Formula: 1 Mark
Application: 1 Mark
Answers: 1 Mark 
We know that the diagonals of a parallelogram bisect each other.
$\therefore$ The coordinates of the mid-point of AC = The coordinates of the mid-point of BD i.e.
If $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ and $D(x_4,y_4)$ are the vertices of parallelogram ABCD, Then
$\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}=\frac{x_2+x_4}{2},\frac{y_2+y_4}{2}$
$\Rightarrow (\frac{-2+4}{2},\frac{-1+b}{2})=(\frac{a+1}{2},\frac{0+2}{2})$
$\Rightarrow (1,\frac{b-1}{2})=(\frac{a+1}{2},1)$
$\Rightarrow \frac{a+1}{2}=1and\frac{b-1}{2}=1$

$\Rightarrow a+1=2andb-1=2$

$\Rightarrow a=1andb=3$

Formula: 1 Mark
Application: 1 Mark
Answer: 1 Mark
Let the given points be A(4, – 1), B(6, 0), C(7, 2) and D(5, 1) respectively. Then,
Coordinates of the mid-point of AC are
$(\frac{4+7}{2},\frac{-1+2}{2})=(\frac{11}{2},\frac{1}{2})$ 
Coordinates of the mid-point of BD are
$(\frac{6+5}{2},\frac{0+1}{2})=(\frac{11}{2},\frac{1}{2})$ 
Thus, AC and BD have the same mid-point.

Hence, ABCD is a parallelogram.

Now
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$ 
So, 
AB = $\sqrt{(6-4{)}^2+(0+1{)}^2}=\sqrt{5}$ 
BC = $\sqrt{(7-6{)}^2+(2-0{)}^2}=\sqrt{5}$ 
$\therefore$ AB = BC 
So, ABCD is a parallelogram whose adjacent sides are equal.
$\Rightarrow$  ABCD is a rhombus.
We have, 

AC = $\sqrt{(7-4{)}^2+(2+1{)}^2}=3\sqrt{2}$   and , 

BD = $\sqrt{(6-5{)}^2+(0-1{)}^2}=\sqrt{2}$ 

 
Clearly, the diagonals AC ≠ BD.
So, ABCD is not a square. 
 

Formula: 1 Mark
Concept: 1 Mark
Answer: 2 Marks
Let D, E, F be the mid-points of the sides BC, CA and AB respectively.
Then, the coordinates of D, E and F are
$D(\frac{5+3}{2},\frac{3-1}{2})=D(4,1),$
$E(\frac{3+7}{2},\frac{-1-3}{2})=E(5,-2)$
and $F(\frac{7+5}{2},\frac{-3+3}{2})=F(6,0)$

Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\therefore AD=\sqrt{(7-4{)}^2+(-3-1{)}^2}$
$\Rightarrow AD=\sqrt{9+16}=5units$
$BE=\sqrt{(5-5{)}^2+(-2-3{)}^2}$
$\Rightarrow BE=\sqrt{0+25}=5units$
and, $CF=\sqrt{(6-3{)}^2+(0+1{)}^2}$
$\Rightarrow CF=\sqrt{9+1}=\sqrt{10}units$

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks
Consider a  parallelogram ABCD .
Let O be the point of intersection of the diagonals AC and BD

We know that diagonals of a parallelogram bisect each other.
$\Rightarrow$ O is the midpoint of both the digonals AC and BD
Now, coordinates of O as mid-point of BD are
$O(x,y)=\left(\frac{x_1+x_2}{2}\right),\left(\frac{y_1+y_2}{2}\right)$
$\Rightarrow O(a,b)=\frac{4+3}{2},\frac{y+5}{2}\dots \dots \left(1\right)$ 
Also, coordinates of O as mid-point of AC are
$O(a,b)=\frac{1+x}{2},\frac{2+6}{2}\dots \dots \left(2\right)$
From (1) and (2), we have
$\frac{4+3}{2}=\frac{1+x}{2}$
$\frac{7}{2}=\frac{1+x}{2}\Rightarrow x=6$
And $\frac{y+5}{2}=\frac{2+6}{2}$
$\frac{y+5}{2}=\frac{8}{2}\Rightarrow y=3$

