3x²+10xy+8y²+14x+22y+15=0
a= 3, h = 5, b = 8, g = 7, f= 11, c = 15
The condition is af²+bg²+ch² abc-2fgh = 0
3(11)² +8(7)² +15(5)² -(3)(8)(15)-2(11)(7)(5)=363+392+375-360-770=0
Hence the equation represents a pair of straight lines
tanθ=\(\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\pm \frac { 2 }{ 11 } \)
tanθ=\(\frac { 2 }{ 11 } \)
⇒ θ=tan-1\(\left( \frac { 2 }{ 11 } \right) \).

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.

9x² +24xy+16y² +21x+28y+6=0
\(\left| \begin{matrix} 2h=24 \\ h=12 \end{matrix} \right| b=16\left| \begin{matrix} 2g=21 \\ g=\frac { 21 }{ 2 } \end{matrix} \right| c=6\left| \begin{matrix} 2f=28 \\ f=14 \end{matrix} \right| \)
h²-ab=(12)² -9(16)=144-144=0
∴ The lines are parallel.
9x²+24xy+16y² =(3x+4y)(3x+4y)
Let 9x² + 24xy+16y²+21x+28y+6 = (3x+4y+l)(3x+4y+m)
Equating the coefficients of x and constant term
3l+ 3m =21
lm= 6
Solving we get, l=1 or 6 m = 6
m = 6 or 1
∴ The separate equations are 3x + 4y + 1 = 0 and 3x + 4y + 6 =0
The distance between the parallel lines are \(\left| \frac { 6-1 }{ \sqrt { 9+16 } } \right| =\frac { 5 }{ 5 } \)=1 unit.

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.

Equation of line through the intersection of straight lines 5x - 6y = 1 and 3x + 2y + 5
5x - 6y - 1 + k (3x + 2y + 5) = 0
x (5 + 3k) + y (-6 + 2k) + (-1 + 5k) = 0
This is perpendicular to 3x - 5y + 11 = 0
That is, the product of their slopes is -1
-\(\left( \frac { 5+3k }{ -6+2k } \right) \left( -\frac { 3 }{ -5 } \right) \)=-1
⇒ \(\frac { 15+9k }{ -30+10k } \)=1
45=k
Required equation is 5x - 6y -1 + 45 (3x + 2y + 5) = 0
140x + 84y + 224 = 0
20x + 12y + 32 = 0
5x+ 3y+ 8 = 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.

Given that:  OM = 4 units.
\(\angle BAX=120^o\)
\(\therefore \angle BAO=180^o-120^o\) or \(\angle MAO =60^o\)
\(\angle MOA+\angle MAO=90^o\)     [\(\therefore OM\bot AB\)]
\(\theta + 60^o=90^o\quad \quad \therefore \theta=30^o\)
So, equation of AB in its normal form
\(x\cos\theta+y\sin\theta=p\)
\(\Rightarrow x\cos30^o+y\sin30^o=4\)
\(\Rightarrow x\times \frac{\sqrt{3}}{2}+y\times \frac{1}{2}=4\Rightarrow \sqrt{3}x+y=8\)

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

Slope of the line joining the points (2, 3) and (3, -1) is
\(\frac { -1-3 }{ 3-2 } \)=-4
Slope of the required line which is perpendicular to it
=\(\frac { -1 }{ -4 } =\frac { 1 }{ 4 } \)   [∵ m₁m₂=-1]
Equation of the line passing through the point (5, 2) is
y-2 =\(\frac { 1 }{ 4 } \)(x-5)    [y-y₁=m(x-x₁)]
⇒ 4y-8=x-5
⇒ x-4y+3=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.