^{2}+5=0$

$\Rightarrow(m^{2}+1)=0$

$\Rightarrow m=\sqrt{-1}$, which is not real

Hence, this case is not posible.

If $\frac{3-m}{1+3 m}=-(\frac{1-2 m}{m+2})$, then

$\Rightarrow(3-m)(m+2)=-(1-2 m)(1+3 m)$

$\Rightarrow-m^{2}+m+6=-(1+m-6 m^{2})$

$\Rightarrow 7 m^{2}-2 m-7=0$

$\Rightarrow m=\frac{2 \pm \sqrt{4-4(7)(-7)}}{2(7)}$

$\Rightarrow m=\frac{2 \pm 2 \sqrt{1+49}}{14}$

$\Rightarrow m=\frac{1 \pm 5 \sqrt{2}}{7}$

Thus, the required value of $m$ is $\frac{1 \pm 5 \sqrt{2}}{7}$.

Answer :

The equations of the given lines are

$x+y$ - $5=0$

$3 x$ - $2 y+7=0$

The perpendicular distances of $P(x, y)$ from lines (1) and (2) are respectively given by

$ d_1=\frac{|x+y-5|}{\sqrt{(1)^{2}+(1)^{2}}} \text{ and } d_2=\frac{|3 x-2 y+7|}{\sqrt{(3)^{2}+(-2)^{2}}} $

i.e., $d_1=\frac{|x+y-5|}{\sqrt{2}}$ and $d_2=\frac{|3 x-2 y+7|}{\sqrt{13}}$

It is given that $d_1+d_2=10$ $\therefore \frac{|x+y-5|}{\sqrt{2}}+\frac{|3 x-2 y+7|}{\sqrt{13}}=10$

$\Rightarrow \sqrt{13}|x+y-5|+\sqrt{2}|3 x-2 y+7|-10 \sqrt{26}=0$

$\Rightarrow \sqrt{13}(x+y-5)+\sqrt{2}(3 x-2 y+7)-10 \sqrt{26}=0$

[Assuming $(x+y-5)$ and $(3 x-2 y+7)$ are positive]

$\Rightarrow \sqrt{13} x+\sqrt{13} y-5 \sqrt{13}+3 \sqrt{2} x-2 \sqrt{2} y+7 \sqrt{2}-10 \sqrt{26}=0$

$\Rightarrow x(\sqrt{13}+3 \sqrt{2})+y(\sqrt{13}-2 \sqrt{2})+(7 \sqrt{2}-5 \sqrt{13}-10 \sqrt{26})=0$

which is the equation of a line.

Similarly, we can obtain the equation of line for any signs of $(x+y-5)$ and $(3 x-2 y+7)$.

Thus, point $P$ must move on a line.

Answer :

The equations of the given lines are

$9 x+6 y - 7=0$

$3 x+2 y+6=0$…

Let $P(h, k)$ be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of $P(h, k)$ from line (1) is given by

$ d_1=\frac{|9 h+6 k-7|}{(9)^{2}+(6)^{2}}=\frac{|9 h+6 k-7|}{\sqrt{117}}=\frac{|9 h+6 k-7|}{3 \sqrt{13}} $

The perpendicular distance of $P(h, k)$ from line (2) is given by

$ d_2=\frac{|3 h+2 k+6|}{\sqrt{(3)^{2}+(2)^{2}}}=\frac{|3 h+2 k+6|}{\sqrt{13}} $

Since $P(h, k)$ is equidistant from lines (1) and (2), $d_1=d_2$

$$ \begin{aligned} & \therefore \frac{|9 h+6 k-7|}{3 \sqrt{13}}=\frac{|3 h+2 k+6|}{\sqrt{13}} \\ & \Rightarrow 9 h+6 k-7|=3| 3 h+2 k+6 \mid \\ & \Rightarrow 9 h+6 k-7 \mid= \pm 3(3 h+2 k+6) \\ & \Rightarrow 9 h+6 k-7=3(3 h+2 k+6) \text{ or } 9 h+6 k-7=-3(3 h+2 k+6) \end{aligned} $$

The case $9 h+6 k-7=3(3 h+2 k+6)$ is not possible as

$9 h+6 k-7=3(3 h+2 k+6) \Rightarrow-7=18$ (which is absurd)

$\therefore h+6 k-7=-3(3 h+2 k+6)$

$9h + 6k -7 = -9h-6k-18$

$\Rightarrow 18 h+12 k+11=0$

Thus, the required equation of the line is $18 x+12 y+11=0$.

Answer :

$9h + 6k -7 = -9h-6k-18$

Let the coordinates of point $A$ be $(a, 0)$.

