d Equate the diagonals' equations and solve for x.
A
a Parallelogram
B
b m∠DAB ≈ 63.43^(∘)
m∠BCD ≈ 63.43^(∘) Relationship: The angles are congruent.
C
c AC: y= 1/2x+1/2
BD: y= - x+5
The lines are not perpendicular.
D
d (3,2)
Practice makes perfect
a Let's plot the points on graph paper and connect them.
Examining the diagram, we see that AD∥ BC as they are both horizontal. If we also can prove that AB∥ DC, we have a quadrilateral with two pairs of parallel sides which means it is a parallelogram. To do that, we should find the segments' slopes.
The slope of AB and DC are both 2 which means these sides are also parallel. With this information, we know that ABCD is a parallelogram.
But, a parallelogram can also be a rhombus if all sides are congruent. Therefore, let's calculate the side's lengths using the Distance Formula. Note that BC and AD are horizontal. These lengths can be found by measuring the number of steps between their endpoints. From the diagram, we see that AD=3 and BC=3.
Segment
Points
sqrt((x_2-x_1)^2+(y_2-y_1)^2)
d
AB
( 1,1), ( 2,3)
sqrt(( 1- 2)^2+( 1- 3)^2)
sqrt(5)
DC
( 4,1), ( 5,3)
sqrt(( 4- 5)^2+( 1- 3)^2)
sqrt(5)
Since sqrt(5)≠3, we know that ABCD is a parallelogram.
b From Part A, we know the slope of AB and DC. Let's reverse the order of the second slope triangle and label the two angles.
Using the tangent ratio, we can find the measure of these angles.
tan θ =opposite/adjacent
Examining our slope triangles, we see that they both have opposite sides of 2 and adjacent sides of 1. By substituting these value in the formula, we can solve for θ. Notice that the triangles are identical so we only have to calculate one of them.
The angles have equal measures which means they are congruent.
c Let's first draw the diagonals AC and BD.
Both of these lines are straight lines which means we can write them in slope-intercept form.
y=mx+bIn this equation, m is the line's slope and b is the y-intercept. We can find the slope by using the Slope Formula.
Segment
Points
y_2-y_1/x_2-x_1
m
AC
A(1,1), C(5,3)
3- 1/5- 1
1/2
BD
B(2,3), D(4,1)
1- 3/4- 2
- 1
With this information, we have half of what we need to write the equations.
AC:& y= 1/2x+b
BD:& y= - x+b
Finally, we must find the y-intercept by substituting any of the points through which the lines passes, into the equations and solving for b. For example, we can substitute A(1,1) in the equation for AC and B(2,3) in the equation for BD.
segment
y=mx+b
substitute point
solve for b
AC
y= 1/2x+b
1= 1/2( 1)+b
b=1/2
BD
y= - x+b
3= - 2+b
b=5
Now we can complete the equations.
AC:& y= 1/2x+1/2
BD:& y= - x+5
If the equations are perpendicular, the product of their slopes equals - 1.
m_1m_2=- 1
By substituting the slopes in this equation, we can find out if the lines are perpendicular.
d From the diagram, we can see that the diagonals intersect at (3,2). However, we can also find this point algebraically by equating the functions we wrote in Part C, and solving for x.
