Solution

Let the intercepts along the $X$ and $Y$-axes are $a$ and a respectively.

$\therefore$ Equation of the line is

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

Since, the point $(1,-2)$ lies on the line,

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

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

$ \Rightarrow a =-1 $

On putting $a=-1$ in Eq. (i), we get

$ \Rightarrow \begin{gathered} \frac{x}{-1}+\frac{y}{-1}=1 \\ x+y=-1 \Rightarrow x+y+1=0 \end{gathered} $

Thinking Process

First of all find the slope, using the formula $=\frac{y_2-y_1}{x_2-x_1}$. Then, slope of perpendicular line is $-\frac{1}{m}$.

Solution

Consider the given points $A(5,2), B(2,3)$ and $C(3,-1)$.

Slope of the line passing through the points $B$ and $C, m_{B C}=\frac{-1-3}{3-2}=-4$

So, the slope of required line is $\frac{1}{4}$.

Since, the equation of a line passing the point $A(5,2)$ and having slope $\frac{1}{4}$ is $y-2=\frac{1}{4}(x-5)$.

$ \begin{matrix} \Rightarrow & 4 y-8=x-5 \\ \Rightarrow & x-4 y+3=0 \end{matrix} $

Thinking Process

If the angle between the lines having the slope $m_1$ and $m_2$ is $\theta$, then $\tan \theta=\Big|\frac{m_1-m_2}{1+m_1 m_2}\Big|$.

Use this formula to solve the above problem.

Solution

$ \begin{aligned} \text{Given lines,}\quad y & =(2-\sqrt{3})(x+5) \\ \text{Slope of this line,}\quad m_1 & =(2-\sqrt{3}) \\ \text{and}\quad y & =(2+\sqrt{3})(x-7) \\ \text{Slope of this line,}\quad m_2 & =(2+\sqrt{3}) \end{aligned} $

Let $\theta$ be the angle between lines (i) and (ii), then

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

$ \Rightarrow \tan \theta =\Big|\frac{(2-\sqrt{3})-(2+\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}\Big| \Rightarrow \tan \theta=\Big|\frac{-2 \sqrt{3}}{1+4-3}\Big| $

$ \Rightarrow \tan \theta =\sqrt{3} $

$ \Rightarrow \tan \theta =\tan \pi / 3 $

$ \therefore \theta =\pi / 3=60^{\circ} $

For obtuse angle $=\pi-\pi / 3=2 \pi / 3=120^{\circ}$

Hence, the angle between the lines are $60^{\circ}$ or $120^{\circ}$.

Solution

Let the intercept along the axes be $a$ and $b$.

Given, $\quad a+b=14 \Rightarrow b=14-a$

Now, the equation of line is $\frac{x}{a}+\frac{y}{b}=1$.

$\Rightarrow \quad \frac{x}{a}+\frac{y}{14-a}=1$

Since, the point $(3,4)$ lies on the line.

$ \therefore \frac{3}{a}+\frac{4}{14-a} =1 $

$ \Rightarrow \frac{42-3 a+4 a}{a(14-a)} =1 \Rightarrow 42+a=14 a-a^{2} $

$ \Rightarrow a(a-7)-6(a-7) =0 \Rightarrow(a-7)(a-6)=0 $

$ \Rightarrow a-7 =0 \text { or } a-6=0 $

$ \Rightarrow a =7 \text { or } a=6 $

$\text { When } a =7, \text { then } b=7 $

$\text { When } a =6, \text { then } b=8$

$\therefore$ The equation of line, when $a=7$ and $b=7$ is

$ \frac{x}{7}+\frac{y}{7}=1 \Rightarrow x+y=7 $

So, the equation of line, when $a=6$ and $b=8$ is $\frac{x}{6}+\frac{y}{8}=1$

Thinking Process

The perpendicular distance of a point $(x_1, y_1)$ from the line $A x+B y+C=0$, is $d^{\prime}$, where $d=\Big|\frac{A x_1+B y_1+C}{\sqrt{A^{2}+B^{2}}}\Big|$.

Solution

Let the required point be $(h, k)$ and point $(h, k)$ lies on the line $x+y=4$ i.e., $\quad h+k=4$

The distance of the point $(h, k)$ from the line $4 x+3 y=10$ is

Taking positive sign,

$ \begin{aligned} \Big|\frac{4 h+3 k-10}{\sqrt{16+9}} \Big|& =1 \\ 4 h+3 k-10 & = \pm 5 \\ 4 h+3 k & =15 \end{aligned} $

From Eq. (i) $h=4-k$ put in Eq. (ii), we get

$ \begin{aligned} & \Rightarrow 4(4-k)+3 k=15 \\ & \Rightarrow 16-4 k+3 k=15 \\ & \text { On putting } k=1 \text { in Eq. (i), we get } \end{aligned} $

So, the point is $(3,1)$.

Taking negative sign,

$ \begin{aligned} & 4 h+3 k-10 =-5 \\ \Rightarrow & 4(4-k)+3 k =5 \\ \Rightarrow & 16-4 k+3 k =5 \\ \therefore & -k=5-16 =-11 \\ \therefore & k =11 \end{aligned} $

On putting $k=11 in \mathrm{Eq.}$ (i), we get

$ h+11=4 \Rightarrow h=-7 $

Hence, the required points are $(3,1)$ and $(-7,11)$.

Solution

Given equation of lines are

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

$ \text { and } \text { Slope, } m_1 =-\frac{b}{a} $

$ \therefore \frac{x}{a}-\frac{y}{b} =1 $

$ \text { Slope, } m_2 =\frac{b}{a} $

Let $\theta$ be the angle between the given lines, then

$ \begin{aligned} & \tan \theta=\Big|\frac{m_1-m_2}{1+m_1 m_2}\Big| \Rightarrow \tan \theta=\Big|\frac{-\frac{b}{a}-\frac{b}{a}}{1+(\frac{-b}{a})(-\frac{b}{a})}\Big| \\ & \Rightarrow \quad \tan \theta=\Big|\frac{\frac{-2 b}{a}}{\frac{a^{2}-b^{2}}{a^{2}}}\Big| \Rightarrow \tan \theta=\frac{2 a b}{a^{2}-b^{2}} \end{aligned} $

Hence proved.

Thinking Process

Equation of a line passing through the point $(x_1, y_1)$ and having slope $m$ is $y-y_1=m(x-x_1)$.

Solution

Given that, angle with $Y$-axis $=30^{\circ}$

and angle with $X$-axis $=60^{\circ}$

$\therefore$ Slope of the line, $m=\tan 60^{\circ}=\sqrt{3}$

So, the equation of a line passing through $(1,2)$ and having slope $\sqrt{3}$, is

$ \begin{matrix} \Rightarrow & y-2 =\sqrt{3}(x-1) \\ \Rightarrow & y-2 =\sqrt{3} x-\sqrt{3} \\ \Rightarrow & y-\sqrt{3} x-2+\sqrt{3} =0 \end{matrix} $

Thinking Process

First of all solve the given equation of lines to get the point of intersection. Then, if a line having slope $m_1$ is parallel to another line having slope, $m_2$, then $m_1=m_2$. Now, use the formula i.e., equation of a line passing through the point $(x_1, y_1)$ with slope $m$ is $y-y_1=m(x-x_1)$

Solution

Given equation of lines

$ \begin{aligned} 2 x+y & =5 \\ x+3 y & =-8 \\ y & =5-2 x \end{aligned} $

From Eq. (i),

Now, put the value of $y$ in Eq. (ii), we get

$ x+3(5-2 x) =-8 $

$ \Rightarrow x+15-6 x =-8 $

$ \Rightarrow -5 x =-23 \Rightarrow x=\frac{23}{5} $

Now, $x=\frac{23}{5}$ put in Eq. (i), we get

$ y=5-\frac{46}{5}=\frac{25-46}{5}=\frac{-21}{5} $

Since, the required line is parallel to the line $3 x+4 y=7$. So, slope of the line is $m=\frac{-3}{4}$.

