Problem 47
Question
Show that the points \(A(-2,9), B(4,6), C(1,0),\) and \(D(-5,3)\) are the vertices of a square.
Step-by-Step Solution
Verified Answer
The points form a square since all sides are equal and diagonals are equal.
1Step 1: Calculate AB, BC, CD, and DA
First, calculate the length of each side of the quadrilateral. Use the distance formula which is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).For \(AB\), \((-2,9)\) and \((4,6)\):\[ AB = \sqrt{(4 - (-2))^2 + (6 - 9)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}\]For \(BC\), \((4,6)\) and \((1,0)\):\[ BC = \sqrt{(1 - 4)^2 + (0 - 6)^2} = \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\]For \(CD\), \((1,0)\) and \((-5,3)\):\[ CD = \sqrt{(-5 - 1)^2 + (3 - 0)^2} = \sqrt{(-6)^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}\]For \(DA\), \((-5,3)\) and \((-2,9)\):\[ DA = \sqrt{(-2 + 5)^2 + (9 - 3)^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\]
2Step 2: Check the Diagonals AC and BD
Now, calculate the lengths of the diagonals using the distance formula to ensure they are equal.For \(AC\), \((-2,9)\) and \((1,0)\):\[ AC = \sqrt{(1 - (-2))^2 + (0 - 9)^2} = \sqrt{3^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}\]For \(BD\), \((4,6)\) and \((-5,3)\):\[ BD = \sqrt{(-5 - 4)^2 + (3 - 6)^2} = \sqrt{(-9)^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}\]
3Step 3: Confirm Square Properties
For the quadrilateral to be a square, all sides must be equal and both diagonals must be equal. From the previous steps, we've found:- \(AB = BC = CD = DA = 3\sqrt{5}\)- \(AC = BD = 3\sqrt{10}\)Since all sides are equal and the diagonals are equal, confirming both diagonals are longer than the sides verifies that the quadrilateral is a square.
Key Concepts
Distance FormulaQuadrilateralVerticesSquare
Distance Formula
The distance formula is a fundamental concept in geometry that helps us find the distance between two points on the coordinate plane. It is derived from the Pythagorean Theorem and can be applied when the coordinates of the points are known.
The distance formula is expressed as:
The distance formula is expressed as:
- \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Quadrilateral
A quadrilateral is a geometric shape with four sides and four vertices. Based on the length of its sides and internal angles, quadrilaterals can be classified as squares, rectangles, trapezoids, and more. Understanding the properties of quadrilaterals is fundamental in geometry.
In any quadrilateral:
In any quadrilateral:
- The sum of the internal angles is always 360 degrees.
- The shape can vary widely, from efficient parallelograms to irregular four-sided polygons.
Vertices
Vertices are crucial in geometry as they represent the points where two or more line segments meet. In the context of a quadrilateral, there are four vertices, and knowing their coordinates allows for precise calculations of distance and angle measurements.
For a shape like a square, each vertex connects two sides of equal length, which can be confirmed using the distance formula. When coordinates are given, marking these points on a graph can help visualize the shape and better understand its properties.
Through vertices:
For a shape like a square, each vertex connects two sides of equal length, which can be confirmed using the distance formula. When coordinates are given, marking these points on a graph can help visualize the shape and better understand its properties.
Through vertices:
- We determine the structure and symmetry of geometric shapes.
- In quadrilaterals, they give us the basis to apply various geometric formulas.
Square
A square is a special type of quadrilateral distinguished by having four equal sides and four right angles. Verifying a square by evaluating vertices and side lengths is a practical application of the distance formula and the properties of quadrilaterals.
Key properties of a square include:
Key properties of a square include:
- All sides are of equal length.
- The diagonals are also equal, which further confirms its properties.
- The diagonals intersect at a right angle and are longer than the sides, offering a clear marker of its square nature.
Other exercises in this chapter
Problem 47
\(43-50\) . Solve the equation graphically in the given interval. State each answer rounded to two decimals. $$ x-\sqrt{x+1}=0 ; \quad[-1,5] $$
View solution Problem 47
\(41-48\) Find the \(x\) - and \(y\) -intercepts of the graph of the equation. $$ x y=5 $$
View solution Problem 48
Find the slope and y-intercept of the line, and draw its graph. $$ 2 x-5 y=0 $$
View solution Problem 48
\(43-50\) . Solve the equation graphically in the given interval. State each answer rounded to two decimals. $$ 1+\sqrt{x}=\sqrt{1+x^{2}} ;[-1,5] $$
View solution