A(-1,-5), B(5,-2) and C(1, 1). ABCD is a trapezium.
AB is parallel to DC and angle BAD is 90°.
Find the coordinates of D.

D(-3, -1)

Step-by-step explanation:

The given coordinates are;

A(-1, -5), B(5, -2) and (1, 1)

The coordinates and the coordinates of the point D form a trapezium

The parallel sides of the trapezium ABCD = AB and DC

The angle ∠BAD = 90°

The coordinates of the point D = Required

Let (x, y) represent the x and y-coordinates of the point D, by the given information, we get;

The slope of the line DC = The slope of the line AB

The slope of AB = (-2 - (-5))/(5 - (-1)) = 3/6 = 1/2

∴ The slope of CD, m = 1/2

From the point C(1, 1),the equation of the line CD is therefore;

y - 1 = (1/2)·(x - 1)

∴ y = x/2 - (1/2) + 1 = x/2 + 1/2

y = x/2 + 1/2

Given that ∠BAD is 90°, therefore, AD is perpendicular to DC and we have;

The slope of AD = -1/m

∴ The slope of AD = -1/(1/2) = -2

From the point A(-1, -5), the equation of the line AD is therefore;

y - (-5) = -2·(x - (-1))

y = -2·x - 2 - 5 = -2·x - 7

y = -2·x - 7

Equating both (simultaneous) values of y to find the value of x gives;

y = y, therefore;

x/2 + 1/2 = -2·x - 7

x/2 + 2·x = 5·x/2 = -7 - (1/2) = -15/2

∴ 5·x/2 = -15/2

x = (-15/2) × (2/5) = -3

x = -3

From y = -2·x - 7, and x = -3, we get;

y = -2 × (-3) - 7 = 6 - 7 = -1

The coordinates of the point D(x, y) = (-3, -1).

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step-by-step explanation: