If two vertices of a parallalogram …

Let coordinates of C be (x₁, y₁) and D be (.x₂, y₂).
So, {tex}\frac { x _ { 1 } + 3 } { 2 }{/tex} = 2 .....(i) and {tex}\frac { y _ { 1 } + 2 } { 2 }{/tex} = -5 .....(ii) [Mid-point theorem]
Also, {tex}\frac { x _ { 2 } - 1 } { 2 }{/tex} = 2 ......(iii) and {tex}\frac { y _ { 2 } + 0 } { 2 }{/tex} = -5 ......(iv) [Mid-point theorem]
From equation (i), we get
{tex}\frac { x _ { 1 } + 3 } { 2 }{/tex} = 2
{tex}\Rightarrow{/tex} x₁ + 3 = 4
{tex}\Rightarrow{/tex} x₁ = 4 - 3 = 1
Solving equation (ii), we get
y₁ + 2 = - 10
{tex}\Rightarrow{/tex} y₁ = - 10 - 2
{tex}\Rightarrow{/tex} y₁ = - 12
Solving equation (iii), we get
x₂ - 1 = 4
x₂ = 4 + 1 = 5
Solving equation (iv), we get
y₂ + 0 = - 10
{tex}\Rightarrow{/tex} y2 = - 10
{tex}\therefore{/tex} Coordinates of C are (1, -12) and D are (5, -10).