Exercise 131 Page 523

a Let's start by plotting the points on a coordinate plane. This will help us determine what type of quadrilateral are we working with.

This quadrilateral appears to be a parallelogram, maybe even a rhombus. It cannot be a square as the consecutive sides are not perpendicular to each other. To check whether the shape is a rhombus, let's find the lengths of all sides of the quadrilateral using the distance formula. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) If we plug the coordinates of the endpoints of a side into the formula, we can find the length of the side. Let's start with A(-2, 6) and B(2, 3).

d_(AB) = sqrt((x_2- x_1)^2+(y_2-y_1)^2)

SubstitutePoints

Substitute ( -2, 6) & ( 2, 3)

d_(AB) = sqrt((( 2 - ( -2))^2+( 3- 6)^2)

▼

Simplify right-hand side

SubNeg

a-(- b)=a+b

d_(AB) = sqrt((2+2)^2+(3-6)^2)

AddSubTerms

Add and subtract terms

d_(AB) = sqrt(4^2+(-3)^2)

CalcPow

Calculate power

d_(AB) = sqrt(16+9)

AddTerms

Add terms

d_(AB) = sqrt(25)

CalcRoot

Calculate root

d_(AB) = 5

As we can see, the length of AB is 5. Let's check the lengths of the other sides.

Side	Points	sqrt((x_2-x_1)^2+(y_2-y_1)^2)	Length
AB	A( -2, 6), B( 2, 3)	sqrt(( 2-( -2))^2+( 3- 6)^2)	5
BC	B( 2, 3), C( 2, -2)	sqrt(( 2- 2)^2+( -2- 3)^2)	5
CD	C( 2, -2), D( -2, 1)	sqrt(( -2- 2)^2+( 1-( -2))^2)	5
DA	D( -2, 1), A( -2, 6)	sqrt(( -2-( -2))^2+( 6- 1)^2)	5

All sides are an equal length. Since the quadrilateral cannot be a square, it is a rhombus.

b In Part A we determined that ABCD is a rhombus. To find its area, we can use the following formula.

A = b h

In this formula, b is the length of any of the sides of the rhombus and h is the height perpendicular to it. We already know that b = 5, so we need to find the height. To do so, let's find the length of the height that connects D to BC.

Looking at the diagram, we see that point D is 4 units away from BC, so h = 4. Now that we know both b and h, we can substitute their values into the formula for the area and calculate it.

A = bh

SubstituteII

b= 5, h= 4

A = 5( 4)

Multiply

A = 20

The area of ABCD is 20 square units.

c Let's use the slope formula to calculate the slopes of the diagonals AC and BD. m = y_2-y_1/x_2-x_1 Let's start calculating the slope of AC by substituting the coordinates of the endpoints A(-2, 6) and C(2, -2) into the formula.

m_(AC) = y_2 - y_1/x_2 - x_1

SubstitutePoints

Substitute ( -2, 6) & ( 2, -2)

m_(AC) = -2 - 6/2 - ( -2)

▼

Simplify right-hand side

SubNeg

a-(- b)=a+b

m_(AC) = -2-6/2+2

AddSubTerms

Add and subtract terms

m_(AC) = -8/4

MoveNegNumToFrac

Put minus sign in front of fraction

m_(AC) = - 8/4

CalcQuot

Calculate quotient

m_(AC) = -2

Now let's substitute the coordinates of points B(2, 3) and D(-2, 1) into the slope formula to calculate the slope of BD.

m_(BD) = y_2 - y_1/x_2 - x_1

SubstitutePoints

Substitute ( 2, 3) & ( -2, 1)

m_(BD) = 1 - 3/-2 - 2

▼

Simplify right-hand side

SubTerms

Subtract terms

m_(BD) = -2/-4

DivNegNeg

- a/- b=a/b

m_(BD) = 2/4

ReduceFrac

a/b=.a /2./.b /2.

m_(BD) = 1/2

We found that m_(AC) = -2 and m_(BD) = 12. Since ABCD is a rhombus, the diagonals BD and AC should be perpendicular. This means that their slopes should be negative reciprocals of each other. Let's see if the product of the two diagonals is - 1.

m_(AC) * m_(BD) ? = -1

SubstituteII

m_(AC)= -2, m_(BD)= 1/2

-2( 1/2) ? = -1

Multiply

-1 = -1 ✓

The product of the slopes is - 1, so the diagonals are perpendicular to each other.

d To find the point of intersection of diagonals AC and BD, let's add them to the diagram of our rhombus.

It looks like the diagonals intersect at (0,2). Since every rhombus is a parallelogram and in parallelograms diagonals bisect each other, the point (0,2) should be the midpoint of both diagonals. Let's check whether this is the case for AC using the midpoint formula. M (x_1 + x_2/2, y_1 + y_2/2) Let's substitute the coordinates of A(-2, 6) and C(2, -2) into the formula above and check whether the result is (0,2).

M (x_1 + x_2/2, y_1 + y_2/2)

SubstitutePoints

Substitute ( -2, 6) & ( 2, -2)

M (-2 + 2/2, 6 + ( -2)/2)

AddNeg

a+(- b)=a-b

M(-2+2/2,6-2/2)

AddSubTerms

Add and subtract terms

M(0/2,4/2)

SimpQuot

Simplify quotient