So, the equation of the line passing through the point $(\frac{23}{5}), (\frac{-21}{5})$ having slope $(\frac{-3}{4})$ is

$ y+\frac{21}{5} =\frac{-3}{4} x\Big(-\frac{23}{5}\Big) $

$ \Rightarrow 4 y+\frac{84}{5} =-3 x+\frac{69}{5} $

$ \Rightarrow 3 x+4 y =\frac{84-69}{5} \Rightarrow 3 x+4 y+\frac{15}{5}=0 $

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

Solution

Given equation of line

$ \begin{matrix} a x+b y+8=0 \\ \\ \Rightarrow\frac{x}{\frac{-8}{a}}+\frac{y}{\frac{-8}{b}}=1 \end{matrix} $

So, the intercepts are $\frac{-8}{a}$ and $\frac{-8}{b}$.

and another given equation of line is $2 x-3 y+6=0$.

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

So, the intercepts are -3 and 2 .

According to the question,

$ \begin{aligned} & \frac{-8}{a} =3 \text { and } \frac{-8}{b}=-2 \\ \therefore & a =-\frac{8}{3}, b=4 \end{aligned} $

Thinking Process

The coordinates of a point which divides the join of $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m_1: m_2$ internally is $\Big(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\Big)$.

Solution

Let intercept of a line are $(h, k)$.

The coordinates of $A$ and $B$ are $(h, 0)$ and $(0, k)$ respectively.

$ \begin{aligned} & -5=\frac{1 \times 0+2 \times h}{1+2} \\ & \therefore \quad-5=\frac{2 h}{3} \Rightarrow+h=-\frac{15}{2} \\ & \text { and } \quad 4=\frac{1 \cdot k+0 \cdot 2}{1+2} \\ & \Rightarrow \quad k=12 \\ & \therefore \quad A=(-\frac{15}{2}, 0) \text { and } B=(0,12) \end{aligned} $

Hence, the equation of a line $A B$ is

$ \begin{aligned} & & y-0=\frac{12-0}{0+15 / 2} \Big(x+\frac{15}{2}\Big) \\ \Rightarrow & y =\frac{12 \cdot 2}{15} \Big(x+\frac{15}{2}\Big) \\ \Rightarrow & 5 y =8 x+60 \Rightarrow 8 x-5 y+60=0 \end{aligned} $

Thinking Process

The equation of the line having normal distance Pfrom the origin and angle $\alpha$ which the normal makes with the positive direction of X-axis is $x \cos \alpha+y \sin \alpha=p$. Use this formula to solve the above problem.

Solution

Given that, $O C=P=4$ units

$ \begin{aligned} & \angle B A X=120^{\circ} \\ & \text { Let } \quad \angle C O A=\alpha, \angle O C A=90^{\circ} \\ & \because \quad \angle B A X=\angle C O A+\angle O C A \quad \text { [exterior angle property] } \\ & \Rightarrow \quad 120^{\circ}=\alpha+90^{\circ} \\ & \therefore \quad \alpha=30^{\circ} \\ & \text{ Now, the equation of required line is} \\ & x \cos 30^{\circ}+y \sin 30^{\circ}=4 \\ & \Rightarrow \quad x \cdot \frac{\sqrt{3}}{2}+y \cdot \frac{1}{2}=4 \\ & \Rightarrow \sqrt 3 x+y=8 \end{aligned} $

Solution

Let slope of line $A C$ be $m$ and slope of line $B C$ is $\frac{-3}{4}$ and let angle between line $A C$ and $B C$ be $\theta$.

$ \therefore \tan \theta=\Big|\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\Big| \Rightarrow \tan 45^{\circ} = \pm \Big[\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}} \Big]$

$ \text { Taking positive sign, } 1 =\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}} $

$ \Rightarrow m+\frac{3}{4} =1-\frac{3 m}{4}$

$ \Rightarrow m+\frac{3 m}{4} =1-\frac{3}{4} $

$ \Rightarrow \frac{7 m}{4} =\frac{1}{4} \Rightarrow m=\frac{1}{7} $

Taking negative sign,

$ \begin{aligned} & \quad 1=-\Big(\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\Big) \Rightarrow 1-\frac{3 m}{4}=-m-\frac{3}{4} \\ & \Rightarrow \quad m-\frac{3 m}{4}=-1-\frac{3}{4} \\ & \Rightarrow \quad \frac{m}{4}=\frac{-7}{4} \Rightarrow m=-7 \\ & \begin{matrix} \therefore \text { Equation of side } A C \text { having slope } (\frac{1}{7}) \text { is } \end{matrix} \\ & \quad y-2=\frac{1}{7}(x-2) \\ & \Rightarrow \quad 7 y-14=x-2 \\ & \text { and equation of side } A B \text { having slope }(-7) \text { is } \\ & \Rightarrow \quad y-2=-7(x-2) \\ & \Rightarrow \quad y-2=-7 x+14 \\ & \Rightarrow \quad 7 x+y-16=0 \end{aligned} $

Thinking Process

Find the length of perpendicular ( $p$ ) from $(2,-1)$ to the line and use $p=l \sin 60^{\circ}$, where $l$ is the length of the side of the triangle.

Solution

Given that, equilateral $\triangle A B C$ having equation of base is $x+y=2$.

$\ln \triangle A B D$,

$ \begin{aligned} \sin 60^{\circ} & =\frac{A D}{A B} \\ A D & =A B \sin 60^{\circ}=A B \frac{\sqrt{3}}{2} \\ \Rightarrow A D & =A B \frac{\sqrt{3}}{2} \end{aligned} $

Now, the length of perpendicular from $(2,-1)$ to the line $x+y=2$ is given by

$AD = \Big| \frac{2+(-1)-2}{\sqrt {1^{2}+1^{2}}}\Big| = \frac{1}{\sqrt 2}$

From Eq. (i),

$ \begin{aligned} & \frac{1}{\sqrt{2}}=A B \frac{\sqrt{3}}{2} \\ & A B=\sqrt{\frac{2}{3}} \end{aligned} $

Thinking Process

Let the slope of the line be $m$. Then, the equation of the line passing through the fixed point $P(x_1, y_1)$ is $y-y_1=m(x-x_1)$. Taking the algebraic sum of perpendicular distances equal to zero, we get $y_1-1=m(x_1-1)$. Thus, $(x_1, y_1)$ is $(1,1)$.

Solution

Let slope of the line be $m$ and the coordinates of fixed point $P$ are $(x_1, y_1)$.

$\therefore \quad$ Equation of line is $y-y_1=m(x-x_1)$

Since, the given points are $A(2,0), B(0,2)$ and $C(1,1)$.

Now, perpendicular distance from $A$, is

$ \frac{0-y_1-m(2-x_1)}{\sqrt{1+m^{2}}} $

Perpendicular distance from $B$, is

$ \frac{2-y_1-m(0-x_1)}{\sqrt{1+m^{2}}} $

Perpendicular distance from $C$, is

$ \frac{1-y_1-m(1-x_1)}{\sqrt{1+m^{2}}}$

$ \text { Now, } \frac{-y_1-2 m+m x_1+2-y_1+m x_1+1-y_1-m+m x_1}{\sqrt{1+m^{2}}}=0 $

$ \Rightarrow -3 y_1-3 m+3 m x_1+3=0 $

$ \Rightarrow -y_1-m+m x_1+1=0 $

Since, $(1,1)$ lies on this line. So, the point $P$ is $(1,1)$.

