Finding Coordinates of Vertex D in Isosceles Trapezium

The problem asks us to find the coordinates of the fourth vertex, D, of an isosceles trapezium ABCD, given the coordinates of three vertices A(2, 3), B(4, 3), and C(5, 1). We are also told that AB is parallel to DC.

Understanding the Properties of an Isosceles Trapezium

An isosceles trapezium has specific geometric properties that we can use:

• One pair of opposite sides are parallel (given AB || DC).
• The non-parallel sides are equal in length (AD = BC).
• The base angles are equal.
• It has an axis of symmetry which is the perpendicular bisector of the parallel sides.

Step 1: Analyze Given Vertices and Parallel Sides

We are given the vertices A(2, 3) and B(4, 3). Notice that the y-coordinates of A and B are the same (y=3). This means the side AB is a horizontal line segment.

Since AB is parallel to DC (AB || DC), the side DC must also be a horizontal line segment. This implies that the y-coordinates of D and C must be the same. We are given C(5, 1), so the y-coordinate of D must also be 1. Let the coordinates of D be (x, 1).

Step 2: Use the Isosceles Property (Equal Non-Parallel Sides)

In an isosceles trapezium, the non-parallel sides have equal lengths. Given AB || DC, the non-parallel sides are AD and BC.

We need to calculate the length of BC using the distance formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).

• Coordinates of B = (4, 3)
• Coordinates of C = (5, 1)
• Length of BC = \(\sqrt{(5-4)^2 + (1-3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}\).

Now, we calculate the length of AD using the distance formula:

• Coordinates of A = (2, 3)
• Coordinates of D = (x, 1)
• Length of AD = \(\sqrt{(x-2)^2 + (1-3)^2} = \sqrt{(x-2)^2 + (-2)^2} = \sqrt{(x-2)^2 + 4}\).

Since AD = BC for an isosceles trapezium:

\(\sqrt{(x-2)^2 + 4} = \sqrt{5}\)

Step 3: Solve for the x-coordinate of D

To solve for x, we square both sides of the equation:

\((x-2)^2 + 4 = 5\)

Subtract 4 from both sides:

\((x-2)^2 = 5 - 4\)

\((x-2)^2 = 1\)

Take the square root of both sides:

\(x-2 = \pm\sqrt{1}\)

\(x-2 = \pm 1\)

This gives two possible values for x:

• Case 1: \(x-2 = 1 \implies x = 1 + 2 \implies x = 3\). This would give D(3, 1).
• Case 2: \(x-2 = -1 \implies x = -1 + 2 \implies x = 1\). This would give D(1, 1).

We have two potential coordinates for D: (3, 1) and (1, 1).

Step 4: Determine the Correct Coordinates using Symmetry

Let's consider the symmetry of the isosceles trapezium ABCD with AB || DC.

The midpoint of the horizontal side AB is \(M_{AB} = (\frac{2+4}{2}, \frac{3+3}{2}) = (3, 3)\).

The midpoint of the horizontal side DC is \(M_{DC} = (\frac{x+5}{2}, \frac{1+1}{2}) = (\frac{x+5}{2}, 1)\).

The line segment connecting the midpoints of the non-parallel sides would be the axis of symmetry if we were considering AD and BC as parallel. However, AB and DC are parallel.

Alternatively, consider the relationship between points A, B, C, and D. In an isosceles trapezium ABCD with AB || DC, the horizontal distance from A to D should be related to the horizontal distance from B to C due to symmetry. The 'shift' from A to B is (2, 0). The 'shift' from D to C should also be horizontal. The vector BC is \((5-4, 1-3) = (1, -2)\). The vector AD is \((x-2, 1-3) = (x-2, -2)\).

For the trapezium to be 'flat' on top (AB) and 'wider' or 'narrower' at the bottom (DC), and to maintain isosceles property, the displacement from A vertically down and then horizontally should correspond to the displacement from B vertically down and horizontally, relative to the center line. A simpler way is to consider the perpendicular bisector of AB, which is the line \(x=3\). Vertex C(5, 1) is 2 units to the right of this line (\(5-3=2\)). For the trapezium to be isosceles, the vertex D must be symmetrically positioned with respect to this line. Therefore, D should be 2 units to the left of the line \(x=3\).

So, the x-coordinate of D should be \(3 - 2 = 1\). This gives D(1, 1).

Let's check the other possibility D(3, 1). If D=(3, 1), then:

• A(2, 3), B(4, 3), C(5, 1), D(3, 1).
• AB length = 2. DC length = \(|5-3| = 2\). This would make ABCD a parallelogram, not just a trapezium. Since AB=DC and AB||DC, it's a parallelogram. Is it an isosceles trapezium? Yes, parallelograms can be isosceles trapeziums only if they are rectangles. This is not a rectangle. Also, AD = \(\sqrt{(3-2)^2+(1-3)^2} = \sqrt{1^2+(-2)^2}=\sqrt{5}\). BC = \(\sqrt{(5-4)^2+(1-3)^2} = \sqrt{1^2+(-2)^2}=\sqrt{5}\). So AD=BC. It *is* an isosceles trapezium, and also a parallelogram. However, typically, when AB and DC are the parallel bases, D(1,1) forms a more standard isosceles trapezium shape where DC is the longer base.

Let's reconsider the symmetry. The vertex C(5, 1) is positioned relative to B(4, 3). The difference is \(C-B = (1, -2)\). If we apply a similar displacement starting from A(2, 3) but ensuring the y-coordinate matches C, we get D(1,1). The difference \(D-A = (1-2, 1-3) = (-1, -2)\). Comparing \((1, -2)\) and \((-1, -2)\), these represent equal length sides with a symmetric horizontal component.

The point D(1, 1) maintains the isosceles property (AD = BC = \(\sqrt{5}\)) and the parallel property (AB horizontal, DC horizontal). The length of DC is \(|5-1|=4\). The length of AB is \(|4-2|=2\). This configuration aligns with the standard definition and visual representation of an isosceles trapezium where AB and DC are the parallel bases.

Step 5: Verification

Let's verify the properties with D(1, 1):

• Vertices: A(2, 3), B(4, 3), C(5, 1), D(1, 1).
• AB is horizontal (y=3). DC is horizontal (y=1). Thus, AB || DC.
• Length AB = 2. Length DC = \(|5-1| = 4\).
• Length AD = \(\sqrt{(1-2)^2 + (1-3)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}\).
• Length BC = \(\sqrt{(5-4)^2 + (1-3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}\).
• Since AD = BC, the trapezium is isosceles.

All conditions are satisfied for D(1, 1).

Conclusion

The coordinates of the vertex D are (1, 1).