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