x - 7y + 19 = 0

√5

2x + 3y - 9 = 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.

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.

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.

\(\frac{-8\pm5\sqrt{3}}{11}\)

\((\sqrt{3}+1)x+(\sqrt{3}-1)y=18\sqrt{2}\)

Here 6x - 8y + 7 = 0
⇒ -6x+ 8y = 7
\(⇒ -{6x\over \sqrt{(-6)^2+(8)^2}}+{8y\over \sqrt{(-6)^2+(8)^2}}\)
\(={7\over \sqrt{(-6)^2+(8)^2}}\)
\(⇒-{6x\over 10}+{8y\over 10}={7\over 10}\)
\(⇒ -{3\over 5}x+{4\over 5}y={7\over 10}\)
Length of perpendicular from origin =  \(7\over 10\)
Now 8x - 6y + 11= 0 ⇒ - 8x + 6y = 11
\(⇒ -{8x\over \sqrt{(-8)^2+(6)^2}}+{6y\over \sqrt{(-8)^2+(6)^2}}\)
\(={11\over \sqrt{(-8)^2+(6)^2}}\)
\(⇒ -{8x\over 10}+{6y\over 10}={11\over 10}\)
\(⇒ -{4\over 5}x+{3\over 5}y={11\over 10}\)
Length of perpendicular from origin = \(11\over 10\)
Now \({7\over 10}<{11\over 10}\)
Thus 6x- 8y + 7 = 0 is nearer to the origin.

The equation of the given lines are
2x + 3y - 1 = 0 and x + 5y + 4 = 0
Equation ofany line passing through the point of intersection of the given lines is in the form
(2x + 3y - 1) + k (x + 5y + 4) = 0 ... (i)
\(\Rightarrow\)  (2 + k) x + (3 + 5k) y - 1 + 4k = 0
\(\Rightarrow (\frac{2+k}{1-4k})x+(\frac{3+5k}{1-4k})y=1\)
\(\Rightarrow (\frac{x}{(\frac{1-4k}{2+k})})+\frac{y}{(\frac{1-4k}{3+5k})}=1\)
As the intercepts made by the line (i) on the axes are same
\(\therefore \frac{1-4k}{2+k}=\frac{1-4k}{3+5k}\)
\(\Rightarrow\)  3 + 5k = 2 + k
\(\Rightarrow\)  4k = -1
\(k=\frac{-1}{4}\)
Now putting the value of k in equation (i) we get
\((2x+3y-1)-\frac{1}{4}(x+5y+4)=0\)
\(\Rightarrow\) 8x + 12y - 4 - x - 5y - 4 = 0
\(\Rightarrow\) 7x + 7y - 8 = 0.

The given equations are x - y - 7 = 0 and 2x + y - 2 = 0
Equation of any line which passes through the intersection of the given lines is
(x - y - 7) + k (2x + y - 2) = 0 ... (i)
If equation (i) also passes through the origin
i.e., (0, 0), we get
-7 -2k= 0
\(∴\ k={7\over 2}\)
Putting the value of k is equation (i), we get
(x - y - 7 ) - \(7\over2\) (2x + y. - 2) = 0
2x - 2y - 14 - 14x - 7y + 14 = 0
⇒ -12x- 9y = 0
⇒ 4x + 3y = 0

The equations of lines are 2x + 3y + 4 + k (6x - y + 12) = 0 and 7x + 5y - 4 = 0.
Let m1 be slope of the line 2x + 3y + 4 + k (6x - y + 12) = 0.
Then \(m_1=-{2+6k\over 3-k}\)
Let m₂ be slope of the line 7x + 5y - 4 = 0
Then  \(m_2=-{7\over 5}\)
Now m₁m₂ =-1
\(⇒ \left(-{2+6k\over 3-k}\right)\left(-{7\over5}\right)=-1\)
⇒ 14 + 42k = - 15 + 5k
⇒ 42k - 5k = - 15 - 14
⇒ 37k =- 29
⇒ \(k=-{29\over 37}\)