Answer :

The given equation of line is $(k - 3) x - (4 - k^{2}) y+k^{2} - 7 k+6=0$

(a): If the given line is parallel to the $x$-axis, then

Slope of the given line $=$ Slope of the $x$-axis

The given line can be written as $( 4 - k^{2}) y=(k -3 ) x+k^{2}- 7 k+6=0$

$y=\dfrac{(k-3)}{(4-k^{2})} x+\dfrac{k^{2}-7 k+6}{(4-k^{2})}$ , which is of the form $y=m x+c$.

$\therefore \quad $ Slope of the given line $=\dfrac{(k-3)}{(4-k^{2})}$

Slope of the $x$-axis $=0$

$\therefore \quad \dfrac{(k-3)}{(4-k^{2})}=0$

$\Rightarrow k-3=0$

$\Rightarrow k=3$

Thus, if the given line is parallel to the $x$-axis, then the value of $k$ is $3 .$

(b): If the given line is parallel to the $y$-axis, it is vertical. Hence, its slope will be undefined.

The slope of the given line is $\dfrac{(k-3)}{(4-k^{2})}$.

Now, $\dfrac{(k-3)}{(4-k^{2})}$ is undefined at $k^{2}=4$

$\Rightarrow k^{2}=4$

$\Rightarrow k= \pm 2$

Thus, if the given line is parallel to the $y$-axis, then the value of $k$ is $\pm 2$.

(c): If the given line is passing through the origin, then point $(0,0)$ satisfies the given equation of line.

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

$k^{2}-7 k+6=0$

$k^{2}-6 k-k+6=0$

$(k-6)(k-1)=0$

$k=1$ or 6

Thus, if the given line is passing through the origin, then the value of $k$ is either $1$ or $6 .$

Answer :

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

It is given that

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

$a b=- 6 \qquad\ldots(2)$

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

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

Case I:

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

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

Case II:

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

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

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 $\dfrac{x}{3}+\dfrac{y}{4}=1$ 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 $C= -12$.

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

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

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

$\quad 4=\dfrac{|4(0)+3(b)-12|}{\sqrt{4^{2}+3^{2}}}$

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

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

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

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

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

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

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

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=\left(\dfrac{\sin \phi-\sin \theta}{\cos \phi-\cos \theta}\right)(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 $d=\dfrac{|A x_1+B y_1+C|}{\sqrt{A^{2}+B^{2}}}$.

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

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

Answer :

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

$x=a\qquad \ldots(1)$

The two given lines are

$x- 7 y+5=0\qquad \ldots(2)$

$3 x+y=0\qquad \ldots(3)$

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

Therefore, $\left(-\dfrac{5}{22}, \dfrac{15}{22}\right)$ is the point of intersection of lines $\left(2\right)$ and $\left(3\right)$

Since line $x=a$ passes through point $\left(-\dfrac{5}{22}, \dfrac{15}{22}\right), a=-\dfrac{5}{22}$

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

Answer :

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

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

$y=\dfrac{-3}{2} x+6$, which is of the form $y=m x+c$

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

$\therefore \quad $ Slope of line perpendicular to the given line $ =-\dfrac{1}{\left(-\dfrac{3}{2}\right)}=\dfrac{2}{3} $

Let the given line intersect the $y$-axis at $\left(0, y\right)$

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

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

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

$\left(y-6\right)=\dfrac{2}{3}\left(x-0\right)$

$3 y-18=2 x$

$2 x-3 y+18=0$

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

Answer :

The equations of the given lines are

$y- x=0\qquad \ldots(1)$

$x+y=0\qquad \ldots(2)$

$x- k=0\qquad \ldots(3)$

The point of intersection of lines $\left(1\right)$ and $\left(2\right)$ is given by

$x=0$ and $y=0$

The point of intersection of lines $\left(2\right)$ and $\left(3\right)$ is given by

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

The point of intersection of lines $\left(3\right)$ and $\left(1\right)$ is given by

$x=k$ and $y=k$

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

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

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

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

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

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

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

$=k^{2}$ square units

Answer :

The equations of the given lines are

$3 x+y-2=0\qquad \ldots(1)$

$p x+2 y-3=0\qquad \ldots(2)$

$2 x-y-3=0\qquad \ldots(3)$

On solving equations $\left(1\right)$ and $\left(3\right),$ we obtain

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

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

$p\left(1\right)+2\left(-1\right)-3=0$

$p-2-3=0$

$p=5$

Thus, the required value of $p$ is 5

Answer :

The equations of the given lines are

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

$y=m_2 x+c_2 \qquad \ldots(2)$

$y=m_3 x+c_3\qquad \ldots(3)$

On subtracting equation $\left(1\right)$ from $\left(2\right),$ we obtain

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

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

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

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

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

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

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

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

Answer :

