The statements which are true are:
(6, 6) is the midpoint of CD
(4, 3) is the intersection point of diagonals of parallelogram
Solution:
The mid point (x,y) = [tex]( \frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]
A(-2, -1) and B(6, 1)
[tex]\text{ midpoint of AB } = (\frac{-2+6}{2} , \frac{-1+1}{2})\\\\\text{ midpoint of AB } = (2, 0)[/tex]
Thus statement 1 is wrong
B(6, 1) and C(10, 7)
[tex]\text{ midpoint of BC } = (\frac{6+10}{2} , \frac{1+7}{2})\\\\\text{ midpoint of BC } = (8, 4)[/tex]
Thus statement 2 is wrong
Here ,
[tex]x_1[/tex] = 10
[tex]x_2[/tex]= 2
[tex]y_1[/tex]= 7
[tex]y_2[/tex]=5
now substituting these values,
mid point of CD = [tex](\frac{10+2}{2},\frac{7+5}{2})[/tex]
mid point of CD = [tex](\frac{12}{2},\frac{12}{2})[/tex]
mid point of CD = [tex](6, 6)[/tex]
Therefore (6, 6) is the midpoint of CD
Statement 3 is correct
A = (-2, -1) and D = (2, 5)
[tex]\text{ mid point of AD } = (\frac{-2+2}{2} , \frac{-1+5}{2})\\\\\text{ mid point of AD } = (0, 2)[/tex]
Thus statement 4 is wrong
Let AC and BD be the diagonals of parallelogram
The diagonals of a parallelogram bisect each other, therefore, the point of intersection is the midpoint of either.
Midpoint of AC:
A = (-2, -1) and C(10, 7)
[tex]\text{ Midpoint of AC } = (\frac{-2+10}{2} , \frac{-1+7}{2})\\\\\text{ Midpoint of AC } = (4,3)[/tex]
Thus statement 5 is correct