Formula for centroid and mid-point: 1 Mark each
Let A$(x_1,y_1)$,B$(x_2,y_2)$ and C$(x_3,y_3)$ be the vertices of triangle ABC.
Let P(1,1), Q(2,-3), R(3,4) be the mid-points of sides AB, BC and CA respectively.
Then, P is the mid-point of BC
$\Rightarrow \frac{x_1+x_2}{2}=1,\frac{y_1+y_2}{2}=1$
$\Rightarrow x_1+x_2=2$ and $y_1+y_2=2\dots \dots \left(1\right)$
Q is the mid-point of AB
$\Rightarrow \frac{x_2+x_3}{2}=2,\frac{y_2+y_3}{2}=-3$
$\Rightarrow x_2+x_3=4$ and $y_2+y_3=-6\dots \dots \left(2\right)$
R is the mid-point of AC
$\Rightarrow \frac{x_1+x_3}{2}=3,\frac{y_1+y_3}{2}=4$
$\Rightarrow x_1+x_3=6$ and $y_1+y_3=8\dots \dots \left(3\right)$
Adding (1),(2) and (3) we get,
$x_1+x_2+x_2+x_3+x_1+x_3=2+4+6$
and $y_1+y_2+y_2+y_3+y_1+y_3=2-6+8$
$\Rightarrow x_1+x_2+x_3=6\dots \dots \left(4\right)$ 
and $y_1+y_2+y_3=2\dots \dots \left(5\right)$
The coordinates of the centroid of $△ABC$ are
$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=\frac{6}{3},\frac{2}{3}=2,\frac{2}{3}$ [Using (4) and (5)]

Formula: 1 Mark
Calculations: 2 Marks
Answer: 1 Mark
For points A,B,C to be collinear, area of triangle ABC formed by these points should be zero.
$\Rightarrow$ The area of  $\Delta ABC=0$
 $\Rightarrow \frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0$
$\Rightarrow \frac{1}{2}\left[k\right(2k-6+2k)+(–k+1\left)\right(6–2k-2+2k)+(–4–k\left)\right(2–2k-2k\left)\right]=0$
$\Rightarrow \frac{1}{2}\left[k\right(4k-6)+(–k+1\left)\right(4)+(–4–k\left)\right(2–4k\left)\right]=0$
$\Rightarrow \frac{1}{2}[4k^2-6k-4k+4-8+16k-2k+4k^2]=0$
$\Rightarrow 8k^2+4k-4=0$
$\Rightarrow 8k^2+4k-4=0$
$\Rightarrow 2k^2+k-1=0$
$\Rightarrow (k+1)(2k-1)=0$
$\Rightarrow k=-1ork=\frac{1}{2}$

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks
Let ABCD be a square and two opposite vertices of it are A(-1, 2) and C(3, 2). ABCD is a square.

Since ABCD is a square.
$\Rightarrow AB=BC$
$\Rightarrow A{B}^2=B{C}^2$
[Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$]
$\Rightarrow {(x+1)}^2+{(y-2)}^2={(x-3)}^2+{(y-2)}^2$
$\Rightarrow x^2+2x+1=x^2-6x+9$
$\Rightarrow 2x+6x=9-1=8$
$\Rightarrow 8x=8\Rightarrow x=1$
ABC is right $\Delta$ at B, then
$A{C}^2=A{B}^2+B{C}^2$ (Pythagoras theorem)
$\Rightarrow (3+1{)}^2+(2-2{)}^2=(x+1{)}^2+(y-2{)}^2+(x-3{)}^2+(y-2{)}^2$
$\Rightarrow 16=2(y-2{)}^2+(1+1{)}^2+(1-3{)}^2$
$\Rightarrow 16=2(y-2{)}^2+4+4\Rightarrow 2(y-2{)}^2=16-8=8$
$\Rightarrow (y-2{)}^2=4\Rightarrow y-2=\pm 2\Rightarrow y=4$ and 0
i.e when x = 1 then y = 4 and 0
Co-ordinates of the opposite vertices are: B(1,0) or D(1,4)

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks
Area of quadrilateral $ABCD=|Areaof\Delta ABC|+|Areaof\Delta ACD|$

We have,

$\therefore$ Area of $\Delta ABC=\frac{1}{2}|(1\times -3+7\times 2+12\times 1)-(7\times 1+12\times (-3)+1\times 2)|$
$\Rightarrow \Delta ABC=\frac{1}{2}|(-3+14+12)-(7-36+2)|$
$\Rightarrow \Delta ABC=\frac{1}{2}|23+27|=25sq.units$

Also, we have

$\therefore$ Area of $\Delta ACD=\frac{1}{2}|(1\times 2+12\times 21+7\times 1)-(12\times 1+7\times 2+1\times 21)|$
 $\Rightarrow \Delta ACD=\frac{1}{2}|(2+252+7)-(12+14+21)|$
$\Rightarrow \Delta ACD=\frac{1}{2}|261-47|=107sq.units$
$\therefore$ Area of quadrilateral ABCD = 25 + 107 = 132 sq. units