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

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

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

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

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

$\tan \theta=\left|\dfrac{m_2-m_1}{1+m_1 m_2}\right|$

$\therefore \quad \tan 45^{\circ}=\dfrac{\big|m_1-m_2\big|}{1+m_1 m_2}$

$\Rightarrow 1=\left|\dfrac{\dfrac{1}{2}-m_1}{1+\dfrac{m_1}{2}}\right|\Rightarrow 1 =\left|\dfrac{\left(\dfrac{1-2 m_1}{2}\right)}{\dfrac{2+m_1}{2}}\right|$

$\Rightarrow 1 =\left|\dfrac{1-2 m_1}{2+m_1}\right|\Rightarrow 1 = \pm\left(\dfrac{1-2 m_1}{2+m_1}\right)$

$\Rightarrow 1 =\dfrac{1-2 m_1}{2+m_1}$ or $1=-\left(\dfrac{1-2 m_1}{2+m_1}\right)$

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

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

Case I:

$m_1=3$

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

$y - 2 = 3\left(x - 3 \right)$

$y - 2=3 x - 9$

$3 x -y=7$

Case II:

$m_1=-\dfrac{1}{3}$

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

$y-2=-\dfrac{1}{3}\left(x-3\right)$

$3 y-6=-x+3$

$x+3 y=9$

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

or $ ~ x+y=a\qquad \ldots(1)$

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

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

Since equation $\left(1\right)$ passes through point $\left(\dfrac{1}{13}, \dfrac{5}{13}\right)$, $\dfrac{1}{13}+\dfrac{5}{13}=a$

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

$\therefore \quad $ Equation $\left(1\right)$ becomes

$ x+y=\dfrac{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

$\therefore \quad \tan \theta=\bigg|\dfrac{m_1-m}{1+m_1 m}\bigg|$

$\Rightarrow\quad \tan \theta=\left|\dfrac{\dfrac{y}{x}-m}{1+\dfrac{y}{x} m}\right|$

$\Rightarrow\quad \tan \theta= \pm\left(\dfrac{\dfrac{y}{x}-m}{1+\dfrac{y}{x} m}\right)$

$\Rightarrow\quad \tan \theta=\left(\dfrac{\dfrac{y}{x}-m}{1+\dfrac{y}{x} m}\right) \ $ or $ \ \tan \theta=-\left(\dfrac{\dfrac{y}{x}-m}{1+\dfrac{y}{x} m}\right)$

Case I:

$ \tan \theta=\left(\dfrac{\dfrac{y}{x}-m}{1+\dfrac{y}{x} m}\right)$

$ \tan \theta=\left(\dfrac{\dfrac{y}{x}-m}{1+\dfrac{y}{x} m}\right) $

$\Rightarrow \quad \left(1+\dfrac{y}{x} m\right) \tan \theta=\dfrac{y}{x}-m$

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

$\Rightarrow\quad m+\tan \theta=\dfrac{y}{x}\left(1-m \tan \theta\right)$

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

Case II:

$ \tan \theta=-\left(\dfrac{\dfrac{y}{x}-m}{1+\dfrac{y}{x} m}\right) $

$\tan \theta=-\left(\dfrac{\dfrac{y}{x}-m}{1+\dfrac{y}{x} m}\right)$

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

$\Rightarrow\quad \dfrac{y}{x}\left(1+m \tan \theta\right)=m-\tan \theta$

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

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

Answer :

The equation of the line joining the points $\left( -1,1\right)$ and $\left(5,7\right)$ is given by

$y-1=\left(\dfrac{7-1}{5+1}\right)\left(x+1\right)$

$y-1=\left(\dfrac{6}{6}\right)\left(x+1\right)$

$x-y+2=0\qquad \ldots(1)$

The equation of the given line is

$x +y -4 =0\qquad \ldots(2)$

The point of intersection of lines $\left(1\right)$ and $\left(2\right)$ is given by

$x=1$ and $y=3$

Let point $\left(1,3\right)$ divide the line segment joining $\left(- 1,1\right)$ and $\left(5,7\right)$ in the ratio $1: k$.

