Math Problem Statement

Are the points (−2, 10), (4, 7), (1, 1), and (−5, 4) the vertices of a square?

Solution

To determine whether the points (2,10)(-2, 10), (4,7)(4, 7), (1,1)(1, 1), and (5,4)(-5, 4) are the vertices of a square, we need to verify two main conditions:

  1. All sides of the quadrilateral formed by these points are equal in length.
  2. The diagonals of the quadrilateral are equal in length.

1. Calculate the Lengths of the Sides

We'll use the distance formula to calculate the length between each pair of points: Distance=(x2x1)2+(y2y1)2\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Calculate all four sides:

  • Between (2,10)(-2, 10) and (4,7)(4, 7): Side 1=(4(2))2+(710)2=62+(3)2=36+9=45=35\text{Side 1} = \sqrt{(4 - (-2))^2 + (7 - 10)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}

  • Between (4,7)(4, 7) and (1,1)(1, 1): Side 2=(14)2+(17)2=(3)2+(6)2=9+36=45=35\text{Side 2} = \sqrt{(1 - 4)^2 + (1 - 7)^2} = \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}

  • Between (1,1)(1, 1) and (5,4)(-5, 4): Side 3=(51)2+(41)2=(6)2+32=36+9=45=35\text{Side 3} = \sqrt{(-5 - 1)^2 + (4 - 1)^2} = \sqrt{(-6)^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}

  • Between (5,4)(-5, 4) and (2,10)(-2, 10): Side 4=(2(5))2+(104)2=32+62=9+36=45=35\text{Side 4} = \sqrt{(-2 - (-5))^2 + (10 - 4)^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}

Since all four sides are equal (353\sqrt{5}), the quadrilateral has equal sides, which is a necessary condition for it to be a square.

2. Calculate the Lengths of the Diagonals

Now, calculate the lengths of the diagonals:

  • Diagonal between (2,10)(-2, 10) and (1,1)(1, 1): Diagonal 1=(1(2))2+(110)2=32+(9)2=9+81=90=310\text{Diagonal 1} = \sqrt{(1 - (-2))^2 + (1 - 10)^2} = \sqrt{3^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}

  • Diagonal between (4,7)(4, 7) and (5,4)(-5, 4): Diagonal 2=(54)2+(47)2=(9)2+(3)2=81+9=90=310\text{Diagonal 2} = \sqrt{(-5 - 4)^2 + (4 - 7)^2} = \sqrt{(-9)^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}

Since both diagonals are equal (3103\sqrt{10}), the quadrilateral also satisfies the diagonal condition for a square.

Conclusion

Because the sides are all equal and the diagonals are equal, the points (2,10)(-2, 10), (4,7)(4, 7), (1,1)(1, 1), and (5,4)(-5, 4) do indeed form a square.

Do you have any questions about this conclusion, or would you like more details?

Here are some related questions:

  1. How can you verify if four points form a rectangle?
  2. What are the conditions for four points to form a rhombus?
  3. How do you find the area of a square given its vertices?
  4. What other shapes can be verified using distance formulas?
  5. Can the distance formula be used to determine if four points form a parallelogram?

Tip: When checking if a quadrilateral is a square, equal sides and equal diagonals are both necessary conditions.

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Math Problem Analysis

Mathematical Concepts

Geometry
Distance Formula
Quadrilaterals

Formulas

Distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Properties of squares

Suitable Grade Level

Grades 9-12