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.
$P_1$ and $P_2$ are points on either of the two lines $y-\sqrt{3}|x|=2$ at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from $P_1, P_2$ on the bisector of the angle between the given lines.
$$\begin{aligned} \text{Given equation of lines are} \quad & y-\sqrt{3} x=2 \quad [\because x\ge0]\\ \text{and}\quad & y+\sqrt{3} x=2 \quad [\because x\le0] \end{aligned}$$
$\because\quad y=\sqrt{3} x+2\quad\text{.... (i)}$
$$\begin{aligned} \text{and}\quad y & =-\sqrt{3} x+2 \\ \Rightarrow \quad \sqrt{3} x+2 & =-\sqrt{3} x+2 \\ \Rightarrow \quad 2 \sqrt{3} x & =0 \Rightarrow x=0 \end{aligned}$$
On putting $x=0$ in Eq. (i), we get
So, the point of intersection of line (i) and (ii) is ( 0,2 ).
Here, $OC=2$
In $\Delta DEC,$ $C D =\cos 30 \Upsilon $
$$\begin{aligned} \frac{O C}{C D} & =2 \\ C D & =5 \cos 30 \Upsilon \\ & =5 \cdot \frac{\sqrt{3}}{2} \\ O D & =O C+C D=2+5 \frac{\sqrt{3}}{2} \end{aligned}$$
So, the coordinates of the foot of perpendiculars are $\left(0,2+\frac{5 \sqrt{3}}{2}\right)$.
If $p$ is the length of perpendicular from the origin on the line $\frac{x}{a}+\frac{y}{b}=1$ and $a^2, p^2$ and $b^2$ are in AP, the show that $a^4+b^4=0$.
Given equation of line is,
$$\frac{x}{a}+\frac{y}{b}=1\quad \text{.... (i)}$$
Perpendicular length from the origin on the line (i) is given by $p$
$$\begin{aligned} \text{i.e.,}\quad & p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{a b}{\sqrt{a^2+b^2}} \\ \therefore \quad & p^2=\frac{a^2 b^2}{a^2+b^2} \end{aligned}$$
Given that, $a^2, p^2$ and $b^2$ are in AP.
$$ \begin{array}{lr} \therefore & 2 p^2=a^2+b^2 \\ \Rightarrow & \frac{2 a^2 b^2}{a^2+b^2}=a^2+b^2 \\ \Rightarrow & 2 a^2 b^2=\left(a^2+b^2\right)^2 \\ \Rightarrow & 2 a^2+b^2=a^4+b^4+2 a^2 b^2 \\ \Rightarrow & a^4+b^4=0 \end{array}$$