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 \( B D \).

To tackle this captivating quadrilateral problem, let's dive into each part step by step! a. The gradient (slope) of line \( AB \) is calculated using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). For points \( A(2,3) \) and \( B(1,8) \): \[ m_{AB} = \frac{8 - 3}{1 - 2} = \frac{5}{-1} = -5 \] Now for line \( DC \) between points \( D(3,6) \) and \( C(6,11) \): \[ m_{DC} = \frac{11 - 6}{6 - 3} = \frac{5}{3} \] Since the gradients \( -5 \) and \( \frac{5}{3} \) are not equal, it indicates that lines \( AB \) and \( DC \) are not parallel and will intersect at some point. b. We need to determine the equation of diagonal \( BD \) using the gradient and point-slope form. First, the gradient \( m_{BD} \) between \( B(1,8) \) and \( D(3,6) \): \[ m_{BD} = \frac{6 - 8}{3 - 1} = \frac{-2}{2} = -1 \] Using point-slope form with point \( B(1,8) \): \[ y - 8 = -1(x - 1) \\ y - 8 = -x + 1 \\ y = -x + 9 \] Thus, the equation for diagonal \( BD \) is \( y = -x + 9 \). c. To find the intersection of diagonals \( AC \) and \( BD \), we set their equations equal: From \( AC: y = x + 5 \) and \( BD: y = -x + 9 \): \[ x + 5 = -x + 9 \\ 2x = 4 \\ x = 2 \] Substituting \( x = 2 \) back into \( y = x + 5 \): \[ y = 2 + 5 = 7 \] Thus, the point of intersection is \( (2, 7) \). d. The midpoint of diagonal \( AC \) can be found with the midpoint formula, taking points \( A(2,3) \) and \( C(6,11) \): \[ M_{AC} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{2 + 6}{2}, \frac{3 + 11}{2} \right) = \left( \frac{8}{2}, \frac{14}{2} \right) = (4, 7) \] Comparing the midpoint \( (4, 7) \) with the intersection point \( (2, 7) \), we see that while they share the same \( y \)-coordinate, their \( x \)-coordinates differ. This shows that the diagonals don’t bisect each other in a symmetric manner, characteristic of some irregular quadrilaterals!