Question

Three vertices of a parallelogram ABCD taken in order are A(3, 6), B (5, 10) and C(3, 2). Then the coordinates of the fourth vertex D is

• A. (1, -2)
• B. (-1, 2)
• C. (2, -1)
• D. (-2, 1)

Answer 1

The correct option is A

(1, -2)

Let the coordinates of the vertex D be (x, y)
Mid-point of AC = Mid -point of BD
[$\because$ Diagonals of a parallelogram bisect each other]

We know that mid-point of the line joining two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$\Rightarrow (\frac{3+3}{2},\frac{6+2}{2})=(\frac{x+5}{2},\frac{y+10}{2})\Rightarrow (\frac{6}{2},\frac{8}{2})=(\frac{x+5}{2},\frac{y+10}{2})\Rightarrow (3,4)=(\frac{x+5}{2},\frac{y+10}{2})\therefore \frac{x+5}{2}=3and\frac{y+10}{2}=4\Rightarrow x+5=6andy+10=8\Rightarrow x=1andy=-2$
$\therefore$ Coordinates of the fourth vertex D are (1, -2)