Draw a line (AL) perpendicular to the $x$-axis.

We know that angle of incidence is equal to angle of reflection. Hence, let

$\angle BAL=\angle CAL=\varnothing$

Let $\angle CAX=\theta$

$\therefore \angle OAB=180^{\circ} a \ell(\theta+2 \Phi)=180^{\circ} a \ell[\theta+2(90^{\circ} a \ell \theta)]$

$ \begin{aligned} & =180^{\circ} a \epsilon \theta a \epsilon 180^{\circ}+2 \theta \\ & =\theta \\ & \therefore \angle B A X=180^{\circ} - \theta \end{aligned} $

Now, slope of line $AC=\frac{3-0}{5-a}$

$\Rightarrow \tan \theta=\frac{3}{5-a}$

Slope of line $AB=\frac{2-0}{1-a}$

$\Rightarrow \tan (180^{\circ}-\theta)=\frac{2}{1-a}$

$\Rightarrow-\tan \theta=\frac{2}{1-a}$

$\Rightarrow \tan \theta=\frac{2}{a-1}$

From equations (1) and (2), we obtain

$ \begin{aligned} & \frac{3}{5-a}=\frac{2}{a-1} \\ & \Rightarrow 3 a-3=10-2 a \\ & \Rightarrow a=\frac{13}{5} \end{aligned} $

Thus, the coordinates of point $A$ are $(\frac{13}{5}, 0)$.

Answer :

The equation of the given line is

$\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$

Or, $b x \cos \theta+a y \sin \theta-a b=0$

Length of the perpendicular from point $(\sqrt{a^{2}-b^{2}}, 0)$ to line (1) is

$p_1=\frac{|b \cos \theta(\sqrt{a^{2}-b^{2}})+a \sin \theta(0)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}$

Length of the perpendicular from point $(-\sqrt{a^{2}-b^{2}}, 0)$ to line (2) is

$p_2=\frac{|b \cos \theta(-\sqrt{a^{2}-b^{2}})+a \sin \theta(0)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}+a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}$

On multiplying equations (2) and (3), we obtain

$ \begin{aligned} & p_1 p_2=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}-a b|(b \cos \theta \sqrt{a^{2}-b^{2}}+a b) \mid}{(\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta})^{2}} \\ & =\frac{|(b \cos \theta \sqrt{a^{2}-b^{2}}-a b)(b \cos \theta \sqrt{a^{2}-b^{2}}+a b)|}{(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta)} \\ & =\frac{|(b \cos \theta \sqrt{a^{2}-b^{2}})^{2}-(a b)^{2}|}{(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta)} \\ & =\frac{|b^{2} \cos ^{2} \theta(a^{2}-b^{2})-a^{2} b^{2}|}{(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta)} \\ & =\frac{|a^{2} b^{2} \cos ^{2} \theta-b^{4} \cos ^{2} \theta-a^{2} b^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} \\ & =\frac{b^{2}|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} \\ & =\frac{b^{2}|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2} \sin ^{2} \theta-a^{2} \cos ^{2} \theta|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} \quad[\sin ^{2} \theta+\cos ^{2} \theta=1] \\ & =\frac{b^{2}|-(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta)|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} \\ & =\frac{b^{2}(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta)}{(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta)} \\ & =b^{2} \end{aligned} $

Hence, proved.

Answer :

The equations of the given lines are

$2 x$ - $3 y+4=0$

$3 x+4 y-=0$

$6 x$ - $7 y+8=0$

The person is standing at the junction of the paths represented by lines (1) and (2).

On solving equations (1) and (2), we obtain

$ x=-\frac{1}{17} \text{ and } y=\frac{22}{17} $

Thus, the person is standing at point $(-\frac{1}{17}, \frac{22}{17})$.

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point $(-\frac{1}{17}, \frac{22}{17})$

Slope of the line $(3)=\frac{6}{7}$

$\therefore$ Slope of the line perpendicular to line (3)

$ =-\frac{1}{(\frac{6}{7})}=-\frac{7}{6} $

The equation of the line passing through $(-\frac{1}{17}, \frac{22}{17})$ and having a slope of $-\frac{7}{6}$ is given by

$ \begin{aligned} & (y-\frac{22}{17})=-\frac{7}{6}(x+\frac{1}{17}) \\ & 6(17 y-22)=-7(17 x+1) \\ & 102 y-132=-119 x-7 \\ & 119 x+102 y=125 \end{aligned} $

Hence, the path that the person should follow is $119 x+102 y=125$.