Question

If $A( - 2, - 1)$, $B(a,0)$ ,$C(4,b)$ and $D(1,2)$ are the vertices of a parallelogram, then the values of $a$ and $b$ are
A. 2,1
B. 1,3
C. -1,-3
D. 2,-3

Answer 1

Hint: In this question, we have to find values of vertices $a$ and $b$ of the parallelogram.
Diagonal is the line segment joining two vertices that are not on the same edges.
The midpoint of a line divides the line segment into two equal parts.
When the diagonals of the parallelogram bisect each other their intersection point is the midpoint of each diagonal.
Find the midpoint of the diagonals whose vertices are given and equate to each other.
Midpoint formula of the line:
$({x_m},{y_m}) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
$({x_m},{y_m})$=midpoint coordinates
$({x_1},{y_1})$=coordinate of first vertex
$({x_2},{y_2})$=coordinate of second vertex

Complete step-by-step answer:
Consider $A( - 2, - 1)$, $B(a,0)$ ,$C(4,b)$ and $D(1,2)$ are the vertices of a parallelogram.
According to the properties of the parallelogram, the diagonal of the parallelogram bisect each other.
$\therefore$ The Midpoint of the diagonals are the same.

Consider the diagonals are $AC$ and $BD$, and the midpoint is $o$.
Determine the midpoint of the diagonal $AC$, given is $A( - 2, - 1)$, and $C(4,b)$ ,
$ \Rightarrow ({x_1},{y_1}) = ( - 2, - 1)$ and $({x_2},{y_2}) = (4,b)$
Midpoints $({x_m},{y_m}) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Substitute $({x_1},{y_1}) = ( - 2, - 1)$ and, $({x_2},{y_2}) = (4,b)$ into $({x_m},{y_m}) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$,
$({x_m},{y_m}) = \left( {\dfrac{{ - 2 + 4}}{2},\dfrac{{ - 1 + b}}{2}} \right)$
$({x_m},{y_m}) = \left( {1,\dfrac{{ - 1 + b}}{2}} \right) \ldots (1)$
Determine the midpoint of the diagonal $BD$, given is $B(a,0)$, and $D(1,2)$
$ \Rightarrow ({x_1},{y_1}) = (a,0)$ and, $({x_2},{y_2}) = (1,2)$
Midpoints $({x_m},{y_m}) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Substitute $({x_1},{y_1}) = (a,0)$ and, $({x_2},{y_2}) = (1,2)$into $({x_m},{y_m}) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$,
$({x_m},{y_m}) = \left( {\dfrac{{a + 1}}{2},\dfrac{{0 + 2}}{2}} \right)$
$({x_m},{y_m}) = \left( {\dfrac{{a + 1}}{2},1} \right) \ldots (2)$
Equate equations $(1)$ and $(2)$,
$\left( {\dfrac{{a + 1}}{2},1} \right) = \left( {1,\dfrac{{ - 1 + b}}{2}} \right)$
$ \Rightarrow \dfrac{{a + 1}}{2} = 1$ and $\dfrac{{ - 1 + b}}{2} = 1$
Solve $\dfrac{{a + 1}}{2} = 1$ by cross multiplication,
$a + 1 = 2$
$ \Rightarrow a = 1$
Solve $\dfrac{{ - 1 + b}}{2} = 1$ by cross multiplication,
$ - 1 + b = 2$
$ \Rightarrow b = 3$

Correct Answer: B. 1,3

Note:
The important step is that When the diagonals of the parallelogram bisect each other their intersection point is the midpoint of each diagonal.
The midpoint of two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is the point \[M\] found by the following formula,
$M = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
You basically are averaging the \[X\] and \[Y\] values.