Perpendicular distance from vertex to the opposite side is \(\frac{2+3-2}{\sqrt{1+1}}=\frac{3}{\sqrt{2}}\)
\(\frac{3}{\sqrt{2}}=\frac{\sqrt{3}}{2} a\)
\(\Rightarrow a =\frac{6}{\sqrt{6}} \)
\(\therefore a =\sqrt{6}\)

Let y = 2x + c be the required line.
Given perpendicular distance from origin'to this line is \(\sqrt5\)
\(\frac{c}{\sqrt{1+4}}=\sqrt{5}\)
c  = 5
This required line is y - 2x + 5
2x - y + 5 = 0

Equation of line joining (1, 2) and (3, 4) is
\(\frac{y-2}{4-2} =\frac{x-1}{3-1} \)
\(x-y+1 =0 \)
Any line perpendicular to this x + y + k = 0
This passes through midpoint of (1, 2) and (3, 4)
That is (2, 3)
k = -5, x+y-5 = 0
x intercept is 5, y intercept is 5.

The point M is (0, 3)
The equation of line joining ('-1, 2) and (0, 3) is
\(\frac{y-2}{3-2} =\frac{x+1}{0+1} \)
\(x-y+3 =0\)

\(\text {Perimeter }=4+2 \sqrt{2}\)
\(a+a+\sqrt{2} =4+2 \sqrt{2} \)
\(2 a+\sqrt{2} a =4+2 \sqrt{2} \)
\(\therefore a =2 \)
\(\text {Equation of line is } \frac{x}{2}+\frac{y}{2}=1\)
\(x+y-2=0\)

\(\text { Slope of } 3 x-y+5=0 \text { is } \frac{-3}{-1}=3=m_{1}\)
Let m₂ be the slope of the second line
\(\text {Given } \tan \theta=\tan 45^{\circ}=1 \Rightarrow \frac{m_{1}-m_{2}}{1+m_{1} m_{2}}=\pm 1\)
\(\frac{3-m_{2}}{1+3 m_{2}}=1 \quad \frac{m_{2}-3}{1+3 m_{2}}=1\)
\(3-m_{2}=1+3 m_{2} \quad m_{2}-3=1+3 m_{2}\)
\(2=4 m_{2} \quad-2 m_{2}=4\)
\(\mathrm{m}_{2}=\frac{1}{2} \quad \mathrm{~m}_{2}=-2\)
\(\left(\frac{1}{2},-2\right)\)

\(\frac{\left(8^{2}\right)}{16}-\frac{(-5)^{2}}{25} =\mathrm{k} \)
\(\frac{64}{16}-\frac{25}{25} =\mathrm{k} \)
\(\mathrm{k} =4-1=3 \)

\((1,2) \text { lies on } 3 x^{2}+3 y^{2}-8 x-12 y+17=0\)
\(\text { Because, } 3(1)^{2}+3(2)^{2}-8(1)-12(2)+17\)
\(=3+12-8-24+17=0\)

\(\left(a t^{2}, 2 a t\right) \Rightarrow x=a t^{2}, \quad y=2 a t\)
\(y^{2} =4 a^{2} t^{2} \)
\(=4 a^{2}\left(\frac{x}{a}\right) \)
\(y^{2} =4 a x \)

Let the point be (x, y)
Its distance from origin is \(\sqrt{x^{2}+y^{2}}\)
Given \(x =\frac{1}{2} \sqrt{x^{2}+y^{2}} \)
\(\Rightarrow 2 x =\sqrt{x^{2}+y^{2}} \)
\(4 x^{2} =x^{2}+y^{2} \)
\(3 x^{2}-y^{2}=0 \)  is the required equation of the locus