The equations of the given lines are 2x - y + 3 = 0 and x + 2y + 1 = 0
Equation of any line that passes through the intersection of the given lines is in the form
(2x - y + 3) + k (x + 2y + 1) = 0 ...(i)
\(\Rightarrow\) (2 + k)x + (-1 + 2k)y + 3 + k = 0
If this line is parallel to y-axis, then its slope is tan 90° i.e., \(\infty\) (infinity)
\(\therefore \frac{-(2+k)}{(-1+2k)}=\frac{1}{0}\)
\(\Rightarrow\)  -1 + 2k = 0
\(\Rightarrow\)  k = \(\frac{1}{2}\)
Now putting the value of k in equation (i) we get
\((2x-y+3)+\frac{1}{2}(x+2y+1)=0\)
\(\Rightarrow\) 4x - 2y + 6 + x + 2y + 1 = 0
\(\Rightarrow\) 5x + 7 = 0.

Let (x', y') be the new coordinates of the point (x, y0.
Origin is shifted to (-1, -3)
\(\therefore\)   h = -1 and k = -3
Now x = x' + h = x'-1
and y = y' + k = y' -3
Substituting these values of x and y in equation of crude x² + y² = 9, we get
(x'-1)²+(y'-3)² = 9
\(\Rightarrow\)  x'²+1-2x'+y'²+9-6y'=9
 \(\Rightarrow\)  x'²+y'²-2x'-6y'+1 = 0
Hence the equation of the given circle in new system is x²+y²-2x-6y+1 = 0.

Here h = b, k = -b
and x = a, y = b
The transformation relation between the old coordinates (x, y) and the new coordintes (x', y') are given by
x = x'+h
i.e., x' = x-h=a-b
and y = y'+k
i.e., y' = y-k = -a + b
Hence the coordinates of the new point after shifting the origin are (a-b, -a + b).

Slope of line
m = tan 15° = tan (45°- 30)
\(={tan45^0-tan30^0\over 1+tan45^0tan30^0}={1-{1\over \sqrt3}\over 1+{1\over \sqrt3}}\)
\(={\sqrt3-1\over \sqrt3+1}={\sqrt3-1\over \sqrt3+1}\times{\sqrt3-1\over \sqrt3+1}=2-\sqrt3\)
Here c=-4
Putting values of m and c in y = mx + c, we have y = (2 - √3 )x - 4
which is required, equation of line.

Origin is shifted to (-3, 2) by a translation
h = -3 and k = 2
Let (x', y') be the new coordinates of the point (x, y)
\(\therefore\) x = x' + h = x' - 3
and y = y' + k = y' + 2
Now substituting the values of x and y in the given of the parabola y² = 4ax we get
(y' + 2)²= 4a (x'-3)
\(\Rightarrow\) y'² + 4 + 4y' = 4ax' -12a
\(\Rightarrow\) y'² + 4y' - 4ax' + 12a + 4 =0
\(\therefore\) the equation of the parabola in new system is y² + 4y'-4ax+12a + 4 = 0.

Let (x',y') be the new coordinates of a gives point (x,y) after shifting the coordinates of origin to (2, -1).
When the new coordinates (x', y') lie on the given line then
3x'+2y'-5 = 0 .........(1)
Now here x' = x - h and y' = y - k
\(\therefore\) x' = x - 2    \(\therefore\) y' = y + 1
Substituting these values of (x', y') in equation (1), we get
3(x-2)+2(y+1)-5=0
\(\Rightarrow\) 3x + 2y - 9 = 0
Required equation.

Let (x', y') be the coordinates of the new point
x = x'-4 and y = y'+1
Substituting the values of x and y in the given equation x²+2xy-y² + x-2 = 0, we get
(x'-4)²+2(x'-4)(y'+1)-(y'+1)-(y'+1)²+(x'-4)+2=0
\(\Rightarrow\) x^'2+16-8x^'+2x'y'+2x'-8y'-8-y'²-1-2y'+x'-4+2 = 0
\(\Rightarrow\) x'²+2x' y' -y'²-5x'-10y'+1 = 0
therefore, the equation of the pair of straight lines in new system is x² + 2xy-y²-5x-10y+1 = 0.

Let A=(1,-1),B(5,2)and C(9,5)
Now,distance between A and B
\(AB=\sqrt { (5-1)^{ 2 }+(2+1)^{ 2 } } \) [By distance formula]

\(=\sqrt { (4)^{ 2 }+(3)^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } =5\)
Distance between B and C,BC=
\(\sqrt { (5-9)^{ 2 }+(2-5)^{ 2 } } \)
\(=\sqrt { (-4)^{ 2 }+(-3)^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } =5\)
Distance between A and C
\(AC=\sqrt { (1-9)^{ 2 }+(-1-5)^{ 2 } } =\sqrt { (-8)^{ 2 }+(-6)^{ 2 } } =\sqrt { 64+36 } =10\)
Clearly,AC=AB+BCHence,A,B and C are collinear points

We know that, the acute angle \(\theta \) between 2 lines with slopes m₁ and m₂ is given by
\(\tan\theta =\left| \frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(\text{Let } { m }_{ 1 }=\frac { 1 }{ 2 } ,{ m }_{ 2 }=m\text{ and} \theta =\frac { \pi }{ 4 }\)
Now, putting these values in (1), we get
\(\tan\frac { \pi }{ 4 } =\left| \frac { m-\frac { 1 }{ 2 } }{ 1+\frac { 1 }{ 2 } m } \right| \Rightarrow 1=\left| \frac { m-\frac { 1 }{ 2 } }{ 1+\frac { 1 }{ 2 } m } \right| \quad \left[ \because tan\frac { \pi }{ 4 } =1 \right] \)
which gives \( \frac { m-\frac { 1 }{ 2 } }{ 1+\frac { 1 }{ 2 } m } =1\text{ or} \frac { m-\frac { 1 }{ 2 } }{ 1+\frac { 1 }{ 2 } m } =-1\)
Therefore m = 3 or m = \(-\frac{1}{3}\).
Hence, slope of the other line is 3 or  \(-\frac{1}{3}\). explains the reason of two answers.

