Respuesta :
Answer:
The Conclusion is
Diagonals AC and BD,
a. Bisect each other
b. Not Congruent
c. Not Perpendicular
Step-by-step explanation:
Given:
[]ABCD is Quadrilateral having Vertices as
A(-1, 1),
B(2, 3),
C(6, 0) and
D(3, -2).
So the Diagonal are AC and BD
To Check
The diagonals AC and BD
a. Bisect each other. B. Are congruent. C. Are perpendicular.
Solution:
For a. Bisect each other
We will use Mid Point Formula,
If The mid point of diagonals AC and BD are Same Then
Diagonal, Bisect each other,
For mid point of AC
[tex]Mid\ point(AC)=(\dfrac{x_{1}+x_{2} }{2},\dfrac{y_{1}+y_{2} }{2})[/tex]
Substituting the coordinates of A and C we get
[tex]Mid\ point(AC)=(\dfrac{-1+6}{2},\dfrac{1+0}{2})=(\dfrac{5}{2},\dfrac{1}{2})[/tex]
Similarly, For mid point of BD
Substituting the coordinates of B and D we get
[tex]Mid\ point(BD)=(\dfrac{2+3}{2},\dfrac{3-2}{2})=(\dfrac{5}{2},\dfrac{1}{2})[/tex]
Therefore The Mid point of diagonals AC and BD are Same
Hence Diagonals,
a. Bisect each other
B. Are congruent
For Diagonals to be Congruent We use Distance Formula
For Diagonal AC
[tex]l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}[/tex]
Substituting A and C we get
[tex]l(AC) = \sqrt{((6-(-1))^{2}+(0-1)^{2} )}=\sqrt{(49+1)}=\sqrt{50}[/tex]
Similarly ,For Diagonal BD
Substituting Band D we get
[tex]l(BD) = \sqrt{((3-2))^{2}+(-2-3)^{2} )}=\sqrt{(1+25)}=\sqrt{26}[/tex]
Therefore Diagonals Not Congruent
For C. Are perpendicular.
For Diagonals to be perpendicular we need to have the Product of slopes must be - 1
For Slope we have
[tex]Slope(AC)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]
Substituting A and C we get
[tex]Slope(AC)=\dfrac{0-1}{6--1}\\\\Slope(AC)=\dfrac{-1}{7}[/tex]
Similarly, for BD we have
[tex]Slope(BD)=\dfrac{-2-3}{3-2}\\\\Slope(BD)=\dfrac{-5}{1}[/tex]
The Product of slope is not -1
Hence Diagonals are Not Perpendicular.
