Which of the following best classifies a quadrilateral with coordinates A(2, 6), B(5, 1), C(10, 4), and D(7, 9)?


Rectangle

Parallelogram

Rhombus

Square

Answer:

Square

Step-by-step explanation:

I took the flvs test, and I actually looked on course hero (a trusted site) and found it was square, then worked the problem out and found out it was actually a square.

Quadrilateral is a polygon with four sides. The specified quadrilateral falls in the category shown by: Option

Any closed figure made by 4 line segments joined end to end in series is called a quadrilateral.

That quadrilateral in which opposite sides are parallel is called a parallelogram.

Thus, a parallelogram is always a quadrilateral but a quadrilateral can or cannot be a parallelogram.

That parallelogram in which adjacent sides are perpendicular to each other is called a rectangle.

A rectangle is always a parallelogram and a quadrilateral but reverse statement may or may not be true.

Its a parallelogram but all sides of same length.  Its diagonals are perpendicular bisector of each other, but might not be equal.

That rectangle whose all sides are equal is called a square.

A square has its diagonals perpendicular bisector of each other and are of same length. (this is actually a characteristic property, which means that if a quadrilateral has its diagonals equal and perpendicular, then it is a square for sure).

A square is always a rectangle, parallelogram, a rhombus and a quadrilateral but its reverse statement may or may not be true.(ie it is not always necessary that a rectangle is a square or a parallelogram is a square or a rhombus is a square).

The shortest distance(straight line segment's length connecting both given points) between points ( p,q) and (x,y) is:

[tex]D = \sqrt{(x-p)^2 + (y-q)^2 }[/tex](in units)

If two lines are perpendicular to each other, then their slopes are negative reciprocal of each other. That means, if one line's slope is 'a', then its perpendicular line's slope would be -1/a

For the given case, the coordinates of  the vertices of the quadrilateral are: A(2, 6), B(5, 1), C(10, 4), and D(7, 9)

The diagonals are AC and BD.

Slope of AC = [tex]m_{AC} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{4-6}{10-2} = -1/4[/tex]

Slope of BD = [tex]m_{BD} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{9-1}{7-5} = 4[/tex]

It is visible that AC and BD's slope are negative reciprocal of each other. Thus, they are perpendicular to each other. Thus, the quadrilateral is either a square or Rhombus

Now, finding lengths of the diagonals:

[tex]|AC| = \sqrt{(4-6)^2 + (10-2)^2 } = \sqrt{4 + 64} = \sqrt{68}[/tex]

[tex]|BD| = \sqrt{(7-5)^2 + (9-1)^2 } = \sqrt{4 + 64} = \sqrt{68}[/tex]

Thus, their lengths are also same,

This is why, the considered quadrilateral is a square.

Learn more about square here:

https://brainly.com/question/5426789

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