Let θ be the angle between the given lines.
We have,
m₁ = Slope of the line joining (0,0) and (2,3) \(=\frac { 3-0 }{ 2-0 } =\frac { 3 }{ 2 } \)
m₂ = Slope of the line joining (2,−2) and (3,5) \(=\frac { 5+2 }{ 3-2 } =7\)
\(Now,\tan\theta =\underset { - }{ + } \left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right) =\underset { - }{ + } \left( \frac { 7-3/2 }{ 1+7(3/2) } \right)\)
\(=\underset { - }{ + } \left( \frac { 11/2 }{ 23/2 } = \right) \underset { - }{ + } \left( \frac { 11 }{ 23 } \right) \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 11 }{ 23 } \right) or{ \pi -tan }^{ -1 }\left( \frac { 11 }{ 23 } \right) \)

\(\text { Let } A(x, y) \equiv A(-2-1), B\left(x_{2}, y_{2}\right) \equiv B(4,0)\)\(C(x, y) \equiv C(33) \text { and } D\left(x, y_{j}\right) \equiv(-3,2)\)
Now, mid-point of AC \(=\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)\)
\(=\left(\frac{-2+3}{2}, \frac{-1+3}{2}\right)=\left(\frac{1}{2}, 1\right)\)...(i)
and mid-point of BD \(=\left(\frac{x_{2}+x_{4}}{2}, \frac{y_{2}+y_{4}}{2}\right)\)
\(=\left(\frac{4-3}{2}, \frac{0+2}{2}\right)=\left(\frac{1}{2}, 1\right)\)  ...(ii)
From Eqs. (i) and (ii), we get Mid-point of AC = Mid-point of BC
Thus, mid-points of both diagonals are coincide each other.
Hence, the points A, B, C and D are vertices of a parallelogram.

Let, P(h,k) be the moving point such that the sum of its distances from A(ae,0) and B(-ae,0) is 2a.
Then, PA+PB=2a
\(\Rightarrow \)\(\sqrt { { (h-ae) }^{ 2 }+{ (k-0) }^{ 2 } } +\sqrt { { (h+ae) }^{ 2 }+{ (k-0) }^{ 2 } } =2a\) [by distance formula]
\(\Rightarrow \)\(\sqrt { { (h-ae) }^{ 2 }+{ k }^{ 2 } } +2a-\sqrt { { (h+ae) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow \)\({ (h-ae) }^{ 2 }+{ k }^{ 2 }={ 4a }^{ 2 }+{ (h+ae) }^{ 2 }+{ k }^{ 2 }-4a\sqrt { { (h+ae) }^{ 2 }+{ k }^{ 2 } } \)
[squaring on both sides]
\({ h }^{ 2 }+{ a }^{ 2 }{ e }^{ 2 }{ -2hae+4a }^{ 2 }{ h }^{ 2 }+{ a }^{ 2 }{ e }^{ 2 }+2hae-4a\sqrt { { (h+ae) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow \)\(-4aeh-{ 4a }^{ 2 }=-4a\sqrt { { (h+ae) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow \)\((eh+a)=\sqrt { { (h+ae) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow \)\({ (eh+a) }^{ 2 }={ (h+ae) }^{ 2 }+{ k }^{ 2 }\)[again squaring on both sides]
\(\Rightarrow \)\({ e }^{ 2 }{ h }^{ 2 }+2aeh+{ a }^{ 2 }={ h }^{ 2 }{ +a }^{ 2 }{ e }^{ 2 }+2aeh+{ k }^{ 2 }\)
\(\Rightarrow \)\({ h }^{ 2 }{ (1-e) }^{ 2 }{ +k }^{ 2 }{ =a }^{ 2 }{ (1-e }^{ 2 })\Rightarrow \frac { { h }^{ 2 } }{ { a }^{ 2 } } +\frac { { k }^{ 2 } }{ { a }^{ 2 }{ (1-e }^{ 2 }) } =1\)
Hence, locus of point P(h,k) is 
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { a }^{ 2 }{ (1-e }^{ 2 }) } =1\)  or \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) where b²-=a²(1-e²)

Let coordinates of a point P changes axes from (x,y) to (X,Y) in new coordinate axes whose origin has the coordinate h=-2,k=3.
Then, x+X-2 and y=Y+3
Given equation of line with respect to old axes is
y+3x=2
On putting x=X-2 and y=Y+3 in Eq.(i) we get
Y+3+3(X-2)=2
Y+3X+3-6=2 \(\Rightarrow \) Y+3X=5
Hence, the required equation of line with respect to new axes y+3x=5.

The coordinates of the new origin are h=1, k=2 and the original coordinates are given point are x = 3,y = -4.
The transformation relation between the old coordinates (x,y) and the new coordinates (X,Y) are given by
x = X + h,i.e.X = x - h ...(i)
y = Y+ k,i.e.Y=y-k. ..(ii)
On substituting the values x=3, y=4,h=1 and k=2 ineqs (i) and (ii), we get
X = 3 = 2 and Y = 42 = 6
Hence the coordinates of point (3,-4) in the new system are (2,-6).

Let B=(-4,1)
Since,x-coordinates of B is negative and its y-coordinates is positive, therefore B lies in the second quadrant