- Read the coordinates of both endpoints from the graph. - The midpoint is halfway between the endpoint coordinates in both directions. - Use the midpoint formula on the x-coordinates and y-coordinates separately.

1. Average the x-coordinates: \(\frac{-2+6}{2}=2\). 2. Average the y-coordinates: \(\frac{1+1}{2}=1\). 3. Therefore, \(M=(2, 1)\).

\(M=(2, 1)\)

- Compare the x-coordinates of the two points first. - For a vertical segment, only the change in the y-coordinate determines its length. - Distance is nonnegative, so use an absolute difference.

1. The points have the same x-coordinate, so \(\overline{PQ}\) is vertical. 2. The distance is the absolute difference of the y-coordinates: \(|4-(-2)|=6\).

\(6\) units

- Which coordinate stays fixed along a vertical segment? - What y-coordinate is the same distance from both endpoint y-coordinates?

1. The endpoints lie on the same vertical line at \(x=3\), so the midpoint has that x-coordinate. 2. The y-coordinates are \(-5\) and \(3\). Their average is \(\frac{-5+3}{2}=-1\). 3. Therefore the midpoint is \((3, -1)\).

\((3, -1)\). The midpoint's y-coordinate is \(\frac{-5+3}{2}=-1\), and its x-coordinate remains \(3\).

- Does a change in coordinates name a destination or a movement? - Try taking the average of each pair of endpoint coordinates.

1. The pair \((6, -8)\) tells how far the point moved: 6 right and 8 down. It is not a location halfway between the endpoints. 2. The midpoint averages the endpoint coordinates: \(M=(\frac{-8+(-2)}{2},\frac{3+(-5)}{2})=(-5, -1)\). 3. The two endpoint changes from \(M\) are \((3, -4)\), so \(M\) is halfway.

No. \((6, -8)\) is the displacement; the midpoint is \((-5, -1)\).

- In an axis-aligned rectangle, which coordinates are shared by vertically or horizontally aligned vertices? - The midpoint coordinates are the averages of the corresponding endpoint coordinates. - Check the x-coordinates and y-coordinates separately.

a) In rectangle \(ABCD\), point \(D\) must have the same x-coordinate as \(A\) and the same y-coordinate as \(C\). Therefore, \(D=(2, 8)\). b) Use the midpoint formula on \(A(2, 2)\) and \(B(10, 2)\): \(M=\left(\frac{2+10}{2},\frac{2+2}{2}\right)=(6, 2)\).

a) \(D=(2, 8)\) b) \(M=(6, 2)\)

- First check whether each pair of points is aligned horizontally or vertically. - For a pair that is not horizontally or vertically aligned, compare both coordinate changes. - Keep the final oblique distance exact rather than converting it to a decimal.

1. Points \(A\) and \(B\) have the same x-coordinate, so \(AB=|9-2|=7\) units. 2. Points \(A\) and \(C\) have the same y-coordinate, so \(AC=|6-1|=5\) units. 3. For \(B(1, 9)\) and \(C(6, 2)\), use the distance formula: \(BC=\sqrt{(6-1)^2+(2-9)^2}=\sqrt{25+49}=\sqrt{74}\) units.

a) \(7\) units b) \(5\) units c) \(\sqrt{74}\) units

- Determine the constant coordinate for each horizontal or vertical side. - A perpendicular distance to a horizontal side comes from the y-coordinates; to a vertical side it comes from the x-coordinates. - Compare the four distances only after computing each one.

1. Side \(AB\) lies on \(y=2\), so the distance from \(P=(6,3)\) to \(AB\) is \(|3-2|=1\) unit. 2. Side \(BC\) lies on \(x=9\), so the distance is \(|9-6|=3\) units. 3. Side \(CD\) lies on \(y=8\), so the distance is \(|8-3|=5\) units. 4. Side \(DA\) lies on \(x=2\), so the distance is \(|6-2|=4\) units. 5. The smallest distance is \(1\) unit, so \(\overline{AB}\) is closest to \(P\).

a) \(AB:1\) unit; \(BC:3\) units; \(CD:5\) units; \(DA:4\) units b) \(\overline{AB}\)

