Answer:
a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus
b. The equation of the diagonal line AC is y = 5 - x
The equation of the diagonal line BD is y = 5 - x
c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other
Step-by-step explanation:
The vertices of the given quadrilateral are;
A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)
a. The length, l, of the sides of the given quadrilateral are given as follows;
![l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%7B%5Cleft%20%28y_%7B2%7D-y_%7B1%7D%20%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%28x_%7B2%7D-x_%7B1%7D%20%20%5Cright%20%29%5E%7B2%7D%7D)
The length of side AB, with A = (1, 4) and B = (6, 6) gives;
![l_{AB} = \sqrt{\left (6-4 \right )^{2}+\left (6-1 \right )^{2}} = \sqrt{29}](https://tex.z-dn.net/?f=l_%7BAB%7D%20%3D%20%5Csqrt%7B%5Cleft%20%286-4%20%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%286-1%20%20%5Cright%20%29%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B29%7D)
The length of side BC, with B = (6, 6) and C = (4, 1) gives;
![l_{BC} = \sqrt{\left (1-6 \right )^{2}+\left (4-6 \right )^{2}} = \sqrt{29}](https://tex.z-dn.net/?f=l_%7BBC%7D%20%3D%20%5Csqrt%7B%5Cleft%20%281-6%20%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%284-6%20%20%5Cright%20%29%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B29%7D)
The length of side CD, with C = (4, 1) and D = (-1, -1) gives;
![l_{CD} = \sqrt{\left (-1-1 \right )^{2}+\left (-1-4 \right )^{2}} = \sqrt{29}](https://tex.z-dn.net/?f=l_%7BCD%7D%20%3D%20%5Csqrt%7B%5Cleft%20%28-1-1%20%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%28-1-4%20%20%5Cright%20%29%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B29%7D)
The length of side DA, with D = (-1, -1) and A = (1,4) gives;
![l_{DA} = \sqrt{\left (4-(-1) \right )^{2}+\left (1-(-1) \right )^{2}} = \sqrt{29}](https://tex.z-dn.net/?f=l_%7BDA%7D%20%3D%20%5Csqrt%7B%5Cleft%20%284-%28-1%29%20%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%281-%28-1%29%20%20%5Cright%20%29%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B29%7D)
Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus
b. The diagonals are AC and BD
The slope, m, of AC is given by the formula for the slope of a straight line as follows;
![Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=Slope%2C%20%5C%2C%20m%20%3D%5Cdfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
Therefore;
![Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1](https://tex.z-dn.net/?f=Slope%2C%20%5C%2C%20m_%7BAC%7D%20%3D%5Cdfrac%7B1-4%7D%7B4-1%7D%20%3D%20-1)
The equation of the diagonal AC in point and slope form is given as follows;
y - 4 = -1×(x - 1)
y = -x + 1 + 4
The equation of the diagonal AC is y = 5 - x
![Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1](https://tex.z-dn.net/?f=Slope%2C%20%5C%2C%20m_%7BBD%7D%20%3D%5Cdfrac%7B-1-6%7D%7B-1-6%7D%20%3D%201)
The equation of the diagonal BD in point and slope form is given as follows;
y - 6 = 1×(x - 6)
y = x - 6 + 6 = x
The equation of the diagonal BD is y = x
c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;
The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs
The point of intersection of the two diagonals is given as follows;
5 - x = x
∴ x = 5/2 = 2.5
y = x = 2.5
The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.