Mathematics, 16.10.2019 10:30 eternelle1182

Parallelogram abcd has vertices at a(1,2) , b(0,9) , c(7,8) , and d(8,1) . which conclusion can be made?

1. ac⊥bd ; therefore, abcd is a rhombus.

2. ac=bd ; therefore, abcd is a rhombus.

3. ac⊥bd ; therefore, abcd is a rectangle.

4. ac=bd ; therefore, abcd is a rectangle.

Answers: 2

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Answer from: Jadaflournoy5

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

$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

2. Slope of a line

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

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

$AB=\sqrt{1^2+7^2}=\sqrt{50}, BC=\sqrt{7^2+1^2}=\sqrt{50}, CD=\sqrt{1^2+7^2}=\sqrt{50}, DA=\sqrt{7^2+1^2}=\sqrt{50}\\\\ {\text{Slope of AB}=\frac{7}{-1}=-7 \\\\{\text{Slope of CB}=\frac{1}{-7},\\\\ {\text{Slope of CD}=\frac{-7}{1}=-7\\\\ {\text{Slope of AD}=\frac{-1}{7}$

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 $=[-7 \times \frac{-1}{7}]=1$

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.

${\text{slope of AC}} \times {\text{Slope of BD}}=\frac{6}{6}\times\frac{-8}{8}=-1$

Shows that diagonals are perpendicular bisector of each other.

Option (1) : AC⊥BD; therefore, ABCD is a rhombus.

Parallelogram abcd has vertices at a(1,2) , b(0,9) , c(7,8) , and d(8,1) . which conclusion can be m

Answer from: belladaviau1338

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 ( $(x_{1},y_{1})$ and ( $(x_{2},y_{2})$ as $\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}$

So, the distance between two points are given by,

AB= $\sqrt{(9-2)^{2}+(0-1)^{2}}=\sqrt{49+1}=\sqrt{50}$ = 7.07

BC= $\sqrt{(8-9)^{2}+(7-0)^{2}}=\sqrt{1+49}=\sqrt{50}$ = 7.07

CD= $\sqrt{(8-1)^{2}+(7-8)^{2}}=\sqrt{49+1}=\sqrt{50}$ = 7.07

DA= $\sqrt{(1-2)^{2}+(8-1)^{2}}=\sqrt{1+49}=\sqrt{50}$ = 7.07

Also,

AC= $\sqrt{(8-2)^{2}+(7-1)^{2}}=\sqrt{36+36}=\sqrt{72}$ = 8.49

BD= $\sqrt{(9-1)^{2}+(0-8)^{2}}=\sqrt{64+64}=\sqrt{138}$ = 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.

Answer from: Quest

answer: sorry i can't you.

step-by-step explanation:

Answer from: Quest

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answer: d.) the total number of protons and neutrons is 15

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