K = 4

To form a triangle, sum of any two sides of a triangle must be greater than its third side.

Given vertices of a \(\Delta \)ABC are A(-2,0), B(2,0) and C(0,2).
\(\therefore \)   Area of \(\Delta \)ABC   = \(\frac { 1 }{ 2 } \)[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
=\(\frac { 1 }{ 2 } \)[4+4] = \(\frac { 8 }{ 2 } \) = 4 sq uints.
Also, given vertices of a ll.PQR are P( -4,0), Q( 4,0) and R(0,4).
\(\therefore \)Area of \(\Delta \)PQR = \(\frac { 1 }{ 2 } \)[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
 =\(\frac { 1 }{ 2 } \)[-4(0-4)+4(4-0)+0(0-0)]
=\(\frac { 1 }{ 2 } \)[16+16] =\(\frac { 32 }{ 2 } \) = 16 sq units
Now, AB = \(\sqrt { { \left( 2+2 \right) }^{ 2 }+{ \left( 0-0 \right) }^{ 2 } } =\sqrt { { 4 }^{ 2 } } \) = 4 sq units
and PQ = \(\sqrt { { \left( 4+4 \right) }^{ 2 }+{ \left( 0-0 \right) }^{ 2 } } =\sqrt { { 8 }^{ 2 } } \) = 8 sq units
We have to verify that,  \(\frac { \Delta ABC }{ \Delta PQR } =\quad \frac { { \left( AB \right) }^{ 2 } }{ { \left( PQ \right) }^{ 2 } } \)
Now,  LHS = \(\frac { \Delta ABC }{ \Delta PQR } =\frac { 4 }{ 16 } =\frac { 1 }{ 4 } \)       ....(i)
and    RHS = \(\frac { { \left( AB \right) }^{ 2 } }{ { \left( PQ \right) }^{ 2 } } =\frac { { 4 }^{ 2 } }{ { 8 }^{ 2 } } =\frac { 16 }{ 64 } =\frac { 1 }{ 4 } \)                          ....(ii)
From Eqs. (i) and (ii) we, get   LHS = RHS           
Hence proved.

Let the coordinates of B and C be (a,b) and (x,y) respectively
Then \(\left( \frac { 1+a }{ 2 } ,\frac { -4+b }{ 2 } \right) \)  =(2.-1)

\(\Rightarrow \) \(\frac { 1+a }{ 2 } =2\quad and\quad \frac { -4+b }{ 2 } =-1\)
\(\Rightarrow \) 1+a=4 and -4+b=-2
\(\Rightarrow \) a=3 and b=2
Also  \(\left( \frac { 1+x }{ 2 } ,\frac { -4+y }{ 2 } \right) \)=(0,-1)
\(\Rightarrow \) \(\frac { 1+x }{ 2 } =0\quad and\quad \frac { -4+y }{ 2 } =-1\)
\(\Rightarrow \)  1+x=0 and -4+y=-2
\(\Rightarrow \)   x=-1 and y=2
Thus the coordinates of the vertices of ABC are A(1,-4),B(3,2) and C(-1,2)
Area of ABC =\(\frac { 1 }{ 2 } \) |1(2-2)+3(2+4)+(-1)(-4-2)|
= \(\frac { 1 }{ 2 } \)|0+18+6|
= \(\frac { 1 }{ 2 } \)(24)=12 sq.units 

P(x, y) is any point on the line segment joining the  points A(a, 0) and B(0, b) 
\(\Rightarrow \)  P,A and B are collinear
\(\therefore \)  Area of the triangle PAB formed by these  ints vanishes
 i.e \(\frac { 1 }{ 2 } \) {x(0-b)+a(b-y)+0(y-0)}=0
\(\Rightarrow \) -bx+ab-ay=0
\(\Rightarrow \) bx+ay=ab
Divide through by ab we have 
  \(\frac { bx }{ ab } +\frac { ay }{ ab } =\frac { ab }{ ab } \)
\(\Rightarrow \) \(\frac { x }{ a } +\frac { y }{ b } =1\)

Here, |AB|= \(\sqrt { (-2-2)^{ 2 }+(2-3)^{ 2 } } \)
= \(\sqrt { (-4)^{ 2 }+(-1)^{ 2 } } \)
=\(\sqrt { 16+1 } =\sqrt { 17 } \)  Units
|BC|=  \(\sqrt { (-1+2)^{ 2 }+(-2-2)^{ 2 } } \)
= \(\sqrt { (1)^{ 2 }+(-4)^{ 2 } } \)
=\(\sqrt { 1+16 } =\sqrt { 17 } \)  units
|CD|= \(\sqrt { (3+1)^{ 2 }+(-1+2)^{ 2 } } \)
=\(\sqrt { 4^{ 2 }+1^{ 2 } } \)
=   \(\sqrt { 17 } \)=units
|DA|=\(\sqrt { (2-3)^{ 2 }+(3+1)^{ 2 } } \)
=  \(\sqrt { (-1)^{ 2 }+4^{ 2 } } \)=\(\sqrt { 1+16 } \)
= \(\sqrt { 17 } \) units
\(\Rightarrow \) AB=BC=CD=DA= units
Now,Diagonal |AC| =\(\sqrt { (-1-2)^{ 2 }+(-2-3)^{ 2 } } \)
= \(\sqrt { (-3)^{ 2 }+(-5)^{ 2 } } \)
=\(\sqrt { 9+25 } =\sqrt { 34 } \)  units
Diagonal |BD|= \(\sqrt { (3+2)^{ 2 }+(-1-2)^{ 2 } } \)
= \(\sqrt { { 5 }^{ 2 }+(-3)^{ 2 } } \)
= \(\sqrt { 25+9 } =\sqrt { 34 } \) units
\(\Rightarrow \) Diagonal AC=Diagonal BD= units 
Hence ABCD is a square

