Equation of line perpendicular to x + y + 7 = 0 is
y - x + k = 0
Ans. y - x - 1 = 0

Given equation can be written as
a(x + y -1) + b(2x + 3y -1) = 0
\(\Rightarrow \) (x + y -1) + b(2x + 3y -1) = 0, where \(\lambda \).= \(\frac { b }{ 2 } \)
This is the form of L₁ + \(\lambda \)L₂ = 0. So, it represents that line
passing through the intersectiori of x + y - 1 = 0 and 2x + 3y -1 = 0.

Given that, angle with Y-axis =30^{\(^{o}\)}
Andle with X - axis = 60^{\(^{o}\)}
Slope of the line, m = tan 60^{\(^{o}\)} = \(\sqrt { 3 } \)
Ans. y - \(\sqrt { 3 } \)x - 2 + \(\sqrt { 3 } \) = 0

Let equation of line is \(\frac { x }{ a } +\frac { y }{ 2a } =1\) . Here, we have \(1=\frac { a+0 }{ 2 } \)  and \(2=\frac { b+0 }{ 2 } \)     
Ans. 2x + y - 4 = 0 

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

On dividing both sides of given equation by
\(\sqrt { (\sqrt { 3 } )^{ 2 }+(1)^{ 2 } } \) we get
\(x.\left( \frac { \sqrt { 3 } }{ 2 } \right) +y.\left( \frac { 1 }{ 2 } \right) =2\)
\(xcos30^{ \circ }+ysin30^{ 0 }=2\)
\(\alpha =30^{ \circ }\quad p=2\quad units\)

The given line is \(3x+4y=12\Rightarrow \cfrac { x }{ 4 } +\cfrac { y }{ 3 } =1\)
Let the line (i) cuts X and Y-axes at A and B, respectively.
then, a=(4,0) and B=(0,3).
Let the line AB be trisected at P and Q then AP :PB =1:2
\(=(\cfrac { 1.0+2.4 }{ 1+2 } ,\cfrac { 1.3+2.0 }{ 1+2 } )\)
\( \Rightarrow p=\left( \cfrac { 8 }{ 3 } ,1 \right) and\quad AQ:QB=2:1\)
\(Also\quad q=\left( \cfrac { 2.0+1.4 }{ 1+2 } ,\cfrac { 2.3+1.0 }{ 2+1 } \right) =\left( \cfrac { 4 }{ 3 } ,2 \right) \)
now equation of line OP passing through (0,0) and (8/3, 1) is given by
\(y-0=\cfrac { 1-0 }{ \cfrac { 8 }{ 3 } -0 } (x-0)\Rightarrow y=\cfrac { 3 }{ 8 } x\Rightarrow 3x-8y=0\)
and equation of the line Oq passing through (0,0) and
\((\cfrac { 4 }{ 3 } ,2)\quad is\quad given\quad by\quad y-0=\cfrac { 2-0 }{ 4 } (x-0)\Rightarrow y=\cfrac { 3 }{ 2 } x\)
\(\Rightarrow 2y=3x\Rightarrow 3x-2y=0\)

Let(h,k) be the coordinates of the foot of the perpendicular from the point(2,3) on the line x+y-11=0
Then, the slope of the perpendicular line is\(\cfrac { k-3 }{ h-2 } \)
Again the slope of the given line x + y - 11= 0 is -1
using the condition of perpendicularity of lines,we have
(\(\cfrac { k-3 }{ h-2 } \))(-1)=-1 0r k-h=1...(i)
since,(h,k) lies on the given line, we have
h + k - 11 = 0 or h + k = 11.....(ii)
On solving eqs. (i) and (ii) we get h=5 and k=6
hence,(5,6) are the required coordinates of the foot of the perpendicular.

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) \(=\frac { 5+2 }{ 3-2 } =7\)
\(Now,\tan\theta =\underset { - }{ + } \left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right) =\underset { - }{ + } \left( \frac { 7-3/2 }{ 1+7(3/2) } \right)\)
\(=\underset { - }{ + } \left( \frac { 11/2 }{ 23/2 } = \right) \underset { - }{ + } \left( \frac { 11 }{ 23 } \right) \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 11 }{ 23 } \right) or{ \pi -tan }^{ -1 }\left( \frac { 11 }{ 23 } \right) \)

Show that AB + BC = AC

\(\text { Let } A(x, y) \equiv A(-2-1), B\left(x_{2}, y_{2}\right) \equiv B(4,0)\)\(C(x, y) \equiv C(33) \text { and } D\left(x, y_{j}\right) \equiv(-3,2)\)
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.

Clearly, the equation of a line parallel to the x-axis and passing through(-3,5) is y = 5
The equation of a line parallel to the y-axis and passing through(-3,5) is x = -3.

The distance between P and Q is
\(PQ=\sqrt { (\alpha cos\beta -acos\alpha )^{ 2 }+(asin\beta -asin\alpha )^{ 2 } } \) [by distance formula]
\(=\sqrt { { a }^{ 2 }(cos^{ 2 }\beta +cos^{ 2 }\alpha -2cos\alpha cos\beta +{ a }^{ 2 }(sin^{ 2 }\beta +sin^{ 2 }\alpha -2sin\beta sin\alpha } \)
\(=a\sqrt { (cos^{ 2 }\beta +sin^{ 2 }\beta )+(cos^{ 2 }\alpha +sin^{ 2 }\alpha -2(cos\alpha cos\beta -sin\alpha sin\beta } \)
=\(=a\sqrt { 1+1-2cos(\alpha -\beta ) } =a\sqrt { 2[1-cos(\alpha -\beta )] } \quad [\because cos^{ 2 }\theta +sin^{ 2 }\theta =1\quad and\quad cosAcosB+sinAsinB=cos(A-B)]\quad \quad \)
\(a\sqrt { 2\times 2{ sin }^{ 2 }\left( \frac { \alpha -\beta }{ 2 } \right) } \quad [\because cos\theta =1-2sin^{ 2 }\theta /2\)
\(=2asin\left( \frac { \alpha -\beta }{ 2 } \right) \)  units

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).

Slope of the required perpendicular line=1/3
\(\therefore \) Equation of the required perpendicular line is
y-2=(1/3)(x-2)
For y-intercept put x=0 ,y = 4/3

We have \(\sqrt{3}x+y=1\)
\(\Rightarrow y=-\sqrt{3}x+1\)
\(\therefore m_1=-\sqrt{3}\)
Also  \(x+\sqrt{3}y=1\)
\(\Rightarrow \sqrt{3}y=-x+1\)
\(\Rightarrow y=\frac{-1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}\)
\(\therefore m_2=\frac{-1}{\sqrt{3}}\)
Let \(\theta\) be the angle between the lines. Then
\(\tan\theta=|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3}(\frac{-1}{\sqrt{3}})}|\)
\(=|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}|=|\frac{-2}{\sqrt{3}}\times \frac{1}{2}|\)
\(=|\frac{-1}{\sqrt{3}}|=\frac{1}{\sqrt{3}}\)
\(\tan\theta=\tan 30^o\) and \(\tan(180^o-30^o)\)
\(\theta=30^o\) and \(150^o\)

First, find out the values of AB,BC,CA by distance formula and prove that \(\triangle ABC\) is an isosceles triangle, then prove that this triangle is also right-angled triangle by Converse of Pythagoras theorem.