In a parallelogram, opposite sides are parallel and equal. If we consider $$MNOP$$ as our parallelogram, then $$\overrightarrow{MN} = \overrightarrow{PO}$$. Vector $$\overrightarrow{MN} = (6-2, 3-1) = (4, 2)$$. For $$\overrightarrow{PO} = (4, 2)$$, we need $$(4-x_P, 7-y_P) = (4, 2)$$, which gives us $$4-x_P = 4$$ and $$7-y_P = 2$$. Solving: $$x_P = 0$$ and $$y_P = 5$$. So $$P = (0, 5)$$. We can verify: $$\overrightarrow{NO} = (4-6, 7-3) = (-2, 4)$$ and $$\overrightarrow{MP} = (0-2, 5-1) = (-2, 4)$$. Since $$\overrightarrow{MN} = \overrightarrow{PO}$$ and $$\overrightarrow{NO} = \overrightarrow{MP}$$, we have a parallelogram. Choice B would work if we used diagonal properties but leads to incorrect coordinates. Choice C assumes perpendicularity which isn't required. Choice D doesn't satisfy parallelogram properties.

When you encounter a problem about finding where diagonals intersect in a quadrilateral, start by identifying what type of quadrilateral you're working with. Looking at the coordinates $$V(3, -2)$$, $$W(3, 4)$$, $$X(-1, 4)$$, and $$Y(-1, -2)$$, notice that points $$V$$ and $$W$$ have the same $$x$$-coordinate (3), and points $$X$$ and $$Y$$ have the same $$x$$-coordinate (-1). Similarly, $$W$$ and $$X$$ share $$y$$-coordinate (4), while $$V$$ and $$Y$$ share $$y$$-coordinate (-2). This means you have a rectangle with sides parallel to the coordinate axes.

In any rectangle, the diagonals intersect at the center point, which you can find by averaging the coordinates of opposite vertices. Taking opposite vertices $$V(3, -2)$$ and $$X(-1, 4)$$: $$x$$-coordinate = $$\frac{3 + (-1)}{2} = 1$$, and $$y$$-coordinate = $$\frac{-2 + 4}{2} = 1$$. This gives you point $$(1, 1)$$.

Choice A incorrectly calls $$(2, 1)$$ the midpoint of diagonal $$VX$$, but the actual midpoint calculation shows this is wrong. Choice C gives $$(1, 2)$$, which reverses the coordinates. Choice D claims diagonals always intersect at the origin, which is completely false—the intersection point depends on the quadrilateral's position.

Study tip: For rectangle problems, remember that the diagonals always intersect at the center, which you can find by averaging the coordinates of any pair of opposite vertices. This shortcut saves time compared to finding equations of diagonal lines.

First, we identify that $$WXYZ$$ forms a rectangle with vertices at $$W(-2, -1)$$, $$X(3, -1)$$, $$Y(3, 3)$$, and $$Z(-2, 3)$$. The rectangle extends from $$x = -2$$ to $$x = 3$$ and from $$y = -1$$ to $$y = 3$$. Point $$P(0, 1)$$ has coordinates within these ranges, so it's inside the rectangle. The center of the rectangle is at $$\left(\frac{-2+3}{2}, \frac{-1+3}{2}\right) = (0.5, 1)$$. Since $$P$$ is at $$(0, 1)$$, it's inside but not at the center. Choice A is wrong because $$P$$ is inside. Choice B is wrong because $$P$$ is not at the center $$(0.5, 1)$$. Choice D is wrong because $$P$$ is not on any side of the rectangle.

When you encounter problems involving right triangles and rectangles on a coordinate plane, start by identifying the key measurements and visualizing how the shapes relate to each other.

Looking at the given vertices, triangle $$RST$$ has its right angle at $$R(0,0)$$, with one leg extending horizontally from $$R$$ to $$S(8,0)$$ and another leg extending vertically from $$R$$ to $$T(0,6)$$. The horizontal leg has length $$8$$ units, and the vertical leg has length $$6$$ units. Since the rectangle shares vertices $$R$$ and $$S$$ with the triangle and uses the triangle's legs as two of its sides, the rectangle must have dimensions $$8 \times 6$$. Therefore, its area is $$8 \times 6 = 48$$ square units.

