4.4 Q-12

Question Statement

Find the interior angles of the quadrilateral whose vertices are $A(5,2), B(-2,3), C(-3,-4), D(4,-5)$.

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Background and Explanation

To solve this problem, we need to use the concept of the slope of a line. The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$

Next, we use the formula for the angle between two lines: $\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Where:

• $m_1$ and $m_2$ are the slopes of the two lines.
• $\theta$ is the angle between them.

If the angle’s tangent tends to infinity, this indicates a right angle ($90^\circ$).

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Solution

We are given the coordinates of the quadrilateral:

• $A(5, 2)$,
• $B(-2, 3)$,
• $C(-3, -4)$,
• $D(4, -5)$.

Step 1: Find the slopes of the sides

• Slope of line $\overline{AB}$:
  $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 2}{-2 - 5} = \frac{1}{-7}$

• Slope of line $\overline{BC}$:
  $m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 3}{-3 - (-2)} = \frac{-7}{-1} = 7$

• Slope of line $\overline{CD}$:
  $m_3 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - (-4)}{4 - 3} = \frac{1}{-7}$

• Slope of line $\overline{DA}$:
  $m_4 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 2}{4 - 5} = \frac{-7}{-1} = 7$

Step 2: Use the formula for the angle between two lines

We will calculate the angles between each pair of adjacent lines using the formula: $\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Angle at A:

$\tan(\angle A) = \left| \frac{m_4 - m_1}{1 + m_4 m_1} \right| = \frac{7 - \left(-\frac{1}{7}\right)}{1 + 7 \times \left(\frac{1}{7}\right)} = \frac{7 + \frac{1}{7}}{1 + 1} = \infty$
Thus, $\angle A = 90^\circ$.

Angle at B:

$\tan(\angle B) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \frac{-\frac{1}{7} - 7}{1 + \left(\frac{1}{7}\right)(7)} = \infty$
Thus, $\angle B = 90^\circ$.

Angle at C:

$\tan(\angle C) = \left| \frac{m_2 - m_3}{1 + m_2 m_3} \right| = \frac{7 - \left(-\frac{1}{7}\right)}{1 + 7 \times \left(\frac{1}{7}\right)} = \infty$
Thus, $\angle C = 90^\circ$.

Angle at D:

$\tan(\angle D) = \left| \frac{m_3 - m_4}{1 + m_3 m_4} \right| = \frac{-\frac{1}{7} - 7}{1 + \left(\frac{1}{7}\right)(7)} = \infty$
Thus, $\angle D = 90^\circ$.

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Key Formulas or Methods Used

• Slope of a line:
  $m = \frac{y_2 - y_1}{x_2 - x_1}$

• 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 all sides of the quadrilateral using the given coordinates.
2. Use the angle formula to calculate the angles between adjacent sides.
3. Determine that each angle is $90^\circ$.
4. Conclude that the quadrilateral is a rectangle.

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Hence, the quadrilateral $ABCD$ is a rectangle as all the interior angles are $90^\circ$.