The length of the diagonal AC can be found using the distance formula, \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). For points A(2, 1) and C(6, 6), the length is \(\sqrt{(6-2)^2 + (6-1)^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}\).

Let the coordinates of B be \((x, y)\). The midpoint M is the average of the coordinates of A and B. So, \(3 = \frac{-1+x}{2}\) and \(4 = \frac{6+y}{2}\). Solving for x: \(6 = -1+x\), so \(x=7\). Solving for y: \(8 = 6+y\), so \(y=2\). The coordinates of point B are (7, 2).

The diagonals connect opposite vertices. The first diagonal connects (1,1) and (5,5). Its slope is \(\frac{5-1}{5-1} = \frac{4}{4} = 1\). The second diagonal connects (1,5) and (5,1). Its slope is \(\frac{1-5}{5-1} = \frac{-4}{4} = -1\). The sum of the slopes is \(1 + (-1) = 0\). This is a property of any rhombus, including a square, whose diagonals are not horizontal and vertical.

First, find the center of the original square. The center is the midpoint of a diagonal. Using the diagonal from (1,1) to (4,4), the midpoint is \((\frac{1+4}{2}, \frac{1+4}{2}) = (2.5, 2.5)\). The translation moves every point \((x, y)\) to \((x-3, y+5)\). Applying this translation to the center's coordinates: \((2.5 - 3, 2.5 + 5) = (-0.5, 7.5)\).

When you encounter a triangle classification problem with given vertices, you need to determine two things: the side lengths (to classify by sides) and whether any angles are 90° (to determine if it's a right triangle).

First, calculate the distances between each pair of vertices using the distance formula $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$:

• Side AB: $$\sqrt{(5-(-1))^2 + (-1-(-1))^2} = \sqrt{36 + 0} = 6$$
• Side BC: $$\sqrt{(2-5)^2 + (3-(-1))^2} = \sqrt{9 + 16} = 5$$
• Side AC: $$\sqrt{(2-(-1))^2 + (3-(-1))^2} = \sqrt{9 + 16} = 5$$

Since two sides have equal length (BC = AC = 5), this is an isosceles triangle.

To check if it's a right triangle, use the Pythagorean theorem. For a right triangle, $$a^2 + b^2 = c^2$$. Testing with the longest side (6) as the hypotenuse: $$5^2 + 5^2 = 25 + 25 = 50$$, but $$6^2 = 36$$. Since $$50 \neq 36$$, this is not a right triangle.

Now examining the answer choices: Choice A is wrong because the triangle isn't equilateral (all sides aren't equal) or right. Choice B is incorrect because while it's isosceles, it's not right. Choice C is wrong because it's not scalene (all different sides) or right. Choice D correctly identifies the triangle as isosceles and not right.

Remember: Always calculate all three side lengths first to classify by sides, then use the Pythagorean theorem to test for right angles.

The side lengths of the rectangle are 'a' and 'b'. The area is \(ab = 24\). The perimeter is \(2(a+b) = 20\), which simplifies to \(a+b = 10\). We need to find two numbers that multiply to 24 and add up to 10. Let's test the options. A) 83=24, 8+3=11 (Incorrect). B) 122=24, 12+2=14 (Incorrect). C) 64=24, 6+4=10 (Correct). D) 55=25 (Incorrect).

The center of the circle is the midpoint of its diameter. The midpoint formula is \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\). Using the endpoints (-5, 8) and (3, -2), the center is \((\frac{-5+3}{2}, \frac{8+(-2)}{2}) = (\frac{-2}{2}, \frac{6}{2}) = (-1, 3)\).

When you encounter a trapezoid with given coordinates, you need to identify the parallel sides and calculate the area using the formula: Area = $$\frac{1}{2}(b_1 + b_2) \times h$$, where $$b_1$$ and $$b_2$$ are the lengths of the parallel bases and $$h$$ is the height.

First, plot or visualize the points: A(1, 2), B(9, 2), C(7, 6), and D(3, 6). Notice that points A and B both have y-coordinate 2, making AB a horizontal line. Similarly, points C and D both have y-coordinate 6, making CD horizontal. These are your parallel bases.

Calculate the base lengths:

• Base AB: distance from (1, 2) to (9, 2) = $$9 - 1 = 8$$ units
• Base CD: distance from (3, 6) to (7, 6) = $$7 - 3 = 4$$ units

The height is the perpendicular distance between the parallel lines: $$6 - 2 = 4$$ units.

Now apply the area formula: Area = $$\frac{1}{2}(8 + 4) \times 4 = \frac{1}{2}(12)(4) = 24$$ square units.

Answer choice A (22) likely results from calculation errors in adding the bases. Answer choice B (48) comes from forgetting the $$\frac{1}{2}$$ in the formula and calculating $$(8 + 4) \times 4 = 48$$. Answer choice C (32) might result from using $$8 \times 4 = 32$$, treating it like a rectangle with only one base.

Remember: Always identify which sides are parallel first, then carefully apply the trapezoid area formula with the $$\frac{1}{2}$$ factor.

A reflection across the y-axis transforms a point \((x, y)\) to \((-x, y)\). The x-coordinate changes sign, and the y-coordinate remains the same. Applying this rule to vertex R(2, 7), the new coordinates will be (-2, 7).

To determine the relationship between the points, we can check the slopes between them. The slope between A and B is \(\frac{8-2}{3-1} = \frac{6}{2} = 3\). The slope between B and C is \(\frac{14-8}{5-3} = \frac{6}{2} = 3\). Since the slope between A and B is the same as the slope between B and C, the three points lie on the same straight line. Therefore, they are collinear and do not form a triangle.