Choice A incorrectly suggests adding areas together, but the question asks for the rectangle's area alone, not a combined area. Choice B makes an error about the relationship between the rectangle and triangle areas - the rectangle actually has twice the triangle's area ($$48$$ vs $$24$$), not half of some other rectangle's area. Choice C miscalculates by seemingly squaring one dimension ($$8^2 = 64$$), but rectangles require multiplying length times width, not squaring a single measurement.

The correct answer is D: $$48$$ square units, calculated by multiplying the lengths of the triangle's legs.

Study tip: When working with coordinate geometry problems, always plot the points or sketch the figure first. This helps you visualize relationships between shapes and avoid calculation errors.

When reflecting across $$x = -1$$, point $$A(-4, 2)$$ reflects to $$A'(2, 2)$$ because it's $$3$$ units left of the line, so it goes $$3$$ units right: $$-1 + 3 = 2$$. Point $$B(2, 2)$$ reflects to $$B'(-4, 2)$$ because it's $$3$$ units right of the line, so it goes $$3$$ units left: $$-1 - 3 = -4$$. Point $$C(-1, 6)$$ reflects to $$C'(-1, 6)$$ because it's on the line of reflection. Now $$A'B'$$ connects $$(2, 2)$$ and $$(-4, 2)$$. Since both points have the same $$y$$-coordinate, the distance is $$|2 - (-4)| = 6$$ units. This is the same as the original length $$AB$$, confirming that reflections preserve distances. Choice B incorrectly suggests distances change. Choice C gives an incorrect calculation. Choice D incorrectly applies the distance formula.

This question tests drawing polygons on a coordinate plane given vertex coordinates, using coordinates to find horizontal and vertical side lengths (|x₂ - x₁| if y-coordinates are the same, |y₂ - y₁| if x-coordinates are the same). To draw the parallelogram, plot the vertices as ordered pairs A(0,0), B(4,0), C(5,3), and D(1,3), connect consecutive vertices with line segments, and close the polygon by connecting back to A. For side lengths of axis-aligned sides, the horizontal side AB has the same y-coordinate of 0, length |4 - 0| = 4 using the x-difference, while vertical sides aren't present but diagonal sides like BC use Pythagorean (not for grade 6); absolute value ensures positive length. For this parallelogram, the length of horizontal side AB is 4 units, matching choice B. A common error is calculating a diagonal like AC with simple difference instead of recognizing it's not axis-aligned, or arithmetic wrong like |4-0|=3. To draw accurately: (1) identify axes and scale, (2) plot each vertex, (3) connect in order, and (4) verify closed. For side lengths: (1) check if horizontal or vertical, (2) use the right difference, (3) calculate correctly, and note grade 6 focuses on axis-aligned, avoiding mistakes like not closing or reversing coordinates.

This question tests drawing polygons on a coordinate plane given vertex coordinates, using coordinates to find horizontal and vertical side lengths (|x₂-x₁| if y-coordinates are the same, |y₂-y₁| if x-coordinates are the same). To draw the rectangle, plot the vertices as ordered pairs A(2,1), B(7,1), C(7,5), and D(2,5), connect consecutive vertices with line segments, and close the polygon by connecting the last back to the first. Finding side lengths for axis-aligned sides: horizontal sides AB and DC have the same y-coordinates (y=1 and y=5), so length = |7-2| = 5 using x-coordinate difference; vertical sides BC and DA have the same x-coordinates (x=7 and x=2), so length = |5-1| = 4 using y-difference, with absolute value ensuring positive length. For example, the perimeter is calculated by adding all sides: 5 + 4 + 5 + 4 = 18 units, or 2 × (length + width) = 2 × (5 + 4) = 18 units. The correct answer is 18 units, as it matches the perimeter calculation for the dog park. A common error is miscalculating the differences, like |7-2| = 4 instead of 5, leading to an incorrect perimeter. To draw accurately: (1) identify axes and scale, (2) plot each vertex by counting x units right or left from the origin and y units up or down, (3) connect in order from A to B to C to D and back to A, (4) verify the shape is closed; for perimeter, sum the calculated side lengths using the appropriate differences and absolute values.