- Determine the equation of line \(g\). - A perpendicular to a horizontal line is vertical. - The closest point on \(g\) has the same x-coordinate as \(S\).

a) Line \(g\) is horizontal with equation \(y=1\). The distance from \(S(6, 5)\) to \(g\) is \(5-1=4\) units. b) A perpendicular to a horizontal line is vertical, so \(F\) has the same x-coordinate as \(S\). Since \(F\) lies on \(g\), \(F=(6, 1)\). c) Points \(A(2, 1)\) and \(F(6, 1)\) lie on a horizontal line, so \(AF=6-2=4\) units.

a) \(4\) units b) \(F=(6, 1)\) c) \(4\) units

- Read the coordinates of the three displayed vertices from the graph. - Determine the missing rectangle vertex before finding the midpoint. - A midpoint coordinate is the average of the corresponding endpoint coordinates.

1. From the graph, \(O=(0, 0)\), \(A=(8, 0)\), and \(C=(0, 6)\). 2. The fourth vertex of the rectangle is \(B=(8, 6)\). 3. Apply the midpoint formula to \(B(8, 6)\) and \(C(0, 6)\): \(M=\left(\frac{8+0}{2},\frac{6+6}{2}\right)=(4, 6)\).

\(M=(4, 6)\)

- Use the fact that the diagonals of a rhombus bisect each other. - Place half of each diagonal length on both sides of the origin along its stated axis. - Check the requested clockwise labeling order before assigning the four names.

1. The diagonals of a rhombus bisect each other, so the origin is the midpoint of both diagonals. 2. Half of the horizontal diagonal is \(6\), giving endpoints \((-6, 0)\) and \((6, 0)\). 3. Half of the vertical diagonal is \(5\), giving endpoints \((0, 5)\) and \((0, -5)\). 4. Following the requested clockwise order gives \(A(-6, 0)\), \(B(0, 5)\), \(C(6, 0)\), and \(D(0, -5)\).

\(A(-6, 0)\), \(B(0, 5)\), \(C(6, 0)\), and \(D(0, -5)\)

- Read the coordinates of both endpoints from the graph. - The distance formula combines the horizontal and vertical coordinate changes. - Simplify the square root only after combining the two squared differences.

1. From the graph, \(A=(1, 2)\) and \(B=(7, 10)\). 2. Use the distance formula: \(AB=\sqrt{(7-1)^2+(10-2)^2}\). 3. Thus, \(AB=\sqrt{36+64}=\sqrt{100}=10\) units.

\(10\) units

- Read both endpoint coordinates from the graph. - Use both the horizontal and vertical coordinate changes because the segment is oblique. - Leave the distance in exact radical form if the square root does not simplify.

1. From the graph, \(A=(-2, 1)\) and \(B=(3, 5)\). 2. The coordinate changes are \(3-(-2)=5\) horizontally and \(5-1=4\) vertically. 3. By the distance formula, \(AB=\sqrt{5^2+4^2}=\sqrt{41}\) units.

\(\sqrt{41}\) units

- Compare what an average of endpoint coordinates represents with what a coordinate difference represents. - If you compute a displacement from \(A\), ask what must be done to that displacement to locate an actual point. - Check a proposed midpoint by seeing whether it lies halfway between the endpoint x-coordinates and halfway between the endpoint y-coordinates.

1. The midpoint coordinates are the averages of the endpoint coordinates: \(\left(\frac{-6+2}{2},\frac{4+10}{2}\right)=(-2,7)\). Thus Leo is correct. 2. The coordinate change from \(A\) to \(B\) is \((8,6)\). Maya's \((4,3)\) is half of that displacement. 3. A displacement is not automatically a point coordinate. Starting at \(A\) and adding the half-displacement gives \((-6,4)+(4,3)=(-2,7)\), the actual midpoint.

Leo is correct: the midpoint is \((-2,7)\). Maya's \((4,3)\) is the half-displacement from \(A\) toward \(B\), not the midpoint coordinate itself.

- Does the definition of midpoint involve position on the segment as well as distance? - What do you get by averaging the endpoint coordinates? - What other points can have equal distances to two fixed endpoints?

