If the line 2x - y = 5 makes an angle \(\theta\) with x - axis. Then, tan \(\theta\) = 2. Let P (a, a) be a point on the line 2x - y = 5. Then, 2 a - a = 5 \(\Rightarrow\) a = 5

So, the coordinates of Pare (5, 5). If the line 2x - y - 5 = 0 is rotated about point P through 45° in anti-clockwise direction, then the line in its new position makes angle 8 + 45° with x -axis. Let m be the slope of the line in its new position. Then,
 \(m'=tan(\theta+45^o)=\frac{\tan\theta+\tan45^o}{1-\tan\theta\tan45^o}=\frac{2+1}{1-2\times 1}=-3\)
Thus, the line in its new position passes through P (5, 5) and has slope m' = -3
So, its equationy -5 = m' (x - 5) or, y -5 = -3 (x - 5) or, 3x + y - 20 = 0.

Intercept form of straight line \(\frac{x}{a}+\frac{y}{b}=1,\) where a and b are the intercepts on the axis
Given that a = b, \(\therefore \frac{x}{a}+\frac{y}{b}=1\) .....(1)
If equation (1) passes through the point (1, -2) we get
\(\frac{1}{a}-\frac{2}{a}=1\Rightarrow -\frac{1}{a}=1\Rightarrow a=-1\)
So, equation of the straight line is
 \(\frac{x}{-1}+\frac{y}{-1}=1\Rightarrow x+y=-1\Rightarrow x+y+1=0\)
Hence, the required equation x + y + 1 = 0.

Let AB be a line passing through a point (-4, 3) and meets x-axis at A (a, 0) andy-axis at B (0, b).
\(\therefore -4=\frac{5\times 0+3a}{5+3}\Rightarrow -4=\frac{3a}{8}\)
\(\Rightarrow 3a=-32\)
\(\therefore a=\frac{-32}{3}\)
and \(3=\frac{5.b+3.0}{5+3}\Rightarrow 3=\frac{5.b}{8}\)
\(\Rightarrow 5b=24 \Rightarrow b=\frac{24}{5}\)
Intercept form of line is \(\frac{x}{\frac{-32}{3}}+\frac{y}{\frac{24}{5}}=1\Rightarrow \frac{-3x}{32}+\frac{5y}{24}=1\)
\(\Rightarrow -9x+20y=96\Rightarrow 9x-20y+96=0\)
Hence, the required equation is 9x - 20y + 96 = 0.

Equation of line through the intersection of straight lines
2x +y = 8 and 3x - 2y + 7 = 0 is
2x + y - 8 + k (3x - 2y + 7) = 0
x (2 + 3k) + Y (1 - 2k) +(-8 + 7k) = 0·
This is parallel to 4x +y -11 = 0
∴ Their slopes are equal \(\left( \frac { 2+3k }{ 1-2k } \right) =-\left( \frac { 4 }{ 1 } \right) \)
⇒ \(\frac { 2+3k }{ 1-2k } \)=4
2+3k=4-8k
11k=2 ⇒ k=\(\frac { 2 }{ 11 } \)
Required equation is x\(\left( 2+\frac { 6 }{ 11 } \right) +y\left( 1-\frac { 4 }{ 11 } \right) +\left( -8+\frac { 14 }{ 11 } \right) \)=0
\(\frac { 28x }{ 11 } +\frac { 7y }{ 11 } -\frac { 74 }{ 11 } \)=0
⇒ 28x+7y-74=0

Let coordinates of a moving point P be (x, y).
Given that the sum of the distances from the axis to the point is always 1.
\(\therefore |x|+|y|=1\Rightarrow x+y=1\)
\(\Rightarrow\) - x - y = 1 \(\Rightarrow \) x + y = 1
\(\Rightarrow\) x - y = 1
Hence, these equations give us the locus of the point P which is a square.