Solution

Let slope of the line be $m$. As, the line passes through the point $A(1,2)$.

$\therefore \quad$ Equation of line is $y-2=m(x-1)$

$ \begin{aligned} m x-y+2-m & =0 \\ x+v-4 & =0 \end{aligned} $

and

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

$ \Rightarrow \quad \frac{x}{2+m}=\frac{y}{3 m+2}=\frac{1}{1+m} $

$\Rightarrow \quad x=\frac{2+m}{1+m}$

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

So, the point of intersection is $B \Big(\frac{m+2}{m+1}, \frac{3 m+2}{m+1}\Big)$.

$ \begin{aligned} & \text { Now, } \\ & A B^{2}=\Big(\frac{m+2}{m+1}-1\Big)^{2}+\Big(\frac{3 m+2}{m+1}-2\Big)^{2} \\ & \because \quad A B=\frac{\sqrt{6}}{3} \\ & \therefore \quad \Big(\frac{m+2-m-1}{m+1}\Big)^{2}+\Big(\frac{3 m+2-2 m-2}{m+1}\Big)^{2}=\frac{6}{9} \\ & \Rightarrow \quad \Big(\frac{1}{m+1}\Big)^{2}+\Big(\frac{m}{m+1}\Big)^{2}=\frac{6}{9} \\ & \Rightarrow \quad \frac{1+m^{2}}{(1+m)^{2}}=\frac{6}{9} \\ & \Rightarrow \quad \frac{1+m^{2}}{1+m^{2}+2 m}=\frac{6}{9} \\ & \Rightarrow \quad 9+9 m^{2}=6+6 m^{2}+12 m \\ & \Rightarrow \quad 3 m^{2}-12 m+3=0 \\ & \Rightarrow \quad m^{2}-4 m+1=0 \\ & \therefore \quad m=\frac{4 \pm \sqrt{16-4}}{2} \\ & =2 \pm \sqrt{3} \\ & =2+\sqrt{3} \text { or } 2-\sqrt{3} \\ & \theta=75^{\circ} \text { or } 15^{\circ} \end{aligned} $

Thinking Process

If a line is $\frac{x}{a}+\frac{y}{b}=1$, where $\frac{1}{a}+\frac{1}{b}=$ constant $=\frac{1}{k}$ (say). This implies that $\frac{k}{a}+\frac{k}{b}=1 \Rightarrow$ line passes through the fixed point $(k, k)$.

Solution

Since, the intercept form of a line is $\frac{x}{a}+\frac{y}{b}=1$.

Given that,

$\because \frac{1}{a}+\frac{1}{b}=\text { constant }$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{1}{k}$

$\Rightarrow \frac{k}{a}+\frac{k}{b}=1$

So, $(k, k)$ lies on $\frac{x}{a}+\frac{y}{b}=1$.

Hence, the line passes through the fixed point.

Thinking Process

If the point $(h, k)$ divides the join of $A(x_1, y_2)$ and $B(x_2, y_2)$ internally, in the ratio $m_1: m_2$. Then, first of all find the coordinates of $A$ and $B$ using section formula for internal division i.e., $h=\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, k=\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$. Then, find the equation of required line.

Solution

Since, the line intersects $X$ and $Y$-axes respectively at $A(x, 0)$ and $B(0, y)$.

$\begin{aligned}-4 & =\frac{5 \times 0+3 x}{5+3} \\ \Rightarrow -4 & =\frac{3 x}{8} \Rightarrow x=\frac{-32}{3} \\ \text{and}\quad 3 & =\frac{5 \cdot y+3 \cdot 0}{5+3} \\ \Rightarrow 3 & =\frac{5 y}{8} \Rightarrow y=\frac{24}{5}\end{aligned}$

Since, the intercept on the $X$ and $Y$-axes respectively are $a=\frac{-32}{3}$ and $b=\frac{24}{5}$.

$\therefore$ Equation of required line is

$ \begin{aligned} & \frac{x}{-32 / 3}+\frac{y}{24 / 5} & =1 \\ \Rightarrow & \frac{-3 x}{32}+\frac{5 y}{24} & =1 \\ \Rightarrow & 9 x-20 y+96 & =0 \end{aligned} $

Solution

Given equation of lines

$ \begin{aligned} x-y+1 & =0 \\ 2 x-3 y+5 & =0 \\ x & =y-1 \end{aligned} $

Now, put the value of $x$ in Eq. (ii), we get

$ \begin{matrix} \Rightarrow & 2(y-1)-3 y+5 =0 \\ \Rightarrow & 2 y-2-3 y+5 =0 \\ y=3 \text { put in Eq. (i), we get } & 3-y=0 \Rightarrow y =3 \\ & x =2 \end{matrix} $

Since, the point of intersection is $(2,3)$.

Let slope of the required line be $m$.

$\therefore$ Equation of line is

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

$\Rightarrow$ $ m x-y+3-2 m=0 $

Since, the distance from $(3,2)$ to line (iii) is $\frac{7}{5}$.

$ \begin{aligned} & \therefore \quad \frac{7}{5}=\Big|\frac{3 m-2+3-2 m}{\sqrt{1+m^{2}}}\Big| \\ & \Rightarrow \quad \frac{49}{25}=\frac{(m+1)^{2}}{1+m^{2}} \\ & \Rightarrow \quad 49+49 m^{2}=25(m^{2}+2 m+1) \\ & \Rightarrow \quad 49+49 m^{2}=25 m^{2}+50 m+25 \\ & \Rightarrow \quad 24 m^{2}-50 m+24=0 \\ & \Rightarrow \quad 12 m^{2}-25 m+12=0 \\ & \therefore \quad m=\frac{25 \pm \sqrt{625-4 \cdot 12 \cdot 12}}{24} \\ & =\frac{25 \pm \sqrt{49}}{24}=\frac{25 \pm 7}{24}=\frac{32}{24} \text { or } \frac{18}{24}=\frac{4}{3} \text { or } \frac{3}{4} \\ & \therefore \text { First equation of a line is } \quad y-3=\frac{4}{3}(x-2) \\ & \begin{matrix} \Rightarrow & 3 y-9=4 x-8 \end{matrix} \\ & \Rightarrow \quad 4 x-3 y+1=0 \\ & \text { and second equation of line is } \quad y-3=\frac{3}{4}(x-2) \\ & \Rightarrow \\ & 4 y-12=3 x-6 \\ & \Rightarrow \\ & 3 x-4 y+6=0 \end{aligned} $

Thinking Process

Given that $|x|+|y|=1$, which gives four sides of a square.

Solution

Let the coordinates of moving point $P$ be $(x, y)$. Given that, the sum of distances of this point in a plane from the axes is 1 .

$ \begin{matrix} \therefore & |x|+|y|=1 \\ \Rightarrow & \pm x \pm y=1 \\ \Rightarrow & x+y=1 \\ \Rightarrow & -x-y=1 \\ \Rightarrow & -x+y=1 \\ \Rightarrow & x-y=1 \end{matrix} $

So, these equations give us locus of the point which is a square.

Thinking Process

Lines are $y=\sqrt{3} x+2$ and $y=-\sqrt{3} x+2$ according as $x \geq 0$ or $x<0$. Y-axis is the bisector of the angles between the lines. $P_1, P_2$ are the points on these lines at a distance of 5 units from the point of intersection of these lines which have a point on $Y$-axis as common foot of perpendiculars drawn from these points. The $y$-coordinate of the foot of the perpendiculars is given by $2+5 \cos 30^{\circ}$.

Solution

Given equation of lines are

$ \begin{aligned} y-\sqrt{3} x & =2 \\ y+\sqrt{3} x & =2 \\ y & =\sqrt{3} x+2 \\ y & =-\sqrt{3} x+2 \\ \sqrt{3} x+2 & =-\sqrt{3} x+2 \end{aligned} $

$ [\because x \geq 0] $

$ \Rightarrow \quad 2 \sqrt{3} x=0 \Rightarrow x=0 $

On putting $x=0$ in Eq. (i), we get

So, the point of intersection of line (i) and (ii) is $(0,2)$.

Here,

$ \begin{aligned} O C & =2 \\ \text{In $\triangle D E C$}\quad \frac{C D}{C E} & =\cos 30^{\circ} \\ C D =5 \cos 30 \\ & =5 \cdot \frac{\sqrt{3}}{2} \end{aligned} $

$ \begin{aligned} \therefore & C D =5 \cos 30^{\circ} \\ & =5 \cdot \frac{\sqrt{3}}{2} \\ \Rightarrow & O D =O C+C D=2+5 \frac{\sqrt{3}}{2} \end{aligned} $

So, the coordinates of the foot of perpendiculars are $\Big(0,2+\frac{5 \sqrt{3}}{2}\Big)$.

Solution

Given equation of line is,

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

Perpendicular length from the origin on the line (i) is given by $p$

$ \begin{matrix} \text { i.e., } & p=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{a b}{\sqrt{a^{2}+b^{2}}} \\ \therefore \quad p^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}} \end{matrix} $

Given that, $a^{2}, p^{2}$ and $b^{2}$ are in AP.

$ \begin{matrix} \therefore & 2 p^{2}=a^{2}+b^{2} \\ \Rightarrow & \frac{2 a^{2} b^{2}}{a^{2}+b^{2}}=a^{2}+b^{2} \\ \Rightarrow & 2 a^{2} b^{2}=(a^{2}+b^{2})^{2} \\ \Rightarrow & 2 a^{2}+b^{2}=a^{4}+b^{4}+2 a^{2} b^{2} \\ \Rightarrow & a^{4}+b^{4}=0 \end{matrix} $

Solution

(a) Given that,

$ c=-3 \text { and } m=\frac{3}{5} $

$\therefore$ Equation of the line is $y=m x+c$

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

$ \Rightarrow 5 y =3 x-15 $

$ \Rightarrow 5 y-3 x+15 =0 $

Solution

(a) Let equation of line be

$ \begin{aligned} & \frac{x}{a}+\frac{y}{a}=1 \\ & x+y=a \end{aligned} $

$ \begin{aligned} \Rightarrow & & x+y & =a \\ \Rightarrow & & y & =-x+a \\ \therefore & & \text { Required slope } & =-1 \end{aligned} $

Solution

(b) Since, line passes through the point $(3,2)$ and perpendicular to the line $y=x$.

$\because$ Slope $(m)=-1 \quad$ [since, line is perpendicular to the line $y=x]$

$\therefore \quad$ Equation of line which passes through $(3,2)$ is

$ \begin{aligned} & y-2=-1(x-3) \\ & y-2=-x+3 \end{aligned} $

$ \Rightarrow x+y=5 $

Solution

(b) Given point is $(1,2)$ and slope of the required line is 1 .

$\because x+y+1=0 \Rightarrow y=-x-1 \Rightarrow m_1=-11$

$\therefore$ slope of the line $=\frac{-1}{-1}=1$

$\therefore$ Equation of required line is

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

$ \Rightarrow y-2 =x-1 $

$ \Rightarrow y-x-1 =0 $

Solution

(c) Since, intercepts on the axes are $a,-b$ then equation of the line is $\frac{x}{a}-\frac{y}{b}=1$.

$ \begin{matrix} \Rightarrow & \frac{y}{b}=\frac{x}{a}-1 \\ \Rightarrow & y=\frac{b x}{a}-b \end{matrix} $

So, the lope of this line i.e., $m_1=\frac{b}{a}$.

Also, for intercepts on the axes as $b$ and $-a$, then equation of the line is

$ \Rightarrow \quad \begin{aligned} \frac{x}{b}-\frac{y}{a} & =1 \\ \frac{y}{a} & =\frac{x}{b}-1 \Rightarrow y=\frac{a}{b} x-a \end{aligned} $

and slope of this line i.e., $m_2=\frac{a}{b}$

$ \therefore \quad \tan \theta=\frac{\frac{b}{a}-\frac{a}{b}}{1+\frac{a}{b} \cdot \frac{b}{a}}=\frac{\frac{b^{2}-a^{2}}{a b}}{2}=\frac{b^{2}-a^{2}}{2 a b} $

Solution

(d) Given, line is $\frac{x}{a}+\frac{y}{b}=1$

Since, the points $(2,-3)$ and $(4,-5)$ lies on this line.

$ \begin{matrix} \therefore & \frac{2}{a}-\frac{3}{b}=1 \\ \\ \text { and } & \frac{4}{a}-\frac{5}{b}=1 \end{matrix} $

On multiplying by 2 in Eq. (ii) and then subtracting Eq. (iii) from Eq. (ii), we get

$ -\frac{6}{b}+\frac{5}{b} =1 $

$ \Rightarrow \frac{-1}{b} =1 $

$ \therefore b =-1 $

On putting $b=-1$ in Eq. (ii), we get

$ \begin{aligned} \Rightarrow & \frac{2}{a}+3 =1 \\ \therefore & \frac{2}{a} =-2 \Rightarrow a=-1 \\ & (a, b) =(-1,-1) \end{aligned} $

Thinking Process

First of all find the point of intersection of the given first two lines, then get the perpendicular distance from this point to the third line. Using formula i.e., distance of a point $(x_1, y_1)$ from the line $a x+b y+c=0$ is $d=\frac{|a x_1+b y_1+c|}{\sqrt{a^{2}+b^{2}}}$.

Solution

(a) Given equation of lines

and $\quad \begin{aligned} 2 x-3 y+5 & =0 \\ 3 x+4 y & =0\end{aligned}$

From Eq. (ii), put the value of $x=\frac{-4 y}{3}$ in Eq. (i), we get

$ 2 \Big(\frac{-4 y}{3}\Big)-3 y+5 =0 $

$ \Rightarrow -8 y-9 y+15 =0 $

$ \Rightarrow y =\frac{15}{17} $

From Eq. (ii),

$ \begin{aligned} & 3 x+4 \cdot \frac{15}{17}=0 \\ & \Rightarrow x=\frac{-60}{17 \cdot 3}=\frac{-20}{17} \end{aligned} $

So, the point of intersection is $\frac{-20}{17}, \frac{15}{17}$

$\therefore$ Required distance from the line $5 x-2 y=0$ is,

$ d=\frac{|-5 \times \frac{20}{17}-2 \Big(\frac{15}{17}\Big)|}{\sqrt{25+4}}=\frac{|\frac{-100}{17}-\frac{30}{17}|}{\sqrt{29}}=\frac{130}{17 \sqrt{29}} $

$ \because \text { distance of a point } p(x_1, y_1) \text { from the line } a x+b y+c=0 \text { is } d=\frac{|a x_1+b y_1+c|}{\sqrt{a^{2}+b^{2}}} $

Solution

(a) So, the given point $A$ is $(3,-2)$.

So, the equation of line $\sqrt{3} x+y=1$.

$ \begin{aligned} \Rightarrow & y =-\sqrt{3} x+1 \\ \therefore & \text { Slope, } m_1 =-\sqrt{3} \end{aligned} $

Let slope of the required line be $m_2$.