This question tests drawing polygons on a coordinate plane given vertex coordinates, using coordinates to find horizontal and vertical side lengths with formulas like |x₂ - x₁| if y-coordinates are the same or |y₂ - y₁| if x-coordinates are the same. To draw the rectangle, plot the vertices as ordered pairs A(-3,2), B(4,2), C(4,-1), and D(-3,-1), connect consecutive vertices with line segments in order, and close the polygon by connecting the last point back to the first. For side lengths, the horizontal side AB from (-3,2) to (4,2) has the same y=2, so length = |4 - (-3)| = 7 units using the x-difference, while vertical sides like BC from (4,2) to (4,-1) have the same x=4, so length = |-1 - 2| = 3 units using the y-difference, with absolute values ensuring positive lengths. For this ramp rectangle, the correct length of side AB is 7 units. A common error is forgetting to add the absolute value for negative coordinates, such as calculating 4 - (-3) as 1 instead of 7, or using y-differences for horizontal sides. To draw accurately: (1) identify the axes and scale, including negative values, (2) plot each vertex carefully, (3) connect in order, and (4) verify closure. For side lengths: (1) identify horizontal or vertical alignment, (2) use the appropriate difference, (3) calculate with absolute value, noting applications like material needs for the ramp, and avoiding mistakes like coordinate reversal or arithmetic errors.

This question tests drawing polygons on a coordinate plane given vertex coordinates, using coordinates to find horizontal and vertical side lengths (|x₂ - x₁| if y-coordinates are the same, |y₂ - y₁| if x-coordinates are the same). To draw the rectangle, plot the vertices as ordered pairs P(1,-4), Q(1,2), R(5,2), and S(5,-4), connect consecutive vertices with line segments in the order P to Q to R to S back to P, and close the polygon. For side lengths of axis-aligned sides, vertical sides like P to Q have the same x-coordinate of 1, length |2 - (-4)| = 6 using the y-difference, and horizontal sides like Q to R have the same y-coordinate of 2, length |5 - 1| = 4 using the x-difference; absolute value ensures positive length. For this rectangle, plot and connect to find lengths 6 (vertical), 4 (horizontal), 6 (vertical), and 4 (horizontal), so the area is base × height = 4 × 6 = 24 square units, matching choice A. A common error is plotting coordinates reversed, like treating (1,-4) as (-4,1), or using x-difference for vertical sides, leading to wrong area like 6 × 6 = 36. To draw accurately: (1) identify axes and scale, (2) plot each vertex carefully, (3) connect in order, and (4) verify closed. For applications like area: calculate side lengths first, then multiply for rectangles, while for perimeter sum them (here 6 + 4 + 6 + 4 = 20 units), avoiding arithmetic errors or forgetting to close the polygon.

This question tests drawing polygons on a coordinate plane given vertex coordinates, using coordinates to find horizontal and vertical side lengths (|x₂-x₁| if y-coordinates are the same, |y₂-y₁| if x-coordinates are the same). To draw the rectangle, plot the vertices as ordered pairs (-4,-1), (2,-1), (2,3), and (-4,3), connect consecutive vertices with line segments, and close the polygon by connecting the last back to the first. Finding side lengths for axis-aligned sides: the horizontal side from (-4,-1) to (2,-1) has the same y-coordinate (y=-1), so length = |2 - (-4)| = |6| = 6 using x-coordinate difference; absolute value ensures positive length. For example, if the points were (1,1) and (5,1), the length would be |5-1| = 4 units. The correct length of the specified horizontal side is 6 units. A common error is forgetting the absolute value and getting a negative length like 2 - (-4) = 6 but misapplying signs, or using y-differences instead of x for horizontal sides. To calculate side lengths: (1) identify if horizontal or vertical by checking coordinates, (2) use the appropriate difference (|x₂-x₁| for horizontal), (3) compute the value, (4) ensure it's positive with absolute value.