Question

Three consecutive vertices of a parallelogram are points (2, 4), (0, 0), and (6, 0). The fourth vertex is point

Answer 1

Task. Three consecutive vertices of a parallelogram are points $A=(2,4)$, $B=(0,0)$, and $C=(6,0)$. Find the fourth vertex $D$.

Solution. Recall that the intersection point $M$ of the diagonals $AC$ and $BD$ of the parallelogram is the middle point of these intervals.

Thus if $A=(x_{1},y_{1})$ and $C=(x_{2},y_{2})$, then the coordinates of the middle point $M=(\bar{x},\bar{y})$ of $AC$ can be computed by the formula:

$\bar{x}=\frac{x_{1}+x_{2}}{2},\qquad\bar{y}=\frac{y_{1}+y_{2}}{2}.$

In our case $A=(2,4)$ and $C=(6,0)$, whence

$\bar{x}=\frac{2+6}{2}=4,\qquad\bar{y}=\frac{4+0}{2}=2,$

so

$M=(4,2).$

Let $D=(x,y)$. Since $M=(4,2)$ is the middle point of $BD$ and $B=(0,0)$ we have that

$4=\frac{0+x}{2},\qquad 2=\frac{0+y}{2},$

whence

$x=2*4=8,\qquad y=2*2=4.$

Thus $D=(8,4)$.

Answer. $D=(8,4)$.