Given lines 5x + 23y + 14 = 0 and
23x - 5y + 9 = 0
a₁ = 5, b₁ = 23 and a₂ = 23, b₂ = - 5 . - '
Now a₁a₂ + b₁b₂ = (5)(23) + (23)(- 5) = 0
The given lines are Perpendicular.

\(7x-\frac { 3 }{ 17 } \) = 0
Comparing with ax + by + c = 0
\(\text { Slope } =-\frac{a}{b} \)
\(=-\frac{7}{0}=\text { undefined } \)

Given, θ = 45⁰, y intercept, c = 11
Slope m = tan θ  = tan 45⁰ = 1
Therefore, equation of a straight line is of the form y = mx + c
Hence we get, y = x + 11 gives x − y + 11 = 0

\(\left( -\frac { 1 }{ 3 } ,\frac { 1 }{ 2 } \right) \) and \(\left( \frac { 2 }{ 7 } ,\frac { 3 }{ 7 } \right) \)
The slope \(\frac { \frac { 3 }{ 7 } -\frac { 1 }{ 2 } }{ \frac { 2 }{ 7 } +\frac { 1 }{ 3 } } =\frac { \frac { 6-7 }{ 14 } }{ \frac { 6+7 }{ 21 } } \)
\(\frac { 1 }{ 14 } \times \frac { 21 }{ 13 } =-\frac { 3 }{ 26 } \)

Slope of the straight line 2x + 3y - 8 = 0 is
m₁ = \(\frac { -coefficient\quad of\quad x }{ cofficient\quad of\quad y } \)
m₂ = \(\frac{-2}{3}\)
Slope of the straight line 4x + 6y + 18 = 0 is
m₂ = \(\frac { -4 }{ 6 } =\frac { -2 }{ 3 } \)
Here, m₁ = m₂
That is, slopes are equal. Hence, the two straight lines are parallel.

Given 6x + 8y + 7 = 0
slope m \(=\frac { -coefficient\quad of\quad x }{ coefficient\quad of\quad y } =\frac { 6 }{ 8 } =-\frac { 3 }{ 4 } \)
Therefore, the slope of the straight line is = - \(\frac { 3 }{ 4 } \)

Given points \(\left(2, \frac{2}{3}\right)\) and \(\left(-\frac{1}{2},-2\right)\) 
Equation of the line passing through (x₁ , y₁) and (x₁ , y₁)
\( \frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-\frac{2}{3}}{-2-\frac{2}{3}}=\frac{x-2}{-\frac{1}{2}-2} \)
\(\frac{3 y-2}{-6-2}=\frac{2 x-4}{-1-4}\)
-5 (3y - 2) = - 8 (2x - 4)
- 15y+ 10 = -  16x + 32
16x - 15y - 22 = 0

The equation of a straight line passing through the two points (x₁, y₁) and (x₂, y₂) is \(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
Substituting the points we get, \(\frac { y+3 }{ -4+3 } =\frac { x-5 }{ 7-5 } \)
gives 2y + 6 = − x + 5
Therefore, x + 2y + 1 = 0

Equation of the straight line, parallel to 3x - 7y - 12 = 0 is 3x - 7y + k = 0
Since it passes through the point (6,4)
3(6) - 7(4) + k = 0
k = 28 - 18 = 10
Therefore, equation of the required straight line is 3x - 7y + 10 = 0.

If a and b are the intercepts then a + b = 7 or b = 7 − a
By intercept form \(\frac { x }{ a } +\frac { y }{ b } =1\)...(1)
We have \(\frac { x }{ a } +\frac { y }{ 7-a } =1\)
As this line pass through the point (-3,8), we have
\(\frac { -3 }{ a } +\frac { 8 }{ 7-a } =1\) gives –3(7–a) + 8a = a(7 – a)
−21 + 3a + 8a = 7a − a²
So, a² + 4a − 21 = 0
Solving this equation (a − 3) (a + 7) = 0
a = 3 or a = −7
Since a is positive, we have a = 3 and b = 7 – a = 7 – 3 = 4.
Hence \(\frac { x }{ 3 } +\frac { y }{ 4 } =1\)
Therefore, 4x + 3y − 12 = 0 is the required equation.