First, we need to show that ABCD is a parallelogram by demonstrating that the opposite sides are parallel.
We know that the slopes of two parallel lines are equal.
The slope of line AB is (6 - (-1))/(-9 - (-5)) = 7/-4 = -7/4
The slope of line DC is (-2 - 5)/(3 - (-1)) = -7/4
Similarly, the slope of line AD is (-2 - (-1))/(3 - (-5)) = -1/8
The slope of line BC is (5 - 6)/(-1 - (-9)) = 1/8
Since the slopes of the opposite sides are equal, we can conclude that ABCD is a parallelogram.
Next, we need to show that the diagonals of ABCD are perpendicular to each other.
The slope of diagonal AC is (5 - (-1))/(-1 - (-5)) = 3
The slope of diagonal BD is (6 - (-2))/(-9 - 3) = -1/2
The product of the slopes of perpendicular lines is -1, so the diagonals AC and BD are perpendicular to each other.
Since ABCD is both a parallelogram with perpendicular diagonals, it is a rhombus.
Verify that parallelogram ABCD with vertices
A(-5, - 1), B(-9, 6), C(-1, 5), and D(3, - 2) is a rhombus by showing that it is a parallelogram with perpendicular diagonals.
(3 points)
1 answer
1 answer