x²+2xy-2y²-2x-14y+15=0

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.

29x + 4y + 5 = 0, 8x - 5y - 21 = 0, 13x + 14y + 47 = 0.

Let P(x, 0) be any point on the x-axis which is equidistant from Q(7, 6) and R(3,4).
Then  \(PQ=\sqrt{(x-7)^2+(0-6)^2}\)
\(=\sqrt{x^2-14x+49+36}\)
\(=\sqrt{x^2-14x+85}\)
\(PR = \sqrt{(x-3)^2+(0-4)^2}\)
\(=\sqrt{x^2-6x+9+16}\)
\(=\sqrt{x^2-6x+25}\)
Since PQ = PR,
\(\therefore \sqrt{x^2-14x+85} = \sqrt{x^2-6x+25}\)
Squaring both sides, we have
x² - 14x + 85 = x² - 6x + 25
\(\Rightarrow\)  -14x + 6x = 25 - 85  \(\Rightarrow\)  -8x = -60
\(\Rightarrow\)   \(x=\frac{15}{2}\)
Thus coordinates of point on the x-axis is \((\frac{15}{2},0)\)

Side of a square, \(a=\frac { \left| 26+65 \right| }{ \sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } } =\frac { 91 }{ \sqrt { 169 } } \) units
Area of square = a²
Ans. 49 sq units

m = tan150^{\(^{o}\)} = tan (180\(^{o}\)-  30\(^{o}\)) = - tan 30\(^{o}\)  = \(-\frac { 1 }{ \sqrt { 3 } } \)  
Ans. \(x+\sqrt { 3 } y+(-3+5\sqrt { 3 } )=0\)

The given equation is y = 0.
It can be written as
y = 0.x + 0
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

Given equation of line is
\(\frac { x }{ a } +\frac { y }{ b } =1\) ....(i)
Perpendicular length from the origin to the line (i) is
\(p=\frac { 1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } =\frac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \Rightarrow { p }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } \)
Since, a² , p² and b² are in AP.
\(\therefore { 2p }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\Rightarrow \frac { 2{ a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } ={ a }^{ 2 }{ +b }^{ 2 }\)
\(\Rightarrow \) 2a²b² = (a² + b²)² \(\Rightarrow \)2a²b² = a⁴+b⁴+2a²b²
\(\Rightarrow \) a⁴ + b⁴ = 0

Since,D,E and the F are the mid-points of sides AB,BC and CA, respectively of a \(\triangle \) ABC.
ar (\(\triangle \)ABC) = 4 x ar(\(\triangle \)DEF)
8 sq units

Let the slope of the other line be m.
It is given that the angle between two lines is 60⁰
\(\tan{ 60 }^{ 0 }=\left| \frac { m-2 }{ 1+2m } \right| \left[ \because \quad tan\theta =\left| \frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \right] \)
\(\sqrt { 3 } =\left| \frac { m-2 }{ 1+2m } \right| \Rightarrow \frac { m-2 }{ 1+2m } =\pm \sqrt { 3 }\)
\(\frac { m-2 }{ 1+2m } =\sqrt { 3 } or\frac { m-2 }{ 1+2m } =-\sqrt { 3 }\)
\( m-2=\sqrt { 3 } +2\sqrt { 3 } m\quad or\quad m-2=-\sqrt { 3 } -2\sqrt { 3 } m\)
\(m(2\sqrt { 3 } -1)=-(2+\sqrt { 3 } )\quad or\quad m(2\sqrt { 3 } +1)=2-\sqrt { 3 } \)
\(m=-\left( \frac { 2+\sqrt { 3 } }{ 2\sqrt { 3 } -1 } \right) or\quad m=\left( \frac { 2-\sqrt { 3 } }{ 2\sqrt { 3 } +1 } \right) \)
\(\text{On substituting} { x }_{ 1 }=2,{ y }_{ 1 }=3 \text{ and the values of m in} y-{ y }_{ 1 }=m(x-{ x }_{ 1 })\)
We obtain that the equation of the erquired line is
\(y-3=-\frac { 2+\sqrt { 3 } }{ 2\sqrt { 3 } -1 } (x-2)\quad or\quad y-3=\frac { 2-\sqrt { 3 } }{ 2\sqrt { 3 } +1 } (x-2)\)

Show that AB + BC = AC

Let the point of X-axis be p(x,0) which is equidistant from(say) A(3,2) and (say) B(-5,-2)
Since, P is equidistant from A and B So,
  \(\therefore PA=PB\Rightarrow P{ A }^{ 2 }=P{ A }^{ 3 }\)
\(\Rightarrow (3-x)^{ 2 }+(3-0)^{ 2 }=(-5-x)^{ 2 }+(-2-0)^{ 2 }\) [by distance formula]
\(\Rightarrow 9+{ x }^{ 2 }-6x+4=25+{ x }^{ 2 }+10x+4\)
\(\Rightarrow 16x+16=0\Rightarrow x=-1\)
Thus, point on X-axis is (-1,0)

