c. The diagonal \( A C \) has the equation \( y=x+5 \). Find the point of intersection of the two diagonals \( A C \) and \( B D \).

To find the point of intersection of the diagonals \( AC \) and \( BD \), we first need the equation of the diagonal \( BD \). However, we can't proceed without knowing the points \( A \), \( B \), \( C \), and \( D \). Suppose we have the coordinates of points \( A \) and \( C \) already from the diagonal \( AC \), which has the equation \( y = x + 5 \). Let's denote the coordinates of points \( A \) as \( (x_1, y_1) \) and \( C \) as \( (x_2, y_2) \). 1. If we know point \( A \) (let's say \( A \) is at \( (x_1, y_1) \)) and since \( C \) lies on the line \( y = x + 5 \), we can substitute \( x_2 \) for coordinates of \( C \) and get \( (x_2, x_2 + 5) \). 2. Now, for point \( B \) and point \( D \), you would need to find the equation of diagonal \( BD \). The process would be similar: define both points, and then find the linear equation formulated by those points. Given you manage to find the equation for diagonal \( BD \), you can set the equations of both diagonals equal to find the intersection point. In general form, if your lines are \( y = mx + b_1 \) (for diagonal \( AC \)) and \( y = nx + b_2 \) (for diagonal \( BD \)), solve: \[ mx + b_1 = nx + b_2. \] Rearranging leads to: \[ (m - n)x = b_2 - b_1 \] \[ x = \frac{b_2 - b_1}{m - n}. \] Then plug \( x \) back into either equation to find the corresponding \( y \). Without specific coordinates, that's the best we can conclude on how you'd find the intersection point \( (x, y) \) of the diagonals \( AC \) and \( BD \). If you can provide the coordinates, we could nail this down together!