Show that the points a(2, 2), b(5, 7), c(-5, 13), and d(-8, 8) are the vertices of a rectangle.
Solution:
The points a(2, 2), b(5, 7), c(-5, 13), and d(-8, 8) are the vertices of a rectangle.
Let us prove that the opposite sides are equal.
Consider a(2, 2), b(5, 7)
x\(_1\) = 2, y\(_1\) = 2, x\(_2\)= 5, y\(_2\) = 7
AB2 = (x\(_2\) - x\(_1\))2 + (y\(_2\) - y\(_1\))2 (using the distance formula between 2 points)
Substituting the values
AB2 = (5 - 2)2 + (7 - 2)2
AB2 = (3)2 + (5)2
By further calculation
AB2 = (9 + 25)
AB2 = 34
AB = √34 units
Consider b(5, 7), c(-5, 13)
x\(_1\) = 5, y\(_1\)= 7, x2 = -5, y2 = 13
BC2 = (x\(_2\) - x\(_1\))2 + (y\(_2\) - y\(_1\))2
Substituting the values
BC2 = (-5 - 5)2 + (13 - 7)2
BC2 = (-10)2 + (6)2
By further calculation
BC2 = 100 +36
BC2 = 136
BC = √136 units
Consider c(-5, 13), d(-8, 8)
x1 = -5, y1 = 13, x2 = -8, y2 = 8
CD2 = (x\(_2\) - x\(_1\))2 + (y\(_2\) - y\(_1\))2
Substituting the values
CD2 = (-8 + 5)2 + (8 - 13)2
CD2 = (-3)2 + (-5)2
By further calculation
CD2 = 9 + 25
CD2 = 34
CD = √34 units
Consider a(2, 2), d(-8, 8)
x1 = 2, y1 = 2, x2 = -8, y2 = 8
AD2 = (x2 - x1)2 + (y2 - y1)2
Substituting the values
AD2 = (-8 - 2)2 + (8 - 2)2
AD2 = (-10)2 + (6)2
By further calculation
AD2 = 100 +36
AD2 = 136
AD = √136 units
Here AB = CD = √34 units and BC = AD = √136 units
As the opposite sides are equal, ABCD is a rectangle.
Therefore, the points a(2, 2), b(5, 7), c(-5, 13), and d(-8, 8) are the vertices of a rectangle.
Show that the points a(2, 2), b(5, 7), c(-5, 13), and d(-8, 8) are the vertices of a rectangle.
Summary:
The points a(2, 2), b(5, 7), c(-5, 13), and d(-8, 8) are the vertices of a rectangle
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