Respuesta :
Solution:
Vertices of parallelogram ABCD is given as A(1,2) , B(0,9) , C(7,8) , and D(8,1) .
We will use the following analytical geometry formulas here
1. Distance between two points
[tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
2. Slope of a line
[tex]=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
3. When two lines are perpendicular, product of their slopes is equal to -1.
Since ABCD is a parallelogram
1. Opposite sides are equal and parallel.
2. Diagonals bisect each other.
3. Opposite angles are equal.
Now, coming to problem
[tex]AB=\sqrt{1^2+7^2}=\sqrt{50}, BC=\sqrt{7^2+1^2}=\sqrt{50}, CD=\sqrt{1^2+7^2}=\sqrt{50}, DA=\sqrt{7^2+1^2}=\sqrt{50}\\\\ {\text{Slope of AB}=\frac{7}{-1}=-7 \\\\{\text{Slope of CB}=\frac{1}{-7},\\\\ {\text{Slope of CD}=\frac{-7}{1}=-7\\\\ {\text{Slope of AD}=\frac{-1}{7}[/tex]
As, you can see that, AB=BC=CD=DA=√ 50
But , slope of AB × Slope of BC =slope of CB × Slope of DC=slope of CD × Slope of DA [tex]=[-7 \times \frac{-1}{7}]=1[/tex]
which is not equal to -1. It means lines which are sides of parallelogram are not perpendicular.
As, all side of parallelogram ABCD are equal, so it is a rhombus.
As, diagonal of rhombus bisect each other at right angles.
[tex]{\text{slope of AC}} \times {\text{Slope of BD}}=\frac{6}{6}\times\frac{-8}{8}=-1[/tex]
Shows that diagonals are perpendicular bisector of each other.
Option (1) : AC⊥BD; therefore, ABCD is a rhombus.
Answer:
1. AC ⊥BD ; therefore, ABCD is a rhombus.
Step-by-step explanation:
We are given,
A parallelogram ABCD with vertices A(1,2), B(0,9), C(7,8) and D(8,1).
Using Distance Formula, which gives the distance between two points ([tex](x_{1},y_{1})[/tex] and ([tex](x_{2},y_{2})[/tex] as [tex]\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}[/tex]
So, the distance between two points are given by,
AB= [tex]\sqrt{(9-2)^{2}+(0-1)^{2}}=\sqrt{49+1}=\sqrt{50}[/tex] = 7.07
BC= [tex]\sqrt{(8-9)^{2}+(7-0)^{2}}=\sqrt{1+49}=\sqrt{50}[/tex] = 7.07
CD= [tex]\sqrt{(8-1)^{2}+(7-8)^{2}}=\sqrt{49+1}=\sqrt{50}[/tex] = 7.07
DA= [tex]\sqrt{(1-2)^{2}+(8-1)^{2}}=\sqrt{1+49}=\sqrt{50}[/tex] = 7.07
Also,
AC= [tex]\sqrt{(8-2)^{2}+(7-1)^{2}}=\sqrt{36+36}=\sqrt{72}[/tex] = 8.49
BD= [tex]\sqrt{(9-1)^{2}+(0-8)^{2}}=\sqrt{64+64}=\sqrt{138}[/tex] = 11.3
Since, we have,
AB = BC = CD = DA.
So, ABCD is a rhombus but not a rectangle.
As, AC ≠ BD. Moreover, two diagonals are perpendicular to each other in a rhombus. Thus, AC ⊥ BD.
Hence, we get that, Option A is correct.