The area of the triangle formed by the given points is equal to
\(\frac{1}{2}[-1.5(-2-4)+6(4-3)+(-3)(3+2)]\) 
\(=\frac{1}{2}(9+6-15)=0\) 
Can we have a triangle of area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.

k=-1

(10,3), (13,1)

Area ABC
=\(\frac { 1 }{ 2 } \) |t(t+2-t)+(t+2)[t-(t-2)]+(t+3)[(t-2)-(t+2)]
=\(\frac { 1 }{ 2 } \) |2t+2t+4-4t-12|
= \(\frac { 1 }{ 2 } \)|-8|=4 sq.units is independent of t

Area of =24 (Given)
 \(\frac { 1 }{ 2 } \)|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|=24
|1(2k+5)-4(-5=1)-k(-1-2k)|=48
\(\Rightarrow \) |2k+5+16+k+2k²|=48
\(\Rightarrow \)   |2k²+3k+21|=48
\(\Rightarrow \) 2k²+3k+21=\(\pm \)48
Consider, 2k²+3k+21=48
\(\Rightarrow \)2k²+3k-27=0
\(\Rightarrow \)2k²+9k-6k-27=0
\(\Rightarrow \)k(2k+9)-3(2k+9)=0
\(\Rightarrow \)(2k+9)(k-3)=0
Also 2k2+3k+21=-48
2k²+3k+69=0
 D= (3)² -4 X 2 X 69
=-ve (No solution)
 \(\therefore \)    k=\(\frac { -9 }{ 2 } \) or k=3

Area of quadrilateral ABCD
=Area of traingle ABC+Area of traingle ACD    ...(i)
Now, ar (\(\triangle \)ABC)
=\(\frac { 1 }{ 2 } \) [-3(-7+8)-2(-8-5)+1(5+7)]
=\(\frac { 1 }{ 2 } \) [-3+26+12] = \(\frac { 35 }{ 2 } \)sq.units ...(ii)
Also, ar ( \(\triangle \)ACD)
=\(\frac { 1 }{ 2 } \) [-3(-8-3)-1(3-5)+6(5+8)]
= \(\frac { 1 }{ 2 } \)[33-2+78]=\(\frac { 109 }{ 2 } \) sq.units    ...(iii)
From (i),(ii),(iii) we get 
ar (ABCD)=\(\frac { 35 }{ 2 } +\frac { 109 }{ 2 } =\frac { 144 }{ 2 } \) =72 sq.units

\(\because \) BD:CD =1:2
\(\therefore \) Coordinates of D are
\(\left( \frac { 1\times 4+2\times -1 }{ 1+2 } ,\frac { 1\times 2+2\times -2 }{ 1+2 } \right) \)   ie \(\left( \frac { 2 }{ 3 } ,\frac { -2 }{ 3 } \right) \)
AD= \(\sqrt { \left( \frac { 2 }{ 3 } -0 \right) ^{ 2 }+\left( \frac { -2 }{ 3 } -3 \right) ^{ 2 } } \)
= \(\sqrt { \frac { 4 }{ 9 } +\frac { 121 }{ 9 } } =\sqrt { \frac { 125 }{ 9 } } =\frac { 5\sqrt { 5 } }{ 3 } \)  units
DP=AD-AP = \(\frac { 5\sqrt { 5 } }{ 3 } -\frac { 2\sqrt { 5 } }{ 3 } \)
=\(\frac { 3\sqrt { 5 } }{ 3 } \)  =\(\sqrt { 5 } \) units
\(\therefore \)  \(\frac { AD }{ AP } =\frac { \frac { 2\sqrt { 5 } }{ 3 } }{ \sqrt { 5 } } =\frac { 2 }{ 3 } \)
P divides AD in the ratio 2 : 3.
\(\therefore \)  x-coordinate of P is
    x= \(\frac { 2\times \frac { 2 }{ 3 } +3\times 0 }{ 2+3 } =\frac { 4 }{ 15 } \)
Similarly ,y-coordinates of P is 
y=  \(\frac { 2\times \frac { -2 }{ 3 } +3\times 3 }{ 2+3 } =\frac { 23 }{ 15 } \)
\(\therefore \)  x-Coordinates of P are \(\left( \frac { 4 }{ 15 } ,\frac { 23 }{ 15 } \right) \)

Are of  \(\triangle \)PAB =10 sq units
\(\Rightarrow \) \(\frac { 1 }{ 2 } \) |x(2-8)+1(8-y)+3(y-2)| = 10
\(\Rightarrow \)  |-6x+8-y+3y-6| =20
\(\Rightarrow \) |-6x+2y+2| = 20 
\(\Rightarrow \) -6x+2y+2 = 20 or  -6x+2y+2 =- 20
\(\Rightarrow \) -6(17-3y)+2y+2 =20 or -6(17-3y)+2y+2=-20    ...(ii)
\(\Rightarrow \) -102+18y+2y+2=20  -102  -102+18y+2y+2=20  -20
 \(\Rightarrow \) 20y=120  20y=80
y=6 y=4
When y = 6, eq. (i) becomes x = 17-18 =-1 
\(\therefore \)  Point is (-1,6)
When y = 4, eq. (i) becomes x = 17-12 =5
\(\therefore \)  Point is (5,4)

find mid point O of A and C
x=(3-6)/2
= -3/2
y=(-4+2)/2 = -1
O=(-3/2,-1)
taking O as mid point find D
-3/2=(-1+x1)/2
x1=-2
-1=(-3+y1)/2
y1=1
D=-2,1