Accordingly, by section formula,

$\left(1,3\right)=\left(\dfrac{k\left(-1\right)+1\left(5\right)}{1+k}, \dfrac{k\left(1\right)+1\left(7\right)}{1+k}\right)$

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

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

$\therefore \quad \dfrac{-k+5}{1+k}=1$

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

$\Rightarrow 2 k=4$

$\Rightarrow k=2$

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

Answer :

The given lines are

$2 x - y=0\qquad \ldots(1)$

$4 x+7 y+5=0\qquad \ldots(2)$

A $\left(1,2\right)$ is a point on line $\left(1\right)$

Let $B$ be the point of intersection of lines $\left(1\right)$ and $\left(2\right)$

On solving equations $\left(1\right)$ and $\left(2\right),$ we obtain

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

$\therefore \quad $ Coordinates of point $B$ are $\left(\dfrac{-5}{18}, \dfrac{-5}{9}\right)$

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

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

$\qquad=\sqrt{\left(\dfrac{23}{18}\right)^{2}+\left(\dfrac{23}{9}\right)^{2}}$ units

$\qquad=\sqrt{\left(\dfrac{23}{2 \times 9}\right)^{2}+\left(\dfrac{23}{9}\right)^{2}}$ units

$\qquad=\sqrt{\left(\dfrac{23}{9}\right)^{2}\left(\dfrac{1}{2}\right)^{2}+\left(\dfrac{23}{9}\right)^{2}}$ units

$\qquad=\sqrt{\left(\dfrac{23}{9}\right)^{2}\left(\dfrac{1}{4}+1\right)}$ units $=\dfrac{23}{9} \sqrt{\dfrac{5}{4}}$ units

$\qquad=\dfrac{23}{9} \times \dfrac{\sqrt{5}}{2}$ units $=\dfrac{23 \sqrt{5}}{18}$ units

Thus, the required distance is $\dfrac{23 \sqrt{5}}{18}$ units

Answer :

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

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

$\Rightarrow 2=- m+c$

$\Rightarrow c=m+2$

$\therefore \quad y=m x+m+2\qquad \ldots(1)$

The given line is $ ~ x+y=4\qquad \ldots(2)$

On solving equations $\left(1\right)$ and $\left(2\right),$ we obtain

$ \begin{aligned} & x=\dfrac{2-m}{m+1} \text{ and } y=\dfrac{5 m+2}{m+1} \\ \\ & \therefore \quad \left(\dfrac{2-m}{m+1}, \dfrac{5 m+2}{m+1}\right) \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 $\left(- 1,2 \right), $ according to distance formula,

$ \begin{aligned} & \sqrt{\left(\dfrac{2-m}{m+1}+1\right)^{2}+\left(\dfrac{5 m+2}{m+1}-2\right)^{2}}=3 \\ \\ \Rightarrow & \left(\dfrac{2-m+m+1}{m+1}\right)^{2}+\left(\dfrac{5 m+2-2 m-2}{m+1}\right)^{2}=3^{2} \\ \\ \Rightarrow & \dfrac{9}{\left(m+1\right)^{2}}+\dfrac{9 m^{2}}{\left(m+1\right)^{2}}=9 \\ \\ \Rightarrow & \dfrac{1+m^{2}}{\left(m+1\right)^{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 $A\left(1,3\right)$ and $B\left(-4,1\right)$ be the coordinates of the end points of the hypotenuse.

Now, plotting the line segment joining the points $A\left(1,3\right)$ and $B\left(-4,1\right)$ 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

$x+3 y=7\qquad \ldots(1)$

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

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

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

Since line $\left(1\right)$ is perpendicular to $A B$, $ \left(\dfrac{b-8}{a-3}\right) \times\left(-\dfrac{1}{3}\right)=-1 $

$ \Rightarrow \dfrac{b-8}{3 a-9}=1 $

$ \Rightarrow b-8=3 a-9 $

$ \Rightarrow 3 a-b=1 \qquad \ldots{(2)}$

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

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

Hence, from equation $\left(1\right),$ we have

$\left(\dfrac{a+3}{2}\right)+3\left(\dfrac{b+8}{2}\right)=7$

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

$\Rightarrow a+3 b=-13\qquad \ldots(3)$

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

Thus, the image of the given point with respect to the given line is $\left(-1, -4\right)$

Answer :

The equations of the given lines are

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

$2 y=x+3\qquad \ldots(2)$

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

Slope of line $\left(1\right),$ $m_1=3$

Slope of line $\left(2\right),$ $m_2=\dfrac{1}{2}$

Slope of line $\left(3\right),$ $m_3=m$

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

$\therefore \quad \ \ \bigg|\dfrac{m_1-m_3}{1+m_1 m_3}\bigg|=\bigg|\dfrac{m_2-m_3}{1+m_2 m_3}\bigg|$

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

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

$\Rightarrow \dfrac{3-m}{1+3 m}= \pm\left(\dfrac{1-2 m}{m+2}\right)$

$\Rightarrow \dfrac{3-m}{1+3 m}=\dfrac{1-2 m}{m+2}$ or $\dfrac{3-m}{1+3 m}=-\left(\dfrac{1-2 m}{m+2}\right)$

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

$\left(3-m\right)\left(m+2\right)=\left(1-2 m\right)\left(1+3 m\right)$

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

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

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

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

Hence, this case is not posible.

