Math Problem Statement

There is a square with vertices 4,1 and -1,4 how much possible vertices are there and what are the vertices coordinate points?

Solution

To find the vertices of the square given two vertices (4, 1) and (-1, 4), we can use the properties of squares. A square has four vertices, and the two given points can be opposite vertices.

  1. Find the length of the diagonal: The distance between the two given vertices is calculated using the distance formula: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} Substituting the coordinates: d=((1)4)2+(41)2=(5)2+(3)2=25+9=34d = \sqrt{((-1) - 4)^2 + (4 - 1)^2} = \sqrt{(-5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34}

  2. Determine the side length of the square: The side length ss of the square can be found from the relationship between the diagonal and the side: d=s2    s=d2=342=17d = s\sqrt{2} \implies s = \frac{d}{\sqrt{2}} = \frac{\sqrt{34}}{\sqrt{2}} = \sqrt{17}

  3. Find the midpoint of the diagonal: The midpoint MM of the diagonal can be found as follows: M=(x1+x22,y1+y22)=(4+(1)2,1+42)=(32,52)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{4 + (-1)}{2}, \frac{1 + 4}{2} \right) = \left( \frac{3}{2}, \frac{5}{2} \right)

  4. Find the direction of the sides: The direction vector from (4, 1) to (-1, 4) is: d=(14,41)=(5,3)\vec{d} = (-1 - 4, 4 - 1) = (-5, 3) A perpendicular vector (representing the sides of the square) can be obtained by switching the components and changing one of the signs: s1=(3,5)ors2=(3,5)\vec{s_1} = (3, 5) \quad \text{or} \quad \vec{s_2} = (-3, -5)

  5. Calculate the remaining vertices: To find the remaining vertices, we add and subtract s2\frac{s}{2} in the direction of the perpendicular vectors from the midpoint:

    • Vertex 3: V3=M+s2(s1s1)=M+172(334,534)=(32+317234,52+517234)V_3 = M + \frac{s}{2}\left( \frac{\vec{s_1}}{|\vec{s_1}|} \right) = M + \frac{\sqrt{17}}{2} \left( \frac{3}{\sqrt{34}}, \frac{5}{\sqrt{34}} \right) = \left( \frac{3}{2} + \frac{3\sqrt{17}}{2\sqrt{34}}, \frac{5}{2} + \frac{5\sqrt{17}}{2\sqrt{34}} \right)
    • Vertex 4: V4=Ms2(s1s1)=M172(334,534)=(32317234,52517234)V_4 = M - \frac{s}{2}\left( \frac{\vec{s_1}}{|\vec{s_1}|} \right) = M - \frac{\sqrt{17}}{2} \left( \frac{3}{\sqrt{34}}, \frac{5}{\sqrt{34}} \right) = \left( \frac{3}{2} - \frac{3\sqrt{17}}{2\sqrt{34}}, \frac{5}{2} - \frac{5\sqrt{17}}{2\sqrt{34}} \right)

Thus, the square's vertices are:

  1. (4, 1)
  2. (-1, 4)
  3. Calculated coordinates from Vertex 3
  4. Calculated coordinates from Vertex 4

To find exact coordinates for the remaining vertices, we simplify those expressions.

Would you like me to simplify these coordinates further or clarify any details? Here are some related questions you might consider:

  1. How can I calculate the area of the square using the vertices?
  2. What properties define a square in a coordinate plane?
  3. Can you explain how to determine if a quadrilateral is a square using its vertices?
  4. What is the significance of the midpoint in geometry?
  5. How do you find the distance between two points in a coordinate system?

Tip: When working with coordinates, drawing a diagram can often help visualize the relationships between points.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Properties of Squares
Distance Formula
Midpoint Formula
Vector Perpendicularity

Formulas

Distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2)
Diagonal to side relationship for square: d = s√2
Midpoint formula: M = ((x1 + x2)/2, (y1 + y2)/2)

Theorems

Properties of squares in a coordinate plane
Perpendicular vector theorem

Suitable Grade Level

Grades 9-12