4.4 Q-13

Question Statement

Show that the points $A(-1,-1)$, $B(-3,0)$, $C(3,7)$, and $D(1,8)$ are the vertices of a trapezium. Additionally, find its interior angles.

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Background and Explanation

To solve this problem, we need to know a few key concepts:

1. Slope of a line: The slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$
2. Trapezium: A trapezium (also called a trapezoid in some regions) is a quadrilateral with at least one pair of opposite sides parallel.

The general approach is to calculate the slopes of the sides of the quadrilateral formed by the points and check for parallel sides. If two opposite sides are parallel, we can confirm it’s a trapezium.

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Solution

Let’s break the solution into steps:

Step 1: Calculate the slopes of the sides

Using the formula for the slope of a line, we calculate the slopes of the four sides of the quadrilateral:

1. Slope of $\overline{AB}$:

$$m_1 = \frac{y_B - y_A}{x_B - x_A} = \frac{0 - (-1)}{-3 - (-1)} = \frac{1}{-2} = -\frac{1}{2}$$

2. Slope of $\overline{BD}$:

$$m_2 = \frac{y_D - y_B}{x_D - x_B} = \frac{8 - 0}{1 - (-3)} = \frac{8}{4} = 2$$

3. Slope of $\overline{CD}$:

$$m_3 = \frac{y_D - y_C}{x_D - x_C} = \frac{8 - 7}{1 - 3} = \frac{1}{-2} = -\frac{1}{2}$$

4. Slope of $\overline{AC}$:

$$m_4 = \frac{y_C - y_A}{x_C - x_A} = \frac{7 - (-1)}{3 - (-1)} = \frac{8}{4} = 2$$

Step 2: Check for parallel sides

We observe that:

• $m_1 = -\frac{1}{2}$ (slope of $\overline{AB}$) is equal to $m_3 = -\frac{1}{2}$ (slope of $\overline{CD}$).
• $m_2 = 2$ (slope of $\overline{BD}$) is equal to $m_4 = 2$ (slope of $\overline{AC}$).

Since the opposite sides are parallel, we can conclude that $ABCD$ is a trapezium.

Step 3: Calculate the interior angles

We now need to find the interior angles of the trapezium. We can use the tangent formula for the angle between two lines:

$$\tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$$

Angle at point A:

To find $\angle A$, we use the slopes $m_3$ and $m_4$ (since they correspond to the lines $\overline{AC}$ and $\overline{AD}$):

$$\tan(\angle A) = \left[\frac{m_3 - m_4}{1 + m_3 m_4}\right] = \frac{-\frac{1}{2} - 2}{1 + \left(\frac{1}{2}\right)(2)} = \frac{-\frac{1}{2} - 2}{1 - 1} = \infty$$

Since the tangent of the angle is infinite, we conclude that:

$$\angle A = 90^\circ$$

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Key Formulas or Methods Used

1. Slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$

2. Angle between two lines: $\tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

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Summary of Steps

1. Calculate the slopes of the four sides of the quadrilateral.
2. Check for parallel sides by comparing slopes.
3. Conclude that the figure is a trapezium because opposite sides are parallel.
4. Use the tangent formula to find the angle at point $A$, which is $90^\circ$.