$ \begin{aligned} & \therefore \quad \tan \theta=\Big|\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\Big| \quad \Big[\because \tan \theta=|\frac{m_1-m_2}{1+m_1 m_2}|\Big] \\ & \Rightarrow \quad \tan 60^{\circ}= \pm \Big(\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\Big) \\ & \begin{matrix} \Rightarrow & \sqrt{3}=\Big(\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\Big) \end{matrix} \quad \text { [taking positive sign] } \\ & \Rightarrow \quad \sqrt{3}-3 m_2=-\sqrt{3}-m_2 \\ & \Rightarrow \quad 2 \sqrt{3}=2 m_2 \\ & \Rightarrow \quad m_2=\sqrt{3} \end{aligned} $

$\therefore \quad$ Equation of line passing through $(3,-2)$ is

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

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

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

$ \Rightarrow m_2 =0 $

$\therefore$ The equation of line is $y+2=0(x-3)$

$\Rightarrow \quad y+2=0$

So, the required equation of lines are $\sqrt{3} x-y-2-3 \sqrt{3}=0$ and $y+2=0$.

Solution

(a) Let slope of the line be $m$.

$\because \quad$ Equation of line passing through $(1,0)$ is

$ \begin{aligned} &\Rightarrow y-0=m(x-1) \\ & y-m x+m=0 \end{aligned} $

Since, the distance from origin is $\frac{\sqrt{3}}{2}$.

Then,

$ \begin{aligned} \Rightarrow & \frac{\sqrt{3}}{2} =\frac{m}{\sqrt{1+m^{2}}} \\ \Rightarrow & \frac{3}{4} =\frac{m^{2}}{1+m^{2}} \\ \Rightarrow & 3+3 m^{2} =4 m^{2} \\ \Rightarrow & m^{2} =3 \\ \Rightarrow & m = \pm \sqrt{3} \end{aligned} $

So, the first equation of line is

$ \begin{aligned} & & y=\sqrt{3}(x-1) \\ \Rightarrow & & \sqrt{3} x-y-\sqrt{3}=0 \end{aligned} $

and the second equation of line is

$ \begin{aligned} & & y=-\sqrt{3}(x-1) \\ \Rightarrow & & \sqrt{3} x+y-\sqrt{3}=0 \end{aligned} $

Solution

(b) Given, equation of the lines are and $\quad \begin{aligned} y & =m x+c_1 \\ y & =m x+c_2\end{aligned}$

$\therefore$ Distance between them is given by

$ d=\frac{|c_1-c_2|}{\sqrt{1+m^{2}}} $

Solution

(b) Given, equation of the line is

$ y=3 x+4 $

$\therefore$ Slope of this line, $m_1=3$

So, the slope of line $O P$ is $-\frac{1}{3}$.

$\therefore$ Equation of line $O P$ is

$ \begin{matrix} & y-3 =-\frac{1}{3}(x-2) \\ \Rightarrow & 3 y-9 =-x+2 \\ \Rightarrow & x+3 y-11 =0 \end{matrix} $

Using the value of $y$ from Eq. (i) in Eq. (ii), we get

$ x+3(3 x+4)-11=0 $

$ \begin{matrix} \Rightarrow & x+9 x+12-11=0 \\ \Rightarrow & 10 x+1=0 \Rightarrow x=-\frac{1}{10} \end{matrix} $

Put $x=\frac{-1}{10}$ in Eq. (i), we get

$ y=\frac{-3}{10}+4=\frac{-3+40}{10}=\frac{37}{10} $

So, the foot of perpendicular is $\Big(-\frac{1}{10}, \frac{37}{10}\Big)$.

Solution

(a) Since, the coordinates of the middle point are $P(3,2)$.

$ \begin{aligned} & \therefore \quad 3=\frac{1 \cdot 0+1 \cdot a}{1+1} \\ & \Rightarrow \quad 3=\frac{a}{2} \Rightarrow a=6 \\ & \text { Similarly, } \quad b=4 \\ & \therefore \quad \text { Equation of the line is } \frac{x}{6}+\frac{y}{4}=1 \\ & \Rightarrow \quad 2 x+3 y=12 \end{aligned} $

Solution

(c) Since, the line passes through $(1,2)$ and parallel to the line $y=3 x-1$. So, slope of the required line $m=3$. Hence, the equation of line is

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

Solution

(a) Equation of $O B$ is

$ y-0 =\frac{1-0}{1-0}(x-0) $

$ \Rightarrow y =x $

$ \text { and equation of } A C \text { is } $

$ \Rightarrow y-0 =\frac{1-0}{0-1}(x-1) $

$ \Rightarrow y+y-1 =0 $

Solution

(b) Equation of straight lines are

$ \begin{aligned} \qquad y & =m x+c, \text { parameter }=2 \\ \frac{x}{a}+\frac{y}{b} & =1, \text { parameter }=2 \\ y-y_1 & =m(x-x_1), \text { parameter }=2 \\ \text { and } \quad x \cos w & +y \sin w=p, \text { parameter }=2 \end{aligned} $

It is clear that from Eqs. (i), (ii), (iii) and (iv), for specifying a straight line clearly two parameters should be known.

Solution

(b) Let the reflection of $A(4,1)$ in $y=x$ is $B(h, k)$.

Now, mid-point of $A B$ is $\Big(\frac{4+h}{2}, \frac{1+k}{2}\Big)$ which lies on $y=x$.

i.e., $\quad \frac{4+h}{2}=\frac{1+k}{2} \Rightarrow h-k=-3$

So, the slope of line $y=x$ is 1 .

$ \text { Slope of } A B =\frac{h-4}{k-1} $

$ \Rightarrow 1 \cdot \frac{h-4}{k-1} =-1$

$ \Rightarrow h-4 =1-k $

$ \Rightarrow h+k =5 $

$ \text { and } 2 h =2 \Rightarrow h=1 $

On putting $h=1$ in Eq. (ii), we get

$ k=4 $

So, the point is $(1,4)$.

Hence, after translation the point is $(1+2,4)$ or $(3,4)$.

Solution

(c) The given equation of lines are

$ 4 x+3 y+10 =0 $

$ \Rightarrow 5 x-12 y+26 =0 $

$ \Rightarrow 7 x+24 y-50 =0 $

Let the point $(h, k)$ which is equidistant from these lines.

Distance from line (i) $=\frac{|4 h+3 k+10|}{\sqrt{16+9}}$

Distance from line (ii) $=\frac{|5 h-12 k+26|}{\sqrt{25+144}}$

Distance from the line (iii) $=\frac{|7 h+24 k-50|}{\sqrt{7^{2}+24^{2}}}$

So, the point $(h, k)$ is equidistant from lines (i), (ii) and (iii).

$ \begin{matrix} \therefore & \frac{4 h+3 k+10}{\sqrt{16+9}}=\frac{5 h-12 k+26}{\sqrt{25+144}}=\frac{7 h+24 k-50}{\sqrt{49+576}} \\ \\ \Rightarrow & \frac{|4 h+3 k+10|}{5}=\frac{|5 h-12 k+26|}{13}=\frac{|7 h+24 k-50|}{25} \end{matrix} $

Clearly, if $h=0, k=0$, then $\frac{10}{5}=\frac{26}{13}=\frac{50}{25}=2$

Hence, the required point is $(0,0)$.

Thinking Process

First of all find the equation of required line using the formulae. i.e., $y-y_1=m(x-x_1)$ then put $x=0$ to get $y$-intercept.

Solution

(d) Given line is $y=3-3 x$.

