The local public pool is a rectangle with two semicircles.
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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;
The length of side AB, with A = (1, 4) and B = (6, 6) gives;
The length of side BC, with B = (6, 6) and C = (4, 1) gives;
The length of side CD, with C = (4, 1) and D = (-1, -1) gives;
The length of side DA, with D = (-1, -1) and A = (1,4) gives;
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;
Therefore;
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
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.