Answer:
The fourth vertex is D(8, 0)
Step-by-step explanation:
Let A(0, 0), B(2, 4), and C(10, 4) be the three vertices of a parallelogram ABCD and let its fourth vertex be D(a, b).
Join AC and BD. Let AC and BD intersect at point O.
- We have known that the diagonals of a parallelogram bisect each other.
So, O would be the midpoint of AC as well as that of BD.
[tex]A\left(0,\:0\right)=\left(x_1,\:x_2\:\right)[/tex]
[tex]C\left(10,\:4\right)=\left(x_2,\:x_2\:\right)[/tex]
The midpoint of AC is:
[tex]\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)[/tex]
[tex]\left(x_1,\:y_1\right)=\left(0,\:0\right),\:\left(x_2,\:y_2\right)=\left(10,\:4\right)[/tex]
[tex]=\left(\frac{10+0}{2},\:\frac{4+0}{2}\right)[/tex]
[tex]=\left(5,\:2\right)[/tex]
The midpoint of BD is:
[tex]=\left(\frac{a+2}{2},\:\frac{b+4}{2}\right)[/tex]
so
∵ [tex]\frac{a+2}{2}=5[/tex]
[tex]a+2=10[/tex]
[tex]a=8[/tex]
∵ [tex]\frac{b+4}{2}=2[/tex]
[tex]b+4=4[/tex]
[tex]b=0[/tex]
Hence, the fourth vertex is D(8, 0)