Then, slope of the required line $=\frac{1}{3}$

$\because \quad$ Equation of the required line is

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

$ \Rightarrow 3 y-6 =x-2 $

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

$ \text { For y-intercept, } \text { put } x =0, $

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

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

Solution

(b) Let point $A(x_1, y_1)$ lies on the line $3 x+4 y+5=0$, then $3 x_1+4 y_1+5=0$

Now, perpendicular distance from $A$ to the line

$ \begin{aligned} 3 x+4 y+2 & =0 \\ \Rightarrow \quad & \frac{|3 x_1+4 y_1+2|}{\sqrt{9+16}}=\frac{|-5-2|}{\sqrt{9+16}}=\frac{-7}{5} \end{aligned} $

Let point $B(x_2, y_2)$ lies on the line $3 x+4 y-5=0$ i.e., $3 x_2+4 y_2-5=0$. Now, perpendicular distance from $B$ to the line $3 x+4 y+2=0$,

$ \frac{|3 x_2+4 y_2+2|}{\sqrt{9+16}}=\frac{|+5-2|}{\sqrt{9+16}}=\frac{3}{5} $

Hence, the required ratio is $\frac{3}{5}: \frac{7}{5}$ i.e., $3: 7$.

Thinking Process

Let $A B C$ be the equilateral triangle with vertex $A(h, k)$ and $D(\alpha, \beta)$ be the point on $B C$. Then, $\frac{2 \alpha+h}{3}=0=\frac{2 \beta+k}{3}$. Also, $\alpha+\beta-2=0$ and $\Big(\frac{k-0}{h-0}\Big) \cdot(-1)=-1$.

Solution

(c) Let $A B C$ be the equilateral triangle with vertex $A(h, k)$. Let the coordinates of $D$ are $(\alpha, \beta)$.

We know that, $2: 1$ from the vertex $A$.

$ \begin{matrix} \therefore & 0 =\frac{2 \alpha+h}{3} \text { and } 0=\frac{2 \beta+k}{3} \\ \\ \Rightarrow & 2 \alpha =-h \\ \\ \text { and } & 2 \beta =-k \\ \\ \text { Also, } D(\alpha, \beta) \text { lies on the line } & x+y-2=0 . \\ \\ \therefore & \alpha+\beta-2=0 \\ \\ & A D \perp B C \end{matrix} $

Since, the slope of line $B C$ i.e., $m_{B C}=-1$

and slope of the line $A G$ i.e., $m_{A G}=\frac{k-0}{h-0}=\frac{k}{h}$

$ \begin{aligned} \Rightarrow & (-1) \cdot \Big(\frac{k}{h}\Big) & =-1 \\ \Rightarrow & h =k \end{aligned} $

From Eqs. (i) and (iii),

$ \begin{aligned} & \qquad \begin{aligned} 2 \alpha & =-h \text { and } 2 \beta=-h \\ \therefore & \alpha=\beta \end{aligned} \\ & \text { From Eq. (ii), } \\ & \text { If } \alpha=1 \text {, then } \beta=1 \\ & \text { From Eq. (i), } h=-2, k=-2 \\ & \text { So, the vertex } A \text { is }(-2,-2) . \end{aligned} $

Thinking Process

If $a, b$ and $c$ are in AP, then $2 b=a+c$. Use this property to solve the above problem.

Solution

Given line is

$ a x+b y+c=0 $

Since, $a, b$ and $c$ are in AP, then

$ b=\frac{a+c}{2} $

$ \Rightarrow \quad a-2 b+c=0 $

On comparing Eqs. (i) and (ii), we get

$ x=1, y=2 $

So, $(1,-2)$ lies on the line.

Solution

Let equation of line is

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

Since, this line passes through $(1,-2)$.

$ \Rightarrow \quad \begin{aligned} \frac{1}{a}-\frac{2}{a} & =1 \\ 1-2 & =a \Rightarrow a=-1 \end{aligned} $

$\therefore$ Required equation of the line is

$ -x-y=1 $

$ \Rightarrow \quad x+y+1=0 $

Solution

Since, the given point $P(3,2)$ and line is $x-2 y=3$.

Slope of this line is $m_1=\frac{1}{2}$

Let the slope of the required line is $m$.

$ \begin{matrix} \text { Then, } & \tan \theta=\Big|\frac{m-\frac{1}{2}}{1+\frac{1}{2} m}\Big| \\ \\ \Rightarrow & 1= \pm \Big(\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\Big) \end{matrix} $

Taking positive sign,

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

$ \begin{aligned} \Rightarrow m-\frac{m}{2} & =1+\frac{1}{2} \\ \Rightarrow\frac{m}{2} & =\frac{3}{2} \Rightarrow m=3 \end{aligned} $

Taking negative sign,

$ \begin{aligned} & & 1=-\Big(\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\Big) \\ \Rightarrow & 1+\frac{m}{2} =-m+\frac{1}{2} \\ \Rightarrow & m+\frac{m}{2} =\frac{1}{2}-1 \\ \Rightarrow & \frac{3 m}{2} =\frac{-1}{2} \Rightarrow m=\frac{-1}{3} \end{aligned} $

$\therefore \quad$ First equation of the line is

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

$ 3 x-y-7 =0 $

and second equation of the line is

$ \begin{matrix} y-2 & =-\frac{1}{3}(x-3) \\ \Rightarrow & 3 y-6 & =-x+3 \\ \Rightarrow & x+3 y-9 & =0 \end{matrix} $

Solution

$ \begin{gathered} \text{Given line is}\quad 3 x-4 y-8=0 \\ \text{For point (3,4),\quad}9-4 \cdot 4-8 \\ \Rightarrow 9-16-8 \\ \Rightarrow 9-24 \\ \Rightarrow -15<0 \end{gathered} $

For point $(2,-6)$,

$ \begin{aligned} & 6+24-8 \\ & 22>0 \end{aligned} $

Since, the value are of opposite sign.

Hence, the points $(3,4)$ and $(2,-6)$ lies on opposite side to the line.

Solution

Let the coordinaters of the point are $(h, k)$,

$\therefore \quad$ Distance between $(3,-2)$ and $(h, k)$,

$ d_1^{2}=(3-h)^{2}+(-2-k)^{2} $

Now, distance of the point $(h, k)$ from the line $5 x-12 y=3$ is,

$ d_2=\Big|\frac{5 h-12 k-3}{\sqrt{25+144}}\Big|=\Big|\frac{5 h-12 k-3}{13}\Big| $

Given that, $ d_1^{2}=d_2 $

$ \begin{aligned} & \Rightarrow \quad(3-h)^{2}+(2+k)^{2}=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad 9-6 h+h^{2}+4+4 k+k^{2}=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad h^{2}+k^{2}-6 h+4 k+13=\frac{5 h-12 k-3}{13} \\ & \Rightarrow \quad 13 h^{2}+13 k^{2}-78 h+52 k+169=5 h-12 k-3 \\ & \Rightarrow \quad 13 h^{2}+13 k^{2}-83 h+64 k+172=0 \end{aligned} $

$\therefore$ Locus of this point is

$ \quad 13 x^{2}+13 y^{2}-83 x+64 y+172=0 $

Solution

Given equation of the line is

$ x \sin \theta+y \cos \theta=p $

Let the mid-point of $A B$ is $p(h, k)$.

