The equation of straight line passing through $P(3,4)$ is $y=\tan \dfrac{\pi}{6}{x}+(4-3\tan \dfrac{\pi}{6})\implies y=\dfrac{x}{\sqrt{3}}+4-\sqrt{3}$

$\cfrac { 7x }{ 2 } =18+\cfrac { 4 }{ 5 } x-45\ \Rightarrow \cfrac { 7x }{ 2 } -\cfrac { 4x }{ 5 } =-27\ \Rightarrow \cfrac { 35x-8x }{ 10 } =-27\ \cfrac { 27x }{ 10 } =-27\ \Rightarrow x=-10$

Point is $(2,1)$

The graph is shown in image 

$ y=2x+5$ is an equation of line which satisfied many value of $x$ and $y$ 

Option D is correct.

The given lines are $y=x+3, y=2x+3, y=3x+2$ Vand $y+x=3$ and $y+x=3$

Let x axis & y axis are the perpendicular lines. The sum of the distances from point $p(x, y)$ is $1$ 

Distance of the line $3x-4y-25=0$ from the origin is 
$\displaystyle d=\frac{|-25|}{\sqrt{25}}$
$\Rightarrow d=5$
Only the point given in option B lies on the given line .
Also, its distance from origin is 5.
So, (3,-4) is the nearest point on the line from the origin.

The lines are $2x+3y=0$......(1)

Put x = 2 and y = 5 in the equation,
3x - y = 1
6 - 5 = 1
1 = 1
Hence, the point lies on the equation.

Comparing the given equation $y=2x+3$ with $y = mx+c$, we get

The given line equation is $\frac{3}{2}x-y+\frac{2}{3}$

When $ a= 0 $ then the line equation becomes $ by + c = 0 $ or $ y = -\frac {c}{b} $

Equations of the form $ y =k $ are parallel to x-axis. This also means that they are perpendicular to $ y - $ axis as $ x-$ axis and $ y- $axis are perpendicular to each other.

Equation of the line parallel to $ 2x-3y+8 = 0 $ will be of the form $ 2x-3y + k = 0 $

Now, since it passes through $ (2,-3) $, on substituting it , we get $ 2(2) -3(-3) + k = 0  $
$ => k = -13 $

So, required eqn of parallel line is $ 2x - 3y - 13 = 0 $

Given eq 

Given that: $'y'$ is directly proportional to $'x'$,

Given, $y$ $=$ $2x$ $+$ $3$ and $x$ $<$ $2$

The equation $ax+by+c=0$ represent straight line under one condition i. e, 

Given straight line is $3x-4y=12$
$\Rightarrow\frac{x}{4}-\frac{y}{3} = 1$
this line have x intercept y & y- intercept (-3)
so the evaluation of hypotenuse s the given straight line $3x-4y=12$
so, distance of O(0, 0) form the line $3x-4y=12$ is given by
$=\frac{|3\times0-4\times0-12|}{\sqrt{(3)^2+(-4)^2}}$
$=\frac{12}{5}$
$=2.4$

Let $y=\cos { x } \cos { \left( x+2 \right)  } -\cos ^{ 2 }{ \left( x+1 \right)  } \ =\cos { \left( x+1-1 \right)  } \cos { \left( x+1+1 \right)  } -\cos ^{ 2 }{ \left( x+1 \right)  }\$
 Using $\cfrac{1}{2} \left[\cos\left(A+B+A-B\right)+\cos\left(A+B-A+B\right)\right]\
           = \cfrac{1}{2}\left[\cos 2A + \cos 2B\right]
           = \cfrac{1}{2}\left[\cos^{2}A -1 +1 -2 \sin^{2}A\right ]
           =  \cos ^{2}\left(x+1\right)-\sin^{2}1 $
$=\cos ^{ 2 }{ \left( x+1 \right)  } -\sin ^{ 2 }{ 1 } -\cos ^{ 2 }{ \left( x+1 \right)  } \ =-\sin ^{ 2 }{ 1 } $
This is a straight line which is parallel to x-axis, it passes through $\left( \cfrac { \pi  }{ 2 } ,-\sin ^{ 2 }{ 1 }  \right) $

The equation of line is $2y=3x+2$

Given equations of lines as
$L _{1}: x - 3y + 7 = 0 $
Slope of $L _{1}=\displaystyle \frac{1}{3}$
$ L _{2} : 2x + y - 3 = 0$ 
Slope of $L _{2}=-2$
$L _{3} : 7x +\dfrac{7y}{2}-\dfrac{21}{2}=0$
Slope of $L _{3}=-2$
$\Rightarrow L _{2}$ and $L _{3}$ are parallel
Hence, lines cannot bound any region.

The given lines form a triangle.
The number of circles are $4$, i.e. incircle and three ex-circles of the triangle. 

From the diagram the points form a square

Equation of any line is $y=mx+c$