k=-1

p=8

Let (x, y) be coordinates of D.  Since we know that diagonals of a  parallogram bisect each other. 
So,mid-point of AC = \(\left( \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right) \)
= \(\left( \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right) \)
Mid point of BD = \(\left( \frac { x+8 }{ 2 } ,\frac { y+2 }{ 2 } \right) \)
Mid-point of AC and BD are 
\(\left( \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right) \) and \(\left( \frac { x+8 }{ 2 } ,\frac { y+2 }{ 2 } \right) \)
By comparison \(\frac { x+8 }{ 2 } =\frac { 15 }{ 2 } ,\frac { y+2 }{ 2 } =\frac { 5 }{ 2 } \)
 x=7,y=3
Coordinates of D are (7, 3).  and coordinates of E are 
\(\left( \frac { 7+9 }{ 2 } +\frac { 3+4 }{ 2 } \right) =\left( 8,\frac { 7 }{ 2 } \right) \)
Area of \(\triangle \) ADE 
=\(\frac { 1 }{ 2 } \) |[x₁(y₂-y₃)+x₂(y₃-y₃)+x₃(y₁-y₂)]|
= \(\frac { 1 }{ 2 } \left| \left[ 6\left( 3-\frac { 7 }{ 2 } \right) +\left( \frac { 7 }{ 2 } -1 \right) +8(1-3) \right] \right| \)
= \(\frac { 1 }{ 2 } \left| 6\left( \frac { -1 }{ 2 } \right) +7\left( \frac { 5 }{ 2 } \right) +8(\times -2) \right| \)
= \(\frac { 1 }{ 2 } \left| -3+\frac { 35 }{ 2 } -16 \right| =\frac { 1 }{ 2 } \left| -19+\frac { 35 }{ 2 } \right| \)
=  \(\frac { 1 }{ 2 } \left| -\frac { 3 }{ 2 } \right| =\frac { 3 }{ 4 } \)

\(\because \) 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 \(\left( \frac { 4 }{ 15 } ,\frac { 23 }{ 15 } \right) \)

For collinear points, Area of \(\Delta ABC=0\)
\({1\over 2}[8(-4+5)+k(-5-1)+2X(1+4)]=0\)
8-6k+10=0
6k=18
k=3

For collinear points, Area of \(\Delta ABC={1\over 2}[x_1(y_1-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\)
\({1\over2}[7(1-k)+5(k+2)+3(-2-1)]=0\)
7-7k+5k+10-9=0
2k-8=0
2k=8
k=4

Let points be A(0,0) and B(36,15)
The distance between two points is
\(AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
=\(\sqrt{(36-0)^2+(15-0)^2}\)
=\(\sqrt{1296+225}=\sqrt{1521}=39\)

1: 3

Let AB be diameter and C be the centre of the circle. Let coordinates of A be (a, b) and C be the mid-point of AB.

\(\therefore C=\left( \frac { a+1 }{ 2 } ,\frac { b+4 }{ 2 } \right) \) \(\left[ \because \ mid-point=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \right] \quad \)
But      C=(3, -4)
On comparing the coordinates of x and y, we get
\(\frac { a+1 }{ 2 } =3\) and \(\frac { b+4 }{ 2 } =-4\)
\(\Rightarrow \)  a+1 = 6 and b+4 = -8
\(\Rightarrow \)  a = 5 and b = -12        
Hence, the coordinates of point A are (5, -12).

\(y=\frac { 2x-12 }{ 3 } \); (6, 0), (0, -4)

(i) (-5, 2)   (ii) (5, -5)  (iii) E  (iv) G  (v) 6  (vi) -3  (vii) (0, 5)  (viii) (-3, 0)

parallelogram

pentagon

(i) trapezium
(ii) rhombus

\(\left( -\sqrt { 3 } ,\ \sqrt { 3 } \right) \ or\ \left( \sqrt { 3 } ,\ -\sqrt { 3 } \right) \)

Yes