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

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

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

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

$\Rightarrow m=\dfrac{2 \pm \sqrt{4-4\left(7\right)\left(-7\right)}}{2\left(7\right)}$

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

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

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

Answer :

The equations of the given lines are

$x+y - 5=0\qquad \ldots(1)$

$3 x- 2 y+7=0\qquad \ldots(2)$

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

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

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

It is given that $d_1+d_2=10$

$\therefore \quad \dfrac{|x+y-5|}{\sqrt{2}}+\dfrac{|3 x-2 y+7|}{\sqrt{13}}=10$

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

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

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

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

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

which is the equation of a line.

Similarly, we can obtain the equation of line for any signs of $\left(x+y-5=0\right)$ and $\left(3 x-2 y+7=0\right)$

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

Answer :

The equations of the given lines are

$9 x+6 y - 7=0\qquad \ldots(1)$

$3 x+2 y+6=0\qquad \ldots(2)$

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

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

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

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

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

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

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

$9 h+6 k-7=3\left(3 h+2 k+6\right) $

$\Rightarrow-7=18$ (which is absurd)

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

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

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

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

Answer :

Let the coordinates of point $A$ be $\left(a, 0\right)$

Draw a line $\left(AL\right)$ 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 \quad \angle OAB=180^{\circ} - \left(\theta+2 \varnothing\right)=180^{\circ} - [\theta+2\left(90^{\circ} - \theta\right)]$

$ \begin{aligned} \therefore \quad \angle OAB & =180^{\circ} - \left(\theta+2 \varnothing\right)=180^{\circ} - \big[\theta+2\left(90^{\circ} - \theta\right)\big] \\ \\ & =180^{\circ} - \theta - 180^{\circ}+2 \theta \\ \\ & =\theta \end{aligned} $

$ \therefore \quad \angle B A X=180^{\circ} - \theta$

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

$\Rightarrow \tan \theta=\dfrac{3}{5-a}\qquad \ldots(1)$

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

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

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

$\Rightarrow \tan \theta=\dfrac{2}{a-1}\qquad \ldots(2)$

From equations $\left(1\right)$ and $\left(2\right),$ we obtain

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

Thus, the coordinates of point $A$ are $\left(\dfrac{13}{5}, 0\right)$

Answer :

The equation of the given line is

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

or, $b x \cos \theta+a y \sin \theta-a b=0\qquad \ldots(1)$

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

$p_1=\dfrac{|b \cos \theta\left(\sqrt{a^{2}-b^{2}}\right)+a \sin \theta\left(0\right)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\dfrac{|b \cos \theta \sqrt{a^{2}-b^{2}}-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\qquad \ldots(2)$

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

$p_2=\dfrac{|b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta\left(0\right)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\dfrac{|b \cos \theta \sqrt{a^{2}-b^{2}}+a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\qquad \ldots(3)$

On multiplying equations $\left(2\right)$ and $\left(3\right),$ we obtain

$ \begin{aligned} p_1 p_2 & =\dfrac{|b \cos \theta \sqrt{a^{2}-b^{2}}-a b|\left(b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right) \mid}{\left(\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\right)^{2}} \\ \\ & =\dfrac{|\left(b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right)\left(b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right)|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)} \\ \\ & =\dfrac{|\left(b \cos \theta \sqrt{a^{2}-b^{2}}\right)^{2}-\left(a b\right)^{2}|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)} \\ \\ & =\dfrac{|b^{2} \cos ^{2} \theta\left(a^{2}-b^{2}\right)-a^{2} b^{2}|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)} \\ \\ & =\dfrac{|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} \\ \\ & =\dfrac{b^{2}|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} \\ \\ & =\dfrac{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} \qquad\big[\because\sin ^{2} \theta+\cos ^{2} \theta=1\big] \\ \\ & =\dfrac{b^{2}|-\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} \\ \\ & =\dfrac{b^{2}\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)} \\ \\ & =b^{2} \end{aligned} $

Hence, proved.

Answer :

The equations of the given lines are

$2 x- 3 y+4=0\qquad \ldots(1)$

$3 x+4 y- 5=0\qquad \ldots(2)$

$6 x- 7 y+8=0\qquad \ldots(3)$

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

On solving equations $\left(1\right)$ and $\left(2\right),$ we obtain

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

Thus, the person is standing at point $\left(-\dfrac{1}{17}, \dfrac{22}{17}\right)$

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

Slope of the line $\left(3\right)=\dfrac{6}{7}$

$\therefore \quad $ Slope of the line perpendicular to line $\left(3\right)$ $ =-\dfrac{1}{\left(\dfrac{6}{7}\right)}=-\dfrac{7}{6} $

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

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

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