The figure created by plotting the points A(-1,-3), B(2, 4), C(-1, 6) and D (-4, 4) is a kite.The length of BC and DC isAnswer 1 Question 9The length of AB and AD isAnswer 2 Question 9

To answer your questions, we first need to calculate the distances between the points.

1. The length of BC and DC:

The distance between two points (x1, y1) and (x2, y2) can be calculated using the formula: √[(x2-x1)² + (y2-y1)²]

For BC: B(2,4) and C(-1,6) = √[(2-(-1))² + (4-6)²] = √[(3)² + (-2)²] = √[9 + 4] = √13

For DC: D(-4,4) and C(-1,6) = √[(-4-(-1))² + (4-6)²] = √[(-3)² + (-2)²] = √[9 + 4] = √13

So, the length of BC and DC is √13.

2. The length of AB and AD:

For AB: A(-1,-3) and B(2,4) = √[(2-(-1))² + (4-(-3))²] = √[(3)² + (7)²] = √[9 + 49] = √58

For AD: A(-1,-3) and D(-4,4) = √[(-4-(-1))² + (4-(-3))²] = √[(-3)² + (7)²] = √[9 + 49] = √58

So, the length of AB and AD is √58.

To find the lengths of the lines BC, DC, AB, and AD, we can use the distance formula, which is sqrt((x2-x1)^2 + (y2-y1)^2).

1. Length of BC: B(2,4) and C(-1,6) BC = sqrt((2-(-1))^2 + (4-6)^2) BC = sqrt((3)^2 + (-2)^2) BC = sqrt(9 + 4) BC = sqrt(13)

2. Length of DC: D(-4,4) and C(-1,6) DC = sqrt(((-4)-(-1))^2 + (4-6)^2) DC = sqrt((-3)^2 + (-2)^2) DC = sqrt(9 + 4) DC = sqrt(13)

So, the answer to Question 9, part 1 is that the lengths of BC and DC are both sqrt(13).

3. Length of AB: A(-1,-3) and B(2,4) AB = sqrt((2-(-1))^2 + (4-(-3))^2) AB = sqrt((3)^2 + (7)^2) AB = sqrt(9 + 49) AB = sqrt(58)

4. Length of AD: A(-1,-3) and D(-4,4) AD = sqrt(((-4)-(-1))^2 + (4-(-3))^2) AD = sqrt((-3)^2 + (7)^2) AD = sqrt(9 + 49) AD = sqrt(58)

So, the answer to Question 9, part 2 is that the lengths of AB and AD are both sqrt(58).