Find the equation of lines passing through $(1,2)$ and making angle $30^{\circ}$ with $Y$-axis.
Given that, angle with $Y$-axis $=30 \Upsilon$
and angle with $X$-axis $=60 \Upsilon$
$\therefore$ Slope of the line, $m=\tan 60 \Upsilon=\sqrt{3}$
So, the equation of a line passing through $(1,2)$ and having slope $\sqrt{3}$, is
$$\begin{aligned} & y-2=\sqrt{3}(x-1) \\ & \Rightarrow \quad y-2=\sqrt{3} x-\sqrt{3} \\ & \Rightarrow \quad y-\sqrt{3} x-2+\sqrt{3}=0 \end{aligned}$$
Find the equation of the line passing through the point of intersection of $2 x+y=5$ and $x+3 y+8=0$ and parallel to the line $3 x+4 y=7$.
Given equation of lines $$2 x+y =5$$ .... (i)
and $$x+3 y =-8$$ .... (ii)
From Eq. (i), $$y =5-2 x$$
Now, put the value of $y$ in Eq. (ii), we get
$$\begin{aligned} x+3(5-2 x) & =-8 \\ \Rightarrow \quad x+15-6 x & =-8 \\ \Rightarrow \quad -5 x & =-23 \Rightarrow x=\frac{23}{5} \end{aligned}$$
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 $\left(\frac{23}{5}, \frac{-21}{5}\right)$ having slope $\frac{-3}{4}$ is
$$\begin{aligned} & y+\frac{21}{5}=\frac{-3}{4}\left(x-\frac{23}{5}\right) \\ & \Rightarrow \quad 4 y+\frac{84}{5}=-3 x+\frac{69}{5} \\ & \Rightarrow \quad 3 x+4 y=\frac{84-69}{5} \Rightarrow 3 x+4 y+\frac{15}{5}=0 \\ & \Rightarrow \quad 3 x+4 y+3=0 \end{aligned}$$
For what values of $a$ and $b$ the intercepts cut off on the coordinate axes by the line $a x+b y+8=0$ are equal in length but opposite in signs to those cut off by the line $2 x-3 y+6=0$ on the axes?
$$\begin{array}{l} \text{Given equation of line} \quad a x+b y+8=0 \\ \Rightarrow \quad \frac{x}{\frac{-8}{a}}+\frac{y}{\frac{-8}{b}}=1 \end{array}$$
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 .<>/p>
According to the question,
$$\begin{aligned} \frac{-8}{a} & =3 \text { and } \frac{-8}{b}=-2 \\ a & =-\frac{8}{3}, b=4 \end{aligned}$$
If the intercept of a line between the coordinate axes is divided by the point $(-5,4)$ in the ratio $1: 2$, then find the equation of the line.
Let intercept of a line are $(h, k)$.
The coordinates of $A$ and $B$ are $(h, 0)$ and $(0, k)$ respectively.
$$\begin{aligned} & -5=\frac{1 \times 0+2 \times h}{1+2} \\ \therefore \quad & -5=\frac{2 h}{3} \Rightarrow+h=-\frac{15}{2} \\ & \text { and } \\ & 4=\frac{1 \cdot k+0 \cdot 2}{1+2} \\ & k=12 \\ & A=\left(-\frac{15}{2}, 0\right) \text { and } B=(0,12) \end{aligned}$$
Hence, the equation of a line $A B$ is
$$\begin{aligned} & y-0 =\frac{12-0}{0+15 / 2}\left(x+\frac{15}{2}\right) \\ \Rightarrow \quad & y =\frac{12 \cdot 2}{15}\left(x+\frac{15}{2}\right) \\ \Rightarrow \quad & 5 y =8 x+60 \Rightarrow 8 x-5 y+60=0 \end{aligned}$$
Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of $120^{\circ}$ with the positive direction of $X$-axis.
$$\begin{aligned} &\begin{aligned} & \text { Given that, } \quad O C=P=4 \text { units } \\ & \angle B A X=120 \Upsilon\\ & \text { Let } \quad \angle C O A=\alpha, \angle O C A=90 \Upsilon \\ & \because \quad \angle B A X=\angle C O A+\angle O C A \quad \text { [exterior angle property] }\\ & \Rightarrow \quad 120 \Upsilon=\alpha+90 \Upsilon \\ & \therefore \quad \alpha=30 \Upsilon \end{aligned}\\ &\text { Now, the equation of required line is }\\ &\begin{array}{l} & x \cos 30 \Upsilon+y \sin 30 \Upsilon =4 \\ \Rightarrow & x \cdot \frac{\sqrt{3}}{2}+y \cdot \frac{1}{2} =4 \\ \Rightarrow & \sqrt{3} x+y =8 \end{array} \end{aligned}$$