So, the mid-point of $A B$ are

$ \quad\Big(\frac{a}{2}, \frac{b}{2}\Big) $

Since, the point $(a, 0)$ lies on the line (i), then

$ a \sin \theta+0=p $

$ \Rightarrow \quad a \sin \theta=p \Rightarrow a=\frac{p}{\sin \theta} $

and the point $(0, b)$ also lies on the line, then

$0+b\cos \theta=p$

$ \Rightarrow \quad b \cos \theta=p \Rightarrow b=\frac{p}{\cos \theta} $

Now $ \text { mid-point of } A B=\Big(\frac{a}{2}, \frac{b}{2}\Big) \text { or } \Big(\frac{p}{2 \sin \theta}, \frac{p}{2 \cos \theta}\Big) $

$\because$ $ \frac{p}{2 \sin \theta}=h \Rightarrow \sin \theta=\frac{p}{2 h} $

and

$ \frac{p}{2 \cos \theta}=k \Rightarrow \cos \theta=\frac{p}{2 k} $

$\therefore \quad \sin ^{2} \theta+\cos ^{2} \theta=\frac{p^{2}}{4 h^{2}}+\frac{p^{2}}{4 k^{2}}$

$ \Rightarrow \quad 1=\frac{p^{2}}{4}\Big( \frac{1}{h^{2}}+\frac{1}{k^{2}}\Big) $

Locus of the mid-point is

$ \begin{aligned} 4 & =p^{2} \Big(\frac{1}{x^{2}}+\frac{1}{y^{2}}\Big) \\ \Rightarrow \quad 4 x^{2} y^{2} & =p^{2}\Big(x^{2}+y^{2}\Big) \end{aligned} $

Solution

True

We know that, if the vertices of a triangle have integral coordinates, then the triangle cannot be equilateral. Hence, the given statement is true.

Since, in equilatteral triangle, we get $\tan 60^{\circ}=\sqrt{3}=$ Slope of the line, so with integral coordinates as vertices, the triangle cannot be equilateral.

Solution

False

Given points are $A(-2,1), B(0,5)$ and $C(1,2)$.

Now, $ \text { slope of } A B=\frac{5-1}{0+2}=2 $

Slope of

$ \begin{aligned} & B C=\frac{2-5}{-1-0}=3 \\ & A C=\frac{2-1}{-1+2}=1 \end{aligned} $

Slope of

Since, the slopes are different.

Hence, $A, B$ and $C$ are not collinear. So, statement is false.

Solution

False

Given point $p(a \cos ^{3} \theta, a \sin ^{3} \theta)$ and the line is $x \sec \theta+y cosec \theta=a$

$\because \quad$ Slope of this line $=\frac{-\sec \theta}{cosec \theta}=-\tan \theta$

and

$ \text { slope of required line }=\frac{1}{\tan \theta}=\cot \theta $

$\therefore \quad$ Equation of the required line is

$ \begin{matrix} \Rightarrow & y-a \sin ^{3} \theta=\cot \theta(x-a \cos ^{3} \theta) \\ \Rightarrow & y \sin \theta-a \sin ^{4} \theta=x \cos \theta-acos^{4} \theta \\ \Rightarrow & x \cos \theta-y \sin \theta=a \cos ^{4} \theta-asin^{4} \theta \\ \Rightarrow & x \cos \theta-y \sin \theta=a[(\cos ^{2} \theta+\sin ^{2} \theta)(\cos ^{2} \theta-\sin ^{2} \theta)] \\ \Rightarrow & x \cos \theta-y \sin \theta=a \cos ^{2} \theta \end{matrix} $

Hence, the given statement is false.

Solution

True

$\begin{matrix} \text { Given that, } & \\ \begin{aligned} & x+2 y-10=0 \\ & \text { and }\end{aligned} \\ 2 x+y+5 =0\end{matrix} $

From Eq. (i), put the value of $x=10-2 y$ in Eq. (ii), we get

$ \begin{aligned} \Rightarrow & 20-4 y+y+5 =0 \\ \Rightarrow & 20-3 y+5 =0 \\ \therefore & x+\frac{50}{3}-10 =0 \\ \Rightarrow & x+\frac{20}{3}=0 \Rightarrow x =\frac{-20}{3} \end{aligned} $

So, the point of intersection is $\Big(-\frac{20}{3}, \frac{25}{3}\Big)$.

If the line $5 x+4 y=0$ passes through the point $\Big(-\frac{20}{3}, \frac{25}{3}\Big)$, then this point should lie on this line.

$ \because \quad 5 \frac{-20}{3}+\frac{4(25)}{3}=\frac{-100}{3}+\frac{100}{3}=0 $

So, this point lies on the given line.

Hence, the statement is true.

Solution

True

Let $A B C$ be an equilateral triangle with vertex $A(2,3)$, and equation of $B C$ is $x+y=2$. i.e., slope $=-1$.

Let slope of line $A B$ is $m$.

Since, the angle between line $A B$ and $B C$ is $60^{\circ}$.

$ \begin{aligned} & \therefore \quad \tan 60^{\circ}=\Big|\frac{m+1}{1-m} \Big|\\ & \Rightarrow \quad \sqrt{3}= \pm \Big(\frac{m+1}{1-m}\Big) \quad \text { [taking positive sign] } \\ & \Rightarrow \quad \sqrt{3}-\sqrt{3} m=m+1 \\ & \Rightarrow \quad \sqrt{3}-1=m+\sqrt{3} m \\ & \Rightarrow \quad \sqrt{3}-1=m(1+\sqrt{3}) \\ & \therefore \quad m=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \\ & =\frac{3+1-2 \sqrt{3}}{3-1}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3} \end{aligned} $

Similarly, slope of $A B=2+\sqrt{3}$

$\therefore$ Equation of other two side is

[taking negative sign]

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

Hence, the statement is true.

Thinking Process

Equation of a line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Solution

True

Given equation of lines are $4x+y-1=0$

$ \text { and } \quad 7 x-3 y-35=0 $

From Eq. (i), on putting $y=1-4 x$ in Eq. (ii), we get

$ 7 x-3+12 x-35=0 $

$ \Rightarrow \quad 19 x-38=0 \Rightarrow x=2 $

On putting $x=2$ in Eq. (i), we get

$ 8+y-1=0 \Rightarrow y=-7 $

Now, the equation of a line passing through $(3,5)$ and $(2,-7)$ is

$ \begin{aligned} y-5 =\frac{-7-5}{2-3}(x-3) \\ \Rightarrow y-5 =12(x-3) \\ \Rightarrow 12 x-y-31 =0 \end{aligned} $

Distance from $(0,0)$ to the line (iii),

$ d_1=\frac{|-31|}{\sqrt{144+1}}=\frac{31}{\sqrt{145}} $

$\therefore$ Distance from $(8,34)$ to the line (iii),

$ \begin{aligned} d_2 & =\frac{|96-34-31|}{\sqrt{145}}=\frac{31}{\sqrt{145}} \\ \because \quad d_1 & =d_2 \end{aligned} $

Hence, the statement is true.

Solution

True

Given that, equation of line is

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

Equation of line passing through origin and perpendicular to line (i) is

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

Now, foot of perpendicular is the point of intersection of lines (i) and (ii). To find its locus we have to eliminate the variable $a$ and $b$.

On squaring and adding Eqs. (i) and (ii), we get

$ \begin{matrix} \Rightarrow & \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{2 x y}{a b}+\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}-\frac{2 x y}{a b} =1 \\ \\ \Rightarrow & x^{2} \Big(\frac{1}{a^{2}}+\frac{1}{b^{2}}\Big)+y^{2} \Big(\frac{1}{a^{2}}+\frac{1}{b^{2}}\Big) =1 \\ \\ \Rightarrow \frac{x^{2}}{c^{2}}+\frac{y^{2}}{c^{2}} =1 \\ \\ x^{2}+y^{2} =c^{2} & [\because \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}] \end{matrix} $

Hence, the statement is true.

Thinking Process

First of all find the intersection point of first two line. Then, if the lines are concurrent then this point should lies on the third line.

