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.
$$\begin{aligned} &\begin{aligned} & \text { Given that, } \quad O C=P=4 \text { units } \\ & \angle B A X=120 \Upsilon\\ & \text { Let } \quad \angle C O A=\alpha, \angle O C A=90 \Upsilon \\ & \because \quad \angle B A X=\angle C O A+\angle O C A \quad \text { [exterior angle property] }\\ & \Rightarrow \quad 120 \Upsilon=\alpha+90 \Upsilon \\ & \therefore \quad \alpha=30 \Upsilon \end{aligned}\\ &\text { Now, the equation of required line is }\\ &\begin{array}{l} & x \cos 30 \Upsilon+y \sin 30 \Upsilon =4 \\ \Rightarrow & x \cdot \frac{\sqrt{3}}{2}+y \cdot \frac{1}{2} =4 \\ \Rightarrow & \sqrt{3} x+y =8 \end{array} \end{aligned}$$
Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by $3 x+4 y=4$ and the opposite vertex of the hypotenuse is $(2,2)$.
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 $\theta$.
$$\begin{aligned} & \therefore \quad \tan \theta=\left|\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\right| \Rightarrow \tan 45 \Upsilon= \pm\left[\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\right] \\ & \text { Taking positive sign, } \\ & 1=\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}} \\ & \Rightarrow \quad m+\frac{3}{4}=1-\frac{3 m}{4} \\ & \Rightarrow \quad m+\frac{3 m}{4}=1-\frac{3}{4} \\ & \Rightarrow \quad \frac{7 m}{4}=\frac{1}{4} \Rightarrow m=\frac{1}{7} \end{aligned}$$
Taking negative sign,
$$\begin{array}{rlrl} & 1 =-\left(\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\right) \Rightarrow 1-\frac{3 m}{4}=-m-\frac{3}{4} \\ \Rightarrow \quad & m-\frac{3 m}{4} & =-1-\frac{3}{4} \\ \Rightarrow \quad & \frac{m}{4} & =\frac{-7}{4} \Rightarrow m=-7 \end{array}$$
$\therefore$ Equation of side $A C$ having slope $\left(\frac{1}{7}\right)$ is
$$\begin{array}{rlrl} & y-2 =\frac{1}{7}(x-2) \\ \Rightarrow \quad & 7 y-14 =x-2 \\ \Rightarrow \quad & x-7 y+12 =0 \end{array}$$
and equation of side $A B$ having slope (-7) is
$$\begin{aligned} & y-2 =-7(x-2) \\ \Rightarrow \quad & y-2 =-7 x+14 \\ \Rightarrow \quad & 7 x+y-16 =0 \end{aligned}$$
If the equation of the base of an equilateral triangle is $x+y=2$ and the vertex is $(2,-1)$, then find the length of the side of the triangle.
Given that, equilateral $\triangle A B C$ having equation of base is $x+y=2$.
$$\begin{aligned} & \text { In } \triangle A B D, \quad \sin 60 \gamma=\frac{A D}{A B} \\ & \Rightarrow \quad A D=A B \sin 60 \Upsilon=A B \frac{\sqrt{3}}{2} \\ \because \quad & A D=A B \frac{\sqrt{3}}{2}\quad \text{... (i)} \end{aligned}$$
Now, the length of perpendicular from $(2,-1)$ to the line $x+y=2$ is given by
$$\begin{aligned} & A D=\left|\frac{2+(-1)-2}{\sqrt{1^2+1^2}}\right|=\frac{1}{\sqrt{2}} \\ & \frac{1}{\sqrt{2}}=A B \frac{\sqrt{3}}{2} \\ \text{From Eq. (i),}\quad & A B=\sqrt{\frac{2}{3}} \end{aligned}$$