(i) Let the intercepts on the axes be a' and '- a' respectively.
Equation of line in intercept from is \(\frac{x}{a}+\frac{y}{-a}=1\)
\(\Rightarrow x-y=a\)
Since, this passes through (3,4) then
a = 3 - 4 = -1
The equation is x - y + 1= 0
(ii) Given sum of the intercepts is 14.
\(\text { i.e., } a+b=14 \Rightarrow b=14-a\)
\(\text { Equation of straight line is } \frac{x}{a}+\frac{y}{b}=1\)
\(\Rightarrow \frac{x}{a}+\frac{y}{14-a}=1\)
\(\text { Since, this passes through }(3,4) \text { then }\)
\(\frac{3}{a}+\frac{4}{14-a}=1\)
\(\Rightarrow a^{2}-13 a+42=0 \)
\(\Rightarrow(a-6)(a-7)=0 \)
\(\Rightarrow a=6,7 \)
When a = 6,b = 8
\(\text { Equation of a straight line is } \frac{x}{6}+\frac{y}{8}=1\)
\(\Rightarrow 4 x+3 y=24\)
\(\text { When } \mathrm{a}=7, \mathrm{~b}=7\)
\(\text { Equation of a straight line is } \frac{x}{7}+\frac{y}{7}=1\)
\(\Rightarrow x+y=7\)

Let D (2, 1), E (- 5,7) and F (- 5, - 5) be the mid points of the sides BC, CA and AB of a \(\Delta\) ABC 
EF is parallel to Y-axis
But EF parallel to BC [\(\because\) line joining mid points of two sides is parallel to third side and half of it]
Equation of any line parallel to Y-axis at a distance of 'a' units is x = a.
If it passes through (2, 1), then 2 = a
Equation of side BC is x = 2
\(\text { Now, Slope of } F D=\frac{1+5}{2+5}=\frac{6}{7}\)
By Geometry, CA is Parallel to FD
\(\therefore \text { Slope of } \mathrm{CA}=\frac{6}{7}\)
CA passes through the Point E (-5,7)
\(\therefore \text { Equation of side } C A \text { is } y-7=\frac{6}{7}(x+5)\)
\(\Rightarrow 6 x-7 y+79=0\)
\(\text { Now, slope of } \mathrm{DE}=\frac{7-1}{-5-2}=\frac{-6}{7}\mathrm{AB} \text { is parallel to } \mathrm{DE}\)
\(\text { Slope of } A B=-\frac{6}{7}\)
\(\text { AB passes through }(-5,-5)\)
\(\therefore \text { Equation of } A B \text { is } y+5=-\frac{6}{7}(x+5)\)
\(\Rightarrow 6 x+7 y+65=0\)

\(\text { Slope of } \mathrm{AB}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
\(=\frac{1-y}{0-x} \)
\(=\frac{1-y}{-x} \)
\(\text { Slope of } \mathrm{BC}=\frac{y+2}{x-5}\)
\(\text { Slope of } C D=\frac{-2+1}{5-2}\)
\(=-\frac{1}{3}\)
\(\text { Slope of } \mathrm{DA}=\frac{-1-1}{2-0}\)
\(=-\frac{2}{2}=-1\)
In parallelogram ABCD, AB is parallel to CD
\(\therefore \text { Slope of } \mathrm{AB}=\text { Slope of } \mathrm{CD}\)
\( \frac{1-y}{-x}=-\frac{1}{3} \)
\(\Rightarrow 3-3 y=x \)
\(\Rightarrow x+3 y=3 \)
\(\mathrm{BC} \text { is parallel to } \mathrm{AD}\)
\(\text { Slope of } \mathrm{BC}=\text { Slope of } \mathrm{AD}\)
\( \frac{y+2}{x-5}=-1 \)
\(\Rightarrow y+2=-x+5 \)
\(\Rightarrow x+y=3 \)
\(\text { Solving (1) and (2) }\)
\(x+3 y=3 \)
\(x+y=3\)
\((1)-(2) \Rightarrow 2 y=0\)
\(y= 0\)
\(\text { Substituting in }(2)\)
\(x+0=3\)
\(x = 3\)
\(\therefore \ x=3 \text { and } y=0 \text { and the point B is }(3,0) \text {. }\)