Question

Question 11
The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2,3), B(6,7) and C(8,3) is
(A) (0, 1)
(B) (0, –1)
(C) (–1, 0)
(D) (1, 0)

Answer 1

Let the fourth vertex of the given parallelogram be $D=(x_4,y_4)$ and, L and M be the mid-points of AC and BD, respectively.
Then,
$L=(\frac{-2+8}{2},\frac{3+3}{2})=(3,3)$
[ Since, mid-point of any line segment which passes through the points
$(x_1,y_1)and(x_2,y_2)$ is $\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}$]
Similarly, M = $(\frac{6+x_4}{2},\frac{7+y_4}{2})$

Since, ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other.
Hence, L and M are the same points.So, equating both coordinates equal to each other
$\therefore$ $3=\frac{6+x_4}{2}$ and $3=\frac{7+y_4}{2}$
$\Rightarrow$ $6=6+x_4$ and $6=7+y_4$
$\Rightarrow$ $x_4=0$ and $y_4=6-7$
$\therefore$ $x_4=0$ and $y_4=-1$
Hence, fourth vertex of the parallelogram is
D = $(x_4,y_4)=(0,-1)$