7. Polygons in the Coordinate Plane
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There can be more than one way to classify ABCD. One way is to classify by the slope of each side. To do so, we will use the Slope Formula.
Side | Points | y_2-y_1/x_2-x_1 | Slope |
---|---|---|---|
AB | A( -4, 4) and B( -1, -2) | -2- 4/-1-( -4) | -2 |
BC | B( -1, -2) and C( -3, -3) | -3-( -2)/-3-( -1) | 1/2 |
CD | C( -3, -3) and D( -6, 3) | 3-( -3)/-6-( -3) | -2 |
AD | A( -4, 4) and D( -6, 3) | 3- 4/-6-( -4) | 1/2 |
Notice that the opposite sides are parallel and that the consecutive sides are perpendicular. This means that ABCD is a parallelogram with four right angles, so it is a rectangle.
Side | Points | y_2-y_1/x_2-x_1 | Slope |
---|---|---|---|
EF | E( 1, -3) and F( 4, 3) | 3-( -3)/4- 1 | 2 |
FG | F( 4, 3) and G( 6, 2) | 2-( 3)/6-( 4) | -1/2 |
GH | G( 6, 2) and H( 3, -4) | -4- 2/3- 6 | 2 |
EH | E( 1, -3) and H( 3, -4) | -4-( -3)/3- 1 | -1/2 |
Because of the same reasons that we stated in Part A, EFGH is also a rectangle.
ABCD | EFGH | ||||||
---|---|---|---|---|---|---|---|
Side | Points | sqrt((x_2-x_1)^2+(y_2-y_1)^2) | Length | Side | Points | sqrt((x_2-x_1)^2+(y_2-y_1)^2) | Length |
AB | A( -4, 4) and B( -1, -2) | sqrt(( -1-( -4))^2+( -2- 4)^2) | s sqrt(45) | EF | E( 1, -3) and F( 4, 3) | sqrt(( 4- 1)^2+( 3- 1)^2) | sqrt(45) |
BC | B( -1, -2) and C( -3, -3) | sqrt(( -3-( -1))^2+( -3-( -2))^2) | s sqrt(5) | FG | F( 4, 3) and G( 6, 2) | sqrt(( 6- 4)^2+( 2- 3)^2) | sqrt(5) |
CD | C( -3, -3) and D( -6, 3) | sqrt(( -6-( -3))^2+( 3-( -3))^2) | sqrt(45) | GH | G( 6, 2) and H( 3, -4) | sqrt(( 3- 6)^2+( -4- 2)^2) | sqrt(45) |
AD | A( -4, 4) and D( -6, 3) | sqrt(( -6-( -4))^2+( 3- 4)^2) | sqrt(5) | EH | E( 1, -3) and H( 3, -4) | sqrt(( 3- 1)^2+( -4-( -3))^2) | sqrt(5) |
Looking at the table above, we see that the lengths and widths of the rectangles are congruent. Therefore, ABCD and EFGH are congruent.