1. For the quadrilateral $A B C D$, the following is true: point $A$ has coordinates $(-2,-3)$, point $B$ has coordinates $(1,-5)$, point $C$ is the mirror image of point $E(-6,1)$ when reflected over the coordinate origin, the abscissa of point $D$ is the solution to the equation $t^{2}-6 t+9=0$, and the ordinate of this point is the opposite value of half the sum of the ordinates of points $A$ and $B$. Draw a sketch and calculate the lengths of the diagonals of the quadrilateral $A B C D$.
1. We reflect the point $E(-6,1)$ over the coordinate origin and get the point $C(6,-1)$. We solve the equation $t^{2}-6 t+9=0$ and calculate the abscissa of point $D$, which is equal to 3. The ordinate of point $D$ is $-\frac{(-3-5)}{2}=4$. Using the formula for calculating the distance between points, we calculate the lengths of both diagonals, specifically $d(A, C)=\sqrt{82+22}=\sqrt{68}$ and $? d(B, D)=\sqrt{22+92}=\sqrt{85}$.

Sketch 1 point

Coordinates of point $C(6,-1)$ 1 point Abscissa of point $D\left(x_{D}=3\right) \ldots \ldots \ldots . . .1$ point Ordinate of point $D\left(y_{D}=4\right) \ldots \ldots . \ldots . .1$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_93a0a0489a8a740fd9e9g-09.jpg?height=617&width=776&top_left_y=300&top_left_x=1071)
Calculation of the length of diagonal $A C d(A, C)=\sqrt{68}$ 1 point Calculation of diagonal $B D d(B, D)=\sqrt{85}$. .1 point