Given A(1, 1), B(0,5), C(4,4), and D(5, 0). Use the fact that if the diagonals of a parallelogram are perpendicular, then it is a rhombus to prove ABCD is a rhombus?

Question

Aurora Ayala · Accepted Answer

Please see below.

The two diagonals are #AC# and #BD#. As product of the slopes of two lines perpendicular to each other is #-1#, we intend to show that.

Now, slope of line joining two points #(x_1,y_1)# and #(x_2,y_2)# is given by #(y_2-y_1)/(x_2-x_1)# and therefore

Slope of #AC# is #(4-1)/(4-1)=3/3=1#

Slope of #BD# is #(0-5)/(5-0)=-5/5=-1#

as product of the slopes is #1xx(-1)=-1#, they form a rhombus.

James Atkins · Answer

undefined To prove that ABCD is a rhombus, we need to show that its diagonals are perpendicular. The diagonals of ABCD are AC and BD.

The slope of AC is given by: 
$ m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{4 - 1}{4 - 1} = \frac{3}{3} = 1 $

The slope of BD is given by:
$ m_{BD} = \frac{y_D - y_B}{x_D - x_B} = \frac{0 - 5}{5 - 0} = \frac{-5}{5} = -1 $

Since the product of the slopes of AC and BD is -1, the diagonals AC and BD are perpendicular.

Therefore, by the property that if the diagonals of a parallelogram are perpendicular, then it is a rhombus, we can conclude that ABCD is a rhombus.

Evelyn Arboleda · Answer

undefined To prove that ABCD is a rhombus, we need to show that the diagonals are perpendicular. The diagonals of a parallelogram bisect each other. We first find the midpoints of AC and BD using the midpoint formula:

Midpoint of AC: $(\frac{1+4}{2}, \frac{1+4}{2}) = (2.5, 2.5)$
Midpoint of BD: $(\frac{0+5}{2}, \frac{5+0}{2}) = (2.5, 2.5)$

Since the midpoints are the same, AC and BD intersect at the point (2.5, 2.5).

Next, we calculate the slopes of AC and BD:

Slope of AC: $\frac{4-1}{4-1} = 1$
Slope of BD: $\frac{0-5}{5-0} = -1$

Since the product of the slopes is -1, AC and BD are perpendicular.

Therefore, since the diagonals of ABCD are perpendicular and bisect each other, ABCD is a rhombus.