Let the intercept along the axes be $a$ and $b$.

Given, $$a+b=14 \Rightarrow b=14-a$$

Now, the equation of line is $\frac{x}{a}+\frac{y}{b}=1\quad \text{.... (i)}$.

$$\Rightarrow \quad \frac{x}{a}+\frac{y}{14-a}=1$$

Since, the point $(3,4)$ lies on the line.

$$\begin{aligned} & \therefore \quad \frac{3}{a}+\frac{4}{14-a}=1 \\ & \Rightarrow \quad \frac{42-3 a+4 a}{a(14-a)}=1 \Rightarrow 42+a=14 a-a^2 \\ & \Rightarrow \quad a^2-13 a+42=0 \Rightarrow a^2-7 a-6 a+42=0 \\ & \Rightarrow \quad a(a-7)-6(a-7)=0 \Rightarrow(a-7)(a-6)=0 \\ & \Rightarrow \quad a-7=0 \text { or } a-6=0 \\ & \therefore \quad a=7 \text { or } a=6 \\ & \text { When } \\ & a=7 \text {, then } b=7 \\ & \text { When } \\ & a=6 \text {, then } b=8 \end{aligned}$$

$\therefore$ The equation of line, when $a=7$ and $b=7$ is

$$\frac{x}{7}+\frac{y}{7}=1 \Rightarrow x+y=7$$

So, the equation of line, when $a=6$ and $b=8$ is $\frac{x}{6}+\frac{y}{8}=1$

Let the required point be $(h, k)$ and point $(h, k)$ lies on the line $x+y=4$

i.e., $$h+k=4\quad \text{.... (i)}$$

The distance of the point $(h, k)$ from the line $4 x+3 y=10$ is

$$\begin{aligned} \left|\frac{4 h+3 k-10}{\sqrt{16+9}}\right| & =1 \\ 4 h+3 k-10 & = \pm 5 \\ \text{Taking positive sign,}\quad 4 h+3 k & =15\quad \text{.... (ii)} \end{aligned}$$

From Eq. (i) $h=4-k$ put in Eq. (ii), we get

$$\begin{aligned} 4(4-k)+3 k & =15 \\ \Rightarrow \quad 16-4 k+3 k & =15 \\ \Rightarrow \quad k & =1 \end{aligned}$$

On putting $k=1$ in Eq. (i), we get

$$h+1=4 \Rightarrow h=3$$

So, the point is $(3,1)$.

Taking negative sign,

$$\begin{array}{ll} & 4 h+3 k-10=-5 \\ \Rightarrow & 4(4-k)+3 k=5 \\ \Rightarrow & 16-4 k+3 k=5 \\ \Rightarrow & -k=5-16=-11 \\ \therefore & k=11 \end{array}$$

On putting $k=11$ in Eq. (i), we get

$$h+11=4 \Rightarrow h=-7$$

Hence, the required points are $(3,1)$ and $(-7,11)$.

$$\begin{aligned} &\text { Given equation of lines are }\\ &\begin{array}{lr} & \frac{x}{a}+\frac{y}{b}=1 \quad \text{... (i)}\\ \therefore & \text { Slope, } m_1=-\frac{b}{a} \\ \text { and } & \frac{x}{a}-\frac{y}{b}=1 \quad \text{... (ii)}\\ \therefore & \text { Slope, } m_2=\frac{b}{a} \end{array} \end{aligned}$$

$$\begin{aligned} &\text { Let } \theta \text { be the angle between the given lines, then }\\ &\begin{aligned} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \Rightarrow \tan \theta=\left|\frac{-\frac{b}{a}-\frac{b}{a}}{1+\left(\frac{-b}{a}\right)\left(-\frac{b}{a}\right)}\right| \\ \Rightarrow \quad & \tan \theta=\left|\frac{\frac{-2 b}{a}}{\frac{a^2-b^2}{a^2}}\right| \Rightarrow \tan \theta=\frac{2 a b}{a^2-b^2} \end{aligned} \end{aligned}$$

Hence proved.

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

$$\begin{aligned} & y-2=\sqrt{3}(x-1) \\ & \Rightarrow \quad y-2=\sqrt{3} x-\sqrt{3} \\ & \Rightarrow \quad y-\sqrt{3} x-2+\sqrt{3}=0 \end{aligned}$$

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

$$\begin{aligned} & y+\frac{21}{5}=\frac{-3}{4}\left(x-\frac{23}{5}\right) \\ & \Rightarrow \quad 4 y+\frac{84}{5}=-3 x+\frac{69}{5} \\ & \Rightarrow \quad 3 x+4 y=\frac{84-69}{5} \Rightarrow 3 x+4 y+\frac{15}{5}=0 \\ & \Rightarrow \quad 3 x+4 y+3=0 \end{aligned}$$