Solution

False

Given lines are

and $\quad \begin{aligned} & a x+2 y+1=0 \\ & b x+3 y+1=0\end{aligned}$

From Eq. (i), on putting $y=\frac{-a x-1}{2}$ in Eq. (ii), we get

$ b x-\frac{3}{2}(a x+1)+1=0 $

$ \Rightarrow 2 b x-3 a x-3+2=0 $

$ \Rightarrow x(2 b-3 a)=1 \Rightarrow x=\frac{1}{2 b-3 a} $

Now, using $x=\frac{1}{2 b-3 a}$ in Eq. (i), we get

$ \frac{a}{2 b-3 a}+2 y+1 =0 $

$ \Rightarrow 2 y =-\Big[\frac{a+2 b-3 a}{2 b-3 a}\Big] $

$ \Rightarrow 2 y =\frac{-(2 b-2 a)}{2 b-3 a} $

$ \Rightarrow y =\frac{(a-b)}{2 b-3 a} $

So, the point of intersection is $\Big(\frac{1}{2 b-3 a}, \frac{a-b}{2 b-3 a}\Big)$.

Since, this point lies on $c x+4 y+1=0$, then

$ \frac{c}{2 b-3 a}+\frac{4(a-b)}{2 b-3 a}+1=0 $

$ \Rightarrow c+4 a-4 b+2 b-3 a=0 $

$ \Rightarrow -2 b+a+c=0 \Rightarrow 2 b=a+c $

Hence, the given statement is false.

Solution

False

Given points are $A(3,-4), B(-2,6), P(-3,6)$ and $Q(9,-18)$.

Now $ \begin{aligned} & \text { slope of } A B=\frac{6+4}{-2-3}=-2 \\ & \text { slope of } P Q=\frac{-18-6}{9+3}=-2 \end{aligned} $

and

So, line $A B$ is parallel to line $P Q$.

Solution

(i) Let the coordinate of point $P(x_1, y_1)$ on the line $x+5 y=13 i$.e.,

$ P(13-5 y_1, y_1) $

$\therefore$ Distance of $P$ from the line $12 x-5 y+26=0$,

$ \begin{aligned} & \qquad 2=\Big|\frac{12(13-5 y_1)-5 y_1+26}{\sqrt{144+25}}\Big| \\ & \Rightarrow \quad 2= \pm \frac{156-60 y_1-5 y_1+26}{13} \\ & \Rightarrow \quad-65 y_1=-156 \quad \text { [taking positive sign] } \\ & \begin{aligned} y_1 & =\frac{156}{65}=\frac{12}{5} \\ x_1 & =13-5 y_1 \\ & =13-12=1 \end{aligned} \\ & \text { So, the coordinate of is } P \begin{aligned} & \Big(1, \frac{12}{5}\Big) . \end{aligned} \\ & \text { Similarly, the coordinates of } Q \text { are }\Big(-3, \frac{16}{5}\Big) . \end{aligned} $

(ii) Let coordinates of the point on the line $x+y=4$ be $(4-y_1, y_1)$. Distance from the line $4 x+3 y-10=0$.

$ 1=\Big|\frac{4(4-y_1)+3 y_1-10}{\sqrt{16+9}}\Big| $

$ \Rightarrow 1= \pm \frac{16-4 y_1+3 y_1-10}{5} $

$ \Rightarrow 5=6-y_1 $

$ \Rightarrow y_1=1 $

$ \text { If } y_1=1 \text {, then } x_1=3 $

So, the point is $(3,1)$.

Similarly, taking negative sign the point is $(-7,11)$.

(iii) Given point $A(-2,5)$ and $B(3,1)$.

Now, the point $P$ divides line joining the point $A$ and $B$ in $1: 2$.

$ \begin{matrix} \because & x_1=\frac{1 \cdot 3+2(-2)}{1+2}=\frac{3-4}{3}=\frac{-1}{3} \\ \\ \text { and } & y_1=\frac{1 \cdot 1+2 \cdot 5}{1+2}=\frac{11}{3} \end{matrix} $

So, the coordinates of $P$ are $\Big(\frac{-1}{3}, \frac{11}{3}\Big)$

Thus, the point $Q$ divided the line joining $A$ to $B$ in $2: 1$.

$ \begin{matrix} \because & x_2=\frac{2 \cdot 3+1(-2)}{2+1}=\frac{4}{3} \\ \\ \text { and } & y_2=\frac{2 \cdot 1+1 \cdot 5}{2+1}=\frac{7}{3} \end{matrix} $

Hence, the coordinates of $Q$ are $\Big(\frac{4}{3}, \frac{7}{3}\Big)$.

Hence, the correct matches are (i) $\rightarrow$ (c), (ii) $\rightarrow$ (a), (iii) $\rightarrow$ (b).

Solution

(i) Given equation of the line is

$ (2 x+3 y+4)+\lambda(6 x-y+12)=0 $

If line is parallel to $Y$-axis i.e., it is perpendicular to $X$-axis

$ \therefore \quad \text { Slope }=m=\tan 90^{\circ}=\infty $

$\begin{aligned} & \text { From line (i), } x(2+6 \lambda)+y(3-\lambda)+4+12 \lambda=0 \\ & \text { and slope } \\ & =\frac{-(2+6 \lambda)}{3-\lambda} \\ & \Rightarrow \quad \frac{-2-6 \lambda}{3-\lambda}=\infty \\ & \Rightarrow \quad \frac{-2-6 \lambda}{3-\lambda}=\frac{1}{0} \Rightarrow \lambda=3 \\ & \end{aligned}$

(ii) If the line (i) is perpendicular to the line $7 x+y-4=0$ or $y=-7 x+4$

$ \begin{matrix} \because & \frac{-(2+6 \lambda)}{(3-\lambda)}(-7) & =-1 \\ \Rightarrow & 14+42 \lambda =-3+\lambda \\ \Rightarrow & 41 \lambda =-17 \\ \Rightarrow & \lambda =-\frac{17}{41} \end{matrix} $

(iii) If the line (i) passes through the point $(1,2)$.

$ \begin{matrix} \text { Then, } & (2+6+4)+\lambda(6-2+12) =0 \\ \Rightarrow & 12+16 \lambda=0 \Rightarrow \lambda =-\frac{3}{4} \end{matrix} $

(iv) If the line is parallel to $X$-axis the slope $=0$.

$ \begin{matrix} \text { Then, } & \frac{-(2+6 \lambda)}{3-\lambda}=0 \\ \\ \Rightarrow & -(2+6 \lambda)=0 \Rightarrow \lambda=-\frac{1}{3} \end{matrix} $

So, the correct matches are (i) $\rightarrow$ (d), (ii) $\rightarrow$ (c), (iii) $\rightarrow$ (a), (iv) $\rightarrow$ (b).

Solution

Given equation of the lines are

$ 2 x-3 y =0 $

and $\quad$

$ 4 x-5 y =2 $

From Eq. (i), put $x=\frac{3 y}{2}$ in Eq. (ii), we get

$\Rightarrow$ $ 4 \cdot \frac{(3 y)}{2}-5 y=2 $

$\Rightarrow$ $ 6 y-5 y=2 $

Now, put $y=2$ in Eq. (i), we get

$ y=2 $

So, the intersection points are $(3,2)$.

$ x=3 $

(i) The equation of the line passes through the point $(3,2)$ and $(2,1)$, is

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

$ \Rightarrow y-y-1 =0 $

$ \Rightarrow x-2 =(x-3) $

(ii) If the required line is perpendicular to the line $x+2 y+1=0$

$\because$ Slope of the required line $=2$

$\therefore$ Equation of the line is

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

$ \Rightarrow 2 x-y-4 =0 $

(iii) If the required line is parallel to the line $3 x-4 y+5=0$, then the slope of the required line $=\frac{3}{4}$

$\therefore$ Equation of the required line is

$ y-2=\frac{3}{4}(x-3) $

$ \Rightarrow 4 y-8=3 x-9 $

$ \Rightarrow 3 x-4 y-1=0 $

(iv) If the line is equally inclined to the $X$-axis, then

$ m= \pm \tan 45^{\circ}= \pm 1 $

$\therefore$ Equation of the line is