If the three consecutive vertices of a parallelogram be
Question:

If the three consecutive vertices of a parallelogram be A(3, 4, -3), B(7, 10, -3) and C(5, -2, 7), find the fourth vertex D.

Solution:

the vertices of the parallelogram be A(3, 4, -3), B(7, 10, -3) and C(5, -2, 7), and the fourth coordinate be D(a,b,c).

the property of parallelogram is the diagonal bisect each other. Therefore,

diagonal $\mathrm{AC}$ and $\mathrm{BD}$ will bisect each other, and the bisecting point will be equal to the two diagnals. By using section formula, we get

$\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}, \frac{m z_{2}+n z_{1}}{m+n}\right)$ where m and n are 1 and 1.

Finding the coordinate of mid point of diagonal AC,

$=\left(\frac{1 \times 5+1 \times 3}{1+1}, \frac{1 \times-2+1 \times 4}{1+1}, \frac{1 \times 7+1 \times-3}{1+1}\right)$

$=(4,1,2)$

Now, Finding the coordinate of mid point of diagonal BD,

$=\left(\frac{1 \times a+1 \times 7}{1+1}, \frac{1 \times b+1 \times 10}{1+1}, \frac{1 \times c+1 \times-3}{1+1}\right)$

$=\left(\frac{a+7}{2}, \frac{b+10}{2}, \frac{c-3}{2}\right)$

Equating the two mid points, we get

$(4,1,2)=\left(\frac{\mathrm{a}+7}{2}, \frac{\mathrm{b}+10}{2}, \frac{\mathrm{c}-3}{2}\right)$. Thus,

$4=\frac{a+7}{2}$

$1=\frac{b+10}{2}$

$2=\frac{c-3}{2}$

Therefore,

$8=a+7$

$a=1$

$b+10=2$

$b=-8$

and

$c=7$

therefore the point is $(1,-8,7)$.

 

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