Question

Show that the points (=4,-1), (-2, -4), (4, 0) and (2, 3) are the vertices points of rectangle

Answer 1

let the rectangle be quadrilateral ABCD

A=(x1,y1)=(-4,-1) , B=(x2,y2)=(-2,-4),

C=(x3,y3)=(4,0) , D=(x4,y4)=(2,3).

d(AB)=√(x2-x1)^2+(y2-y1)^2

=√ 13_(1)

d(BC)=√52_(2)

d(CD)=√13_(3)

d(AD)=√52_(4)

Therefore ,the diagonal are also congruent of a rectangle

d(AC)=√65_(5)

d(BD)=√65_(6)

Therefore from 1,3and 2,4

d(AB)=d(CD)....the opposite side of

d(BC)=d(AD)....rectangle are congruent

tgerefore diagonal are also congruent

dAC=dBD

Answer 2

Given:-

• The point (4,-1), (-2, -4), (4, 0) and (2, 3). are the vertices points of rectangle.

To find:-

• Show that the point are vertices point rectangle.

Solutions:-

The distance d between two points (x1, y1) and (x2, y2) is given by the formula.

The opposite sides are equal in length.

The diagonal of a rectangle are also equal in length.

Here, the four points are A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3)

Let the length of the opposite sides of the quadrilateral that is formed by point.

=> AB = √(-4 + 2)² + (-1 +4)²

=> AB = √(-2)² + (3)²

=> AB = √4 + 9

=> AB = √13

=> CD = √(4 - 2)² + (0 - 3)²

=> CD = √(2)² + (-3)²

=> CD = √4 + 9

=> CD = √13

So one pair of opposite sides Equal.

Now, let other pair opposite sides

=> BC = √(-2 - 4)² + (-4 - 0)²

=> BC = √(-6)² + (-4)²

=> BC = √36 + 16

=> BC = √52

=> AD = √(-4 - 2)² + (-1 - 3)²

=> AD = √(-6)² + (-4)²

=> AD = √36 + 16

=> AD = √52

The other pair of opposite sides are also equal.

So, the quadrilateral formed by these four point is a parallelogram.

The diagonal are also equal is length.

=> AC = √(-4 - 4)² + (-1 - 0)²

=> AC = √(-8)² + (-1)²

=> AC = √64 + 1

=> AC = √65

=> BD = √(-2 - 2)² + (-4 - 3)²

=> BD = √(-4)² + (-7)²

=> BD = √16 + 49

=> BD = √65

Now, since the diagonals are also equal. The parallelogram is definitely a rectangle.

Hence, the quadrilateral formed by the four given point is a rectangle.