The equation of any line which is at a distance \(\frac { 1 }{ 2 } \) from origin is \(x \ cos\alpha +y \ sin\alpha =\frac { 1 }{ 2 } \)
It passes through the point (0,1)
\(0\{ cos\alpha )+1(sin\alpha )=\frac { 1 }{ 2 } \Longrightarrow sin\alpha =\frac { 1 }{ 2 } \)
Now,  \(cos\alpha =\pm \sqrt { 1-sin^{ 2 }\alpha } =\pm \sqrt { 1-(\frac { 1 }{ 2 } )^{ 2 } } =\pm \frac { \sqrt { 3 } }{ 2 } \)
\(\sqrt { 3x } +y-1=0and\sqrt { 3 } x-y+1=0\)

The given equations are x-7y+5=0 and 3x+y-7 = 0
Equation of any line passing through the point of intersection of the given lines is
(x-7y+5)+k(3x+y-7)=0....(i)
\(\Rightarrow\) x-7y+5+3kx+ky-7k=0  \(\Rightarrow\) (1+3k)x+(k-7)y+5-7k=0
Slope of this equation = \(\frac { -(1+3k) }{ k-7 } ={ m }_{ 1 }\) (Say)
and the slope of the given equation 2x-5y+1=0=\(\frac { -2 }{ -5 } \) i.e.,  \(\frac { 2 }{ 5 } ={ m }_{ 2 }\) (Say)
as the two lines are perpendicular to each other
\(\therefore\)  m₁m₂ = -1
\(\therefore \frac { -(1+3k) }{ k-7 } \times \frac { 2 }{ 5 } =-1\Rightarrow \frac { 1+3k }{ k-7 } =\frac { 5 }{ 2 } \)
\(\Rightarrow\)  5k-35=2+6k  \(\Rightarrow\)  k = -37
Now substituting the value of k in equation (i)
(x-7y5)-37(3x+y-7)=0  \(\Rightarrow\) x-7y+5-11x-37y+259=0
\(\Rightarrow\) - 110x-44y+264 = 0
\(\Rightarrow\) 5x+2y-12 = 0 Required Equation.

The equations of lines AB, BC and AC are 
x + y - 6 = 0   ...(i)
x - 3y - 2 = 0  ...(ii)
5x- 3y + 2 = 0 ...(iii)
On solving the equation (i) and (ii), we have coordinates of point B (5,1).
On solving the equation (ii) and (iii), we have coordinates of point C (-1, -1).
On solving the equation (iii) and (i), we have coordinates of point A (2, 4).
Let D, E, F are mid points of BC, AC and AB respectively.
Coordinates of D are \(\left({5-1\over 2},{1-1\over 2}\right)\ i.e.(2,0)\)
Coordinates of E are \(\left({2-1\over 2},{4-1\over 2}\right)\ i.e.\left({1\over 2},{3\over 2}\right)\)
Coordinates of D are  \(\left({2+5\over 2},{4+1\over 2}\right)\ i.e.\left({7\over 2},{5\over 2}\right)\)
Equation of median AD is
\(y-4={(0-4)\over (2-2)}(y-2)\)
⇒  x - 2 = 0
Equation of median BE is
\(y-1={\left({3\over2}-1\right)\over \left({1\over2}-5\right)}(x-5)\)
\(⇒\ y-1={{1\over 2}\over -{9\over 2}}(x-5)\)
⇒ 9y - 9 = -x + 5
⇒ x + 9y - 14 = 0
Equation of CF is
\(y+1={\left({5\over 2}+1\right)\over \left({7\over 2}+5\right)}(x+1)\)
\(⇒\ y+1={{7\over 2}\over {9\over 2}}(x+1)\)
\(⇒\ y+1={7\over 9}(x+1)\)
⇒ 9y + 9 = 7x + 7 ⇒ 7x - 9y - 2 = 0
Thus equations of medians are x - 2 = 0, x + 9y - 14 = 0 and 7x - 9y - 2 = O.

Let (x', y') be the new coordinates of the point (x, y) in new system.
Origin is shifted to a new point (-3, 2) by a translation.
\(\therefore\) h = -3, k = 2
x = x' + h = x' - 3
and y = y' + k = y' + 2
Substituting these vales of x and y in the given equation  \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) , we get
\(\frac { ({ x }^{ ' }-3)^{ 2 } }{ { a }^{ 2 } } +\frac { { (y'+2) }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow \frac { { x' }^{ 2 }+9-6x' }{ { a }^{ 2 } } +\frac { { y' }^{ 2 }+4+{ 4y }^{ ' } }{ { b }^{ 2 } } =1\)
\(\Rightarrow\) b²x'²+9b²-6b²x'+a²y'2+4a²+4a²y'=a²b²
\(\Rightarrow\) b²x'²+a²y'²-6b²x'+4a²y'+4a²+9b²-a²b²=0
Hence the new equation of the ellipse is b²x²+a²y²-6b²x+4a²y+4a²+9b²-a²b² = 0.