In what direction should a line be drawn through the point $(1,2)$, so that its point of intersection with the line $x+y=4$ is at a distance $\frac{\sqrt{6}}{3}$ from the given point?
Let slope of the line be $m$. As, the line passes through the point $A(1,2)$.
$\therefore \quad$ Equation of line is $y-2=m(x-1)$
$$m x-y+2-m=0\quad \text{... (i)}$$
$$\begin{aligned} &\begin{aligned} \text { and }\quad & x+y-4=0 \\ & \frac{x}{(4-2+m)}=\frac{y}{2-m+4 m}=\frac{1}{1+m} \\ \Rightarrow \quad & x=\frac{2+m}{1+m} \\ \Rightarrow \quad & y=\frac{3 m+2}{1+m} \end{aligned}\\ &\Rightarrow \quad \frac{x}{2+m}=\frac{y}{3 m+2}=\frac{1}{1+m} \end{aligned}$$
So, the point of intersection is $B\left(\frac{m+2}{m+1}, \frac{3 m+2}{m+1}\right)$.
$$\begin{aligned} & \quad \text { Now, } \quad A B^2=\left(\frac{m+2}{m+1}-1\right)^2+\left(\frac{3 m+2}{m+1}-2\right)^2 \\ & \because \quad A B=\frac{\sqrt{6}}{3} \quad \text{[given]}\\ & \therefore \quad\left(\frac{m+2-m-1}{m+1}\right)^2+\left(\frac{3 m+2-2 m-2}{m+1}\right)^2=\frac{6}{9} \end{aligned}$$
$$\begin{array}{rlrl} \Rightarrow & \left(\frac{1}{m+1}\right)^2+\left(\frac{m}{m+1}\right)^2 =\frac{6}{9} \\ \Rightarrow & \frac{1+m^2}{(1+m)^2} =\frac{6}{9} \\ \Rightarrow & \frac{1+m^2}{1+m^2+2 m} =\frac{6}{9} \\ \Rightarrow & 9+9 m^2 =6+6 m^2+12 m \\ \Rightarrow & 3 m^2-12 m+3 =0 \\ \Rightarrow & m^2-4 m+1 =0 \\ \therefore & m =\frac{4 \pm \sqrt{16-4}}{2} \\ & =2 \pm \sqrt{3} \\ & =2+\sqrt{3} \text { or } 2-\sqrt{3} \\ \therefore \theta& =75 \Upsilon 15^{\circ} \end{array}$$
A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.
Since, the intercept form of a line is $\frac{x}{a}+\frac{y}{b}=1$.
$$\begin{array}{ll} \text { Given that, } & \frac{1}{a}+\frac{1}{b}=\text { constant } \quad =\frac{1}{k}\\ \because & \frac{1}{a}+\frac{1}{b}=\frac{1}{k} \\ \Rightarrow & \frac{k}{a}+\frac{k}{b}=1 \end{array}$$
So, $(k, k)$ lies on $\frac{x}{a}+\frac{y}{b}=1$.
Hence, the line passes through the fixed point.
Find the equation of the line which passes through the point $(-4,3)$ and the portion of the line intercepted between the axes is divided internally in the ratio $5: 3$ by this point.
Since, the line intersects $X$ and $Y$-axes respectively at $A(x, 0)$ and $B(0, y)$.
$$\begin{aligned} -4 & =\frac{5 \times 0+3 x}{5+3} \\ \Rightarrow \quad -4 & =\frac{3 x}{8} \Rightarrow x=\frac{-32}{3} \\ \text{and}\quad 3 & =\frac{5 \cdot y+3 \cdot 0}{5+3} \\ \Rightarrow \quad 3 & =\frac{5 y}{8} \Rightarrow y=\frac{24}{5} \end{aligned}$$
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
$$\begin{aligned} & \frac{x}{-32 / 3}+\frac{y}{24 / 5}=1 \\ \Rightarrow \quad & \frac{-3 x}{32}+\frac{5 y}{24}=1 \\ \Rightarrow \quad & -9 x+20 y=96 \\ \Rightarrow \quad & 9 x-20 y+96=0 \end{aligned}$$
Find the equations of the lines through the point of intersection of the lines $x-y+1=0$ and $2 x-3 y+5=0$ and whose distance from the point $(3,2)$ is $\frac{7}{5}$.
$$\begin{aligned} \text{Given equation of lines}\quad x-y+1 & =0 \quad \text{.... (i)}\\ \text{and}\quad 2 x-3 y+5 & =0 \quad \text{.... (ii)}\\ \text{From Eq. (i),}\quad x & =y-1 \end{aligned}$$
Now, put the value of $x$ in Eq. (ii), we get
$$\begin{array}{lr} & 2(y-1)-3 y+5=0 \\ \Rightarrow & 2 y-2-3 y+5=0 \\ \Rightarrow & 3-y=0 \Rightarrow y=3 \end{array}$$
$y=3$ put in Eq. (i), we get
$$x=2$$
Since, the point of intersection is $(2,3)$.
Let slope of the required line be $m$.
$\therefore$ Equation of line is
$$\begin{aligned} y-3 & =m(x-2) \\ m x-y+3-2 m & =0\quad \text{.... (iii)} \end{aligned}$$
Since, the distance from $(3,2)$ to line (iii) is $\frac{7}{5}$.
$$\begin{aligned} & \therefore \quad \frac{7}{5}=\left|\frac{3 m-2+3-2 m}{\sqrt{1+m^2}}\right| \\ & \Rightarrow \quad \frac{49}{25}=\frac{(m+1)^2}{1+m^2} \\ & \Rightarrow \quad 49+49 m^2=25\left(m^2+2 m+1\right) \\ & \Rightarrow \quad 49+49 m^2=25 m^2+50 m+25 \\ & \Rightarrow \quad 24 m^2-50 m+24=0 \\ & \Rightarrow \quad 12 m^2-25 m+12=0 \\ & \therefore \\ & m=\frac{25 \pm \sqrt{625-4 \cdot 12 \cdot 12}}{24} \\ & =\frac{25 \pm \sqrt{49}}{24}=\frac{25 \pm 7}{24}=\frac{32}{24} \text { or } \frac{18}{24}=\frac{4}{3} \text { or } \frac{3}{4} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} & \therefore \text { First equation of a line is } \quad y-3=\frac{4}{3}(x-2) \\ & \Rightarrow \quad 3 y-9=4 x-8 \\ & \Rightarrow \quad 4 x-3 y+1=0 \end{aligned}\\ &\text { and second equation of line is } y-3=\frac{3}{4}(x-2)\\ &\begin{array}{lr} \Rightarrow & 4 y-12=3 x-6 \\ \Rightarrow & 3 x-4 y+6=0 \end{array} \end{aligned}$$
If the sum of the distance of a moving point in a plane from the axes is 1, then find the locus of the point.
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 .
$$\begin{array}{lr} \therefore & |x|+|y|=1 \\ \Rightarrow & \pm x \pm y=1 \\ \Rightarrow & x+y=1 \\ \Rightarrow & -x-y=1 \\ \Rightarrow & -x+y=1 \\ \Rightarrow & x-y=1 \end{array}$$
So, these equations give us locus of the point which is a square.