Given:
The three vertices of parallelogram are (0,6),(5,-4),(-1,-5).
To find:
The fourth vertex of the parallelogram.
Solution:
Consider the given vertices of parallelogram are A(0,6), B(5,-4), C(-1,-5).
Let the fourth vertex be D(a,b).
Midpoint formula:
[tex]Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]
We know that the diagonal of parallelogram bisect each other. It means their midpoints are same.
Midpoint of AC = Midpoint of BD
[tex]\left(\dfrac{0+(-1)}{2},\dfrac{6+(-5)}{2}\right)=\left(\dfrac{5+a}{2},\dfrac{-4+b}{2}\right)[/tex]
[tex]\left(\dfrac{-1}{2},\dfrac{1}{2}\right)=\left(\dfrac{5+a}{2},\dfrac{-4+b}{2}\right)[/tex]
On comparing both sides, we get
[tex]\dfrac{5+a}{2}=-\dfrac{1}{2}[/tex]
[tex]5+a=-1[/tex]
[tex]a=-1-5[/tex]
[tex]a=-6[/tex]
And,
[tex]\dfrac{-4+b}{2}=\dfrac{1}{2}[/tex]
[tex]-4+b=1[/tex]
[tex]b=1+4[/tex]
[tex]b=5[/tex]
Therefore, the coordinates of fourth vertex are (-6,5).
