One vertex of the equilateral triangle with centroid at the origin and one side as $x+y-2=0$ is
If $a, b$ and $c$ are in AP, then the straight lines $a x+b y+c=0$ will always pass through .............. .
Given line is $$a x+b y+c=0\quad \text{.... (i)}$$
Since, $a, b$ and $c$ are in AP, then
$$\begin{aligned} b & =\frac{a+c}{2} \\ a-2 b+c & =0\quad \text{.... (ii)} \end{aligned}$$
On comparing Eqs.(i) and (ii), we get
$$x=1, y=2\quad \text{[using value of b in Eq. (i)]}$$
So, $(1,-2)$ lies on the line.
The line which cuts off equal intercept from the axes and pass through the point $(1,-2)$ is ............ .
Let equation of line is
$$\frac{x}{a}+\frac{y}{a}=1\quad\text{.... (i)}$$
Since, this line passes through $(1,-2)$.
$$\begin{aligned} & \frac{1}{a}-\frac{2}{a}=1 \\ \Rightarrow \quad & 1-2=a \Rightarrow a=-1 \end{aligned}$$
$\therefore$ Required equation of the line is
$$\begin{aligned} -x-y & =1 \\ \Rightarrow \quad x+y+1 & =0 \end{aligned}$$
Equation of the line through thes point $(3,2)$ and making an angle of $45^{\circ}$ with the line $x-2 y=3$ are ........... .
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 $m$.
Then,
$$\begin{aligned} \tan \theta & =\left|\frac{m-\frac{1}{2}}{1+\frac{1}{2} m}\right| \\ 1 & = \pm\left(\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right) \quad [\because \tan45^\circ=1] \text{... (i)} \end{aligned}$$
Taking positive sign, $$1+\frac{m}{2}=m-\frac{1}{2}$$
$$\begin{aligned} &\begin{array}{ll} \Rightarrow & m-\frac{m}{2}=1+\frac{1}{2} \\ \Rightarrow & \frac{m}{2}=\frac{3}{2} \Rightarrow m=3 \end{array}\\ &\text { Taking negative sign, } \end{aligned}$$
$$\begin{array}{ll} 1 & =-\left(\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right) \\ \Rightarrow & 1+\frac{m}{2}=-m+\frac{1}{2} \\ \Rightarrow & m+\frac{m}{2}=\frac{1}{2}-1 \\ \Rightarrow & \frac{3 m}{2}=\frac{-1}{2} \Rightarrow m=\frac{-1}{3} \end{array}$$
$\therefore$ First equation of the line is
$$\begin{aligned} & y-2=3(x-3) \\ & \Rightarrow \quad 3 x-y-7=0 \end{aligned}$$
and second equation of the line is
$$\begin{aligned} & y-2=-\frac{1}{3}(x-3) \\ & \Rightarrow \quad 3 y-6=-x+3 \\ & \Rightarrow \quad x+3 y-9=0 \end{aligned}$$
The points $(3,4)$ and $(2,-6)$ are situated on the ............. of the line $3 x-4 y-8=0$.
$$\begin{array}{lc} \text { Given line is } & 3 x-4 y-8=0 \quad \text{... (i)}\\ \text { For point }(3,4), & 9-4 \cdot 4-8 \\ \Rightarrow & 9-16-8 \\ \Rightarrow & 9-24 \\ \Rightarrow & -15<0\quad \text{.... (i)} \end{array}$$
For point $(2,-6)$,
$$\begin{aligned} 6 & +24-8 \\ 22 & >0\quad \text{... (ii)} \end{aligned}$$
Since, the value are of opposite sign.
Hence, the points $(3,4)$ and $(2,-6)$ lies on opposite side to the line.