Mid point = \(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
Mid point of AB=\(\left( \frac { 8+10 }{ 2 } ,\frac { 6+10 }{ 2 } \right) \)
\(=\left( \frac { 18 }{ 2 } ,\frac { 6 }{ 2 } \right) =(9,3)\)
Mid point M(x,y) = Mid point of OD
\(=\left( \frac { 0+9 }{ 2 } ,\frac { 0+3 }{ 2 } \right) =\left( \frac { 9 }{ 2 } ,\frac { 3 }{ 2 } \right) =(4.5,1.5) \)

Mid point
\(M(x,y)=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
Mid point AB(2, 4)=\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =2\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=4\ \ \ ...(1)\)
\(\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =4\Rightarrow { y }_{ 1 }+{ y }_{ 2 }\ \ \ ...(2)\)
Mid point of BC (-2, 3) =\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) \)
\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } \right) =-2\Rightarrow { x }_{ 2 }+{ x }_{ 3 }=-4 \quad ...(3)\)
\(\left( \frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) =3\Rightarrow { y }_{ 2 }+y_{ 3 }=6\ \quad ...(4)\)
Mid point of AC (5, 2) =\(\left( \frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } \right) \)
\(\left( \frac { x_{ 1 }+{ x }_{ 3 } }{ 2 } \right) =3\Rightarrow x_{ 1 }+x_{ 3 }=10 \quad ...(5)\)
\(\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } =3\Rightarrow { y }_{ 1 }+y_{ 3 }=4 \quad ...(6)\)


Substitue x₁ = 9 in (5)
         9 +X₃  = 10
              X₃  = 1
Substitue X₁ = 1 in(3)
x₂ + 1 = -4 ⇒ x₂ = -5

substitute  y₁ = 3 in (6)
3 + y₃ = 4
y₃ = 1
substitute y₃ = 1 in (4)
1 + y₂ = 6
y₃ = 5
ஃ The vertices of the triangle A(x₁ y₁) = (9, 3)
B (x₂ y₂) = (-5, 5)
 C(x₃y₃) = (1, 1)

AB =\(\sqrt{(3-3)^2+(8-4)^2}=\sqrt{0^2+4^2}=\sqrt{16}=4 \ units\)

Take a graph sheet and plot all the points and join the points to form a rectangle.
Take a mirror and see the image of the rectangle formed.

The shape of the geometrical figure Trapezium.

Straight line parellel to x-axis

(i) P is (-4, 4) [-4 is abscissa and 4 is ordinate]
(ii) Q is (3, 3) [3 is abscissa and 3 is ordinate]
(iii) R is (4, -2) [4 is abscissa and -2 is ordinate]
(iv) S is (-5, -3) [-5 is abscissa and -3 is ordinate]

(i) P (-7, 6) lies in the II quadrant because the x-coordinate is negative and y-coordinate is positive
(ii) Q (7, -2) lies in the IV quadrant because the x-coordinate is positive and y-coordinate is negative
(iii) R (-6, -7) lies in the III quadrant because the x-coordinate is negative and y-coordinate is negative
(iv) S (3, 5) lies in the I quadrant because the x-coordinate is positive and y-coordinate is also positive
(v) T (3, 9) lies in the I quadrant because the x-coordinate is positive and y-coordinate is also positive

The x-coordinate is negative and y – coordinate is positive. So, Point(–7,3) lies in the II quadrant

The x- coordinate is positive and y – coordinate is negative. So, Point(3,–8) lies in the IV quadrant.