Parallelogram ABCD has vertices A(8, 2), B(6, –4), and C(–5, –4). Find the coordinates of D.
Note that BC is a horizontal line (y = -4) of length 11
So, AD must also be horizontal (y = 2) of length 11.
So, D = (-3,2)
This is quite easy if you actually get out some graph paper and plot points A,B,C !
I see we're playing "Find the Missing Vertex"! Let's put on our detective hats and solve this mystery.
To find the coordinates of vertex D, let's take a look at what we know.
The opposite sides of a parallelogram are parallel, which means that the slope of AB is equal to the slope of CD, and the slope of BC is equal to the slope of AD.
So, let's calculate the slopes first.
The slope of AB = (y2 - y1) / (x2 - x1) = (-4 - 2) / (6 - 8) = -6 / -2 = 3
The slope of BC = (y2 - y1) / (x2 - x1) = (-4 - (-4)) / (-5 - 6) = 0 / -11 = 0
Since the slopes of AB and CD are equal, the slope of CD is also 3.
Now, let's find the equation of line CD using the point-slope form of a linear equation:
y - y1 = m(x - x1), where (x1, y1) is any point on the line
Using point C(–5, –4), we have:
y - (-4) = 3(x - (-5))
Simplifying this equation, we get:
y + 4 = 3x + 15
Rearranging the equation for y, we have:
y = 3x + 11
So, the equation of line CD is y = 3x + 11.
Now we need to find the intersection point of lines AB and CD to determine the coordinates of point D. To do that, we'll equate the equations of the two lines:
3x + 11 = -4
Solving for x, we get:
3x = -4 - 11
3x = -15
x = -15/3
x = -5
Now that we have the x-coordinate, we can substitute it into the equation of line CD to find the y-coordinate:
y = 3(-5) + 11
y = -15 + 11
y = -4
So, the coordinates of vertex D are (-5, -4).
To find the coordinates of point D in parallelogram ABCD, we need to use the properties of a parallelogram.
1. Find the vector AB by finding the difference between the x-coordinates and the y-coordinates of points A and B:
AB = (6 - 8, -4 - 2)
= (-2, -6)
2. Find the coordinates of point D by adding the vector AB to point C:
D = C + AB
= (-5, -4) + (-2, -6)
= (-5 - 2, -4 - 6)
= (-7, -10)
Therefore, the coordinates of point D in parallelogram ABCD are (-7, -10).
To find the coordinates of vertex D in the parallelogram ABCD, we need to use the fact that opposite sides of a parallelogram are parallel and equal in length.
First, let's find the length and slope of side AB.
Length of AB = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((6 - 8)^2 + (-4 - 2)^2)
= sqrt((-2)^2 + (-6)^2)
= sqrt(4 + 36)
= sqrt(40)
Slope of AB = (y2 - y1) / (x2 - x1)
= (-4 - 2) / (6 - 8)
= (-6) / (-2)
= 3
Since opposite sides of a parallelogram are parallel, side DC should also have a slope of 3. Now, we have to find the equation of the line passing through C(-5, -4) with a slope of 3.
Using the point-slope form of a linear equation, we have:
y - y1 = m(x - x1)
y - (-4) = 3(x - (-5))
y + 4 = 3(x + 5)
y + 4 = 3x + 15
y = 3x + 15 - 4
y = 3x + 11
Now, we need to find the coordinates of the intersection point of the lines AB and CD. To find this point, we need to solve the system of equations formed by the equations of the lines AB and CD.
The equations are:
1) y = 3x + 11 (equation of CD)
2) y - 2 = 3(x - 8) (equation of AB, using point A(8, 2))
Let's solve this system of equations:
Substituting equation 1) into equation 2), we have:
3x + 11 - 2 = 3(x - 8)
3x + 9 = 3x - 24
We notice that the x terms cancel out. This indicates that the lines AB and CD are parallel, meaning they will never intersect.
However, we know that opposite sides of a parallelogram are equal in length. Therefore, the length of BC should be equal to the length of AD.
Using the distance formula, we can calculate the length of BC:
Length of BC = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((6 - (-5))^2 + (-4 - (-4))^2)
= sqrt((6 + 5)^2 + (0)^2)
= sqrt(11^2)
= 11
Since opposite sides of a parallelogram are equal in length, the length of AD is also 11.
Now, we need to find the slope of AD, using the slope of BC. Since opposite sides of a parallelogram are parallel, the slope of BC will be equal to the slope of AD.
Slope of AD = slope of BC = 3
Lastly, we need to find the equation of the line passing through A(8, 2) with a slope of 3.
Using the point-slope form of a linear equation, we have:
y - y1 = m(x - x1)
y - 2 = 3(x - 8)
y - 2 = 3x - 24
y = 3x - 24 + 2
y = 3x - 22
Therefore, the coordinates of vertex D are (x, y) = (x, 3x - 22), where x can be any value.