Pregunta
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2. The points \( A(-2,3), B(1,8), C(6,11) \) and \( D(3,6) \) are the vertices of a quadrilateral. a. Find the gradients of \( A B \) and \( D C \). What does this tell you about the lines \( A B \) and \( D C \) ? b. Find the equation of the diagonal joining \( B \) to \( D \). c. The diagonal \( A C \) has the equation \( y=x+5 \). Find the point of intersection of the two diagonals \( A C \) and \( B D \). d. Find the midpoint of \( A C \) and compare this point with the point of intersection of the diagonals \( A C \) and \( B D \).

Ask by Hall Fuentes. in New Zealand
Jan 21,2025

Solución de tutoría real

Respuesta verificada por el tutor

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- **a.** Lines \( AB \) and \( DC \) are parallel because they have the same slope. - **b.** The equation of diagonal \( BD \) is \( y = -x + 9 \). - **c.** The diagonals \( AC \) and \( BD \) intersect at the point \( (2, 7) \). - **d.** The midpoint of \( AC \) is \( (2, 7) \), which is the same as the intersection point of the diagonals, showing that the diagonals bisect each other.

Solución

Let's solve each part of the problem step by step. --- ### **a. Find the gradients of \( AB \) and \( DC \). What does this tell you about the lines \( AB \) and \( DC \)?** **Given Points:** - \( A(-2, 3) \) - \( B(1, 8) \) - \( C(6, 11) \) - \( D(3, 6) \) **Gradient of \( AB \):** \[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 3}{1 - (-2)} = \frac{5}{3} \] **Gradient of \( DC \):** \[ m_{DC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 11}{3 - 6} = \frac{-5}{-3} = \frac{5}{3} \] **Conclusion:** Both lines \( AB \) and \( DC \) have the same gradient of \( \frac{5}{3} \). This implies that **lines \( AB \) and \( DC \) are parallel**. --- ### **b. Find the equation of the diagonal joining \( B \) to \( D \) (Line \( BD \)).** **Given Points:** - \( B(1, 8) \) - \( D(3, 6) \) **Step 1: Calculate the Gradient of \( BD \)** \[ m_{BD} = \frac{6 - 8}{3 - 1} = \frac{-2}{2} = -1 \] **Step 2: Use the Point-Slope Form to Find the Equation** Using point \( B(1, 8) \): \[ y - y_1 = m(x - x_1) \\ y - 8 = -1(x - 1) \\ y - 8 = -x + 1 \\ y = -x + 9 \] **Equation of Diagonal \( BD \):** \[ \boxed{y = -x + 9} \] --- ### **c. The diagonal \( AC \) has the equation \( y = x + 5 \). Find the point of intersection of the two diagonals \( AC \) and \( BD \).** **Equations of Diagonals:** - \( AC: y = x + 5 \) - \( BD: y = -x + 9 \) **Find Intersection Point:** \[ x + 5 = -x + 9 \\ 2x = 4 \\ x = 2 \] \[ y = 2 + 5 = 7 \] **Point of Intersection:** \[ \boxed{(2,\ 7)} \] --- ### **d. Find the midpoint of \( AC \) and compare this point with the point of intersection of the diagonals \( AC \) and \( BD \).** **Given Points:** - \( A(-2, 3) \) - \( C(6, 11) \) **Midpoint Formula:** \[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] \[ \text{Midpoint of } AC = \left( \frac{-2 + 6}{2}, \frac{3 + 11}{2} \right) = \left( \frac{4}{2}, \frac{14}{2} \right) = (2,\ 7) \] **Comparison:** The midpoint of \( AC \) is **\( (2,\ 7) \)**, which is **exactly the same as the point of intersection of the diagonals \( AC \) and \( BD \)**. This indicates that the diagonals bisect each other at the midpoint, a property characteristic of **parallelograms**. --- **Summary:** - **a.** Both \( AB \) and \( DC \) have a gradient of \( \frac{5}{3} \), indicating they are parallel. - **b.** The equation of diagonal \( BD \) is \( y = -x + 9 \). - **c.** The diagonals \( AC \) and \( BD \) intersect at the point \( (2, 7) \). - **d.** The midpoint of \( AC \) is \( (2, 7) \), which coincides with the intersection point of the diagonals, confirming that the diagonals bisect each other.

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Extra Insights

For part (a), let's calculate the gradients (slopes) of \( AB \) and \( DC \). The gradient of line segment \( AB \) is given by: \[ \text{Gradient of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 3}{1 - (-2)} = \frac{5}{3} \] For line segment \( DC \): \[ \text{Gradient of } DC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - 6}{6 - 3} = \frac{5}{3} \] Since both gradients are equal (\( \frac{5}{3} \)), the lines \( AB \) and \( DC \) are parallel. --- Now for part (b), we need to find the equation of the diagonal joining \( B \) to \( D \). First, we calculate the gradient of \( BD \): \[ \text{Gradient of } BD = \frac{6-8}{3-1} = \frac{-2}{2} = -1 \] Using point-slope form, the equation can be formed. The equation with point \( B(1,8) \) is: \[ y - 8 = -1(x - 1) \] This rearranges to: \[ y = -x + 9 \] --- For part (c), to find the point of intersection of the diagonals \( AC \) (given as \( y = x + 5 \)) and \( BD \) (which we found to be \( y = -x + 9 \)), we set the equations equal to each other: \[ x + 5 = -x + 9 \] Solving this: \[ 2x = 4 \Rightarrow x = 2 \] Substituting \( x = 2 \) into \( y = x + 5 \): \[ y = 2 + 5 = 7 \] Thus, the point of intersection is \( (2, 7) \). --- Now for part (d), we need to find the midpoint of \( AC \). The midpoint \( M \) of segment \( AC \) with points \( A(-2,3) \) and \( C(6,11) \) is calculated as follows: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-2 + 6}{2}, \frac{3 + 11}{2} \right) = \left( \frac{4}{2}, \frac{14}{2} \right) = (2, 7) \] Comparing this with the intersection point of the diagonals \( (2, 7) \), we find that they are the same! So, the midpoint of \( AC \) bisects the intersection point of \( BD \). How cool is that? Geometry has its little magic moments!

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