1. The midpoint must lie on \(AB\) as well as being equidistant from \(A\) and \(B\). 2. The coordinate averages give \((\frac{-1+7}{2},\frac{2+8}{2})=(3, 5)\), which is \(Q\). 3. Point \(Q\) is \(5\) units from either endpoint. Point \(P\) is \(\sqrt{50}\) units from either endpoint but is not on the segment, so equal distances do not guarantee a midpoint.

\(Q=(3, 5)\) is the midpoint. Although \(P=(0, 9)\) is equidistant from the endpoints, it is not on segment \(AB\).

- Which two numbers have the midpoint's x-coordinate as their average? - Work backward separately in the x- and y-directions. - Check by finding the midpoint of your two endpoints.

1. The midpoint coordinates are the averages of the two endpoints, so \(-2=\frac{-7+x_B}{2}\) and \(-3=\frac{4+y_B}{2}\). 2. Solving gives \(x_B=3\) and \(y_B=-10\). 3. Checking gives \(\frac{-7+3}{2}=-2\) and \(\frac{4+(-10)}{2}=-3\), as required.

\(B=(3, -10)\). Check: \(\frac{-7+3}{2}=-2\) and \(\frac{4+(-10)}{2}=-3\).

- What coordinate rule locates a point halfway along a straight segment? - First find the change from the starting entrance in grid units. - Convert each coordinate change using the map's scale.

1. The midpoint is \((\frac{-4+8}{2},\frac{2+10}{2})=(2, 6)\). 2. From \(P\) to the midpoint is a change of \(6\) grid units east and \(4\) north. 3. At \(20\,\text{yd}\) per unit, that is \(120\,\text{yd}\) east and \(80\,\text{yd}\) north.

a) \((2, 6)\) b) \(120\,\text{yd}\) east and \(80\,\text{yd}\) north

- Determine the equation of the horizontal line. - Points on the same vertical grid line have the same x-coordinate. - A point can be above or below a horizontal line and have the same distance from it.

a) Line \(h\) is horizontal with equation \(y=5\). The distance from \(R(6, 8)\) to \(h\) is \(|8-5|=3\) units. b) Because \(S\) lies on the same vertical line as \(R\), its x-coordinate is \(6\). A point \(2\) units from \(y=5\) can have y-coordinate \(5+2=7\) or \(5-2=3\). Thus, the two points are \(S_1(6, 7)\) and \(S_2(6, 3)\).

a) \(3\) units b) \(S_1(6, 7)\) and \(S_2(6, 3)\)

- Identify the equations and directions of \(g\) and \(h\). - Find the x-coordinate halfway between the two parallel lines. - A fixed positive perpendicular distance from a line produces two parallel loci, one on each side.

1. Line \(g\) is \(x=3\), and line \(h\) is \(x=7\). Both are vertical and parallel. 2. Points equidistant from two parallel lines lie on the parallel line halfway between them. The halfway x-coordinate is \((3+7)\div2=5\), so the locus is \(x=5\). 3. Points \(2\,\text{cm}\) from \(g\) lie on two vertical lines. Their x-coordinates are \(3-2=1\) and \(3+2=5\), so the lines are \(x=1\) and \(x=5\).

a) The points lie on the vertical line \(x=5\), halfway between \(g\) and \(h\). b) The points lie on the vertical lines \(x=1\) and \(x=5\).

- Read the three given vertex coordinates from the graph before calculating. - Opposite sides of a parallelogram have the same direction vector. - For each midpoint, average the corresponding endpoint coordinates.

1. From the graph, \(A=(2, 1)\), \(B=(8, 1)\), and \(C=(10, 5)\). 2. The vector from \(B\) to \(C\) is \((2, 4)\). Apply the same vector to \(A\): \(D=(2+2, 1+4)=(4, 5)\). 3. The midpoint of \(A(2, 1)\) and \(B(8, 1)\) is \(P=\left(\frac{2+8}{2},\frac{1+1}{2}\right)=(5, 1)\). 4. The midpoint of \(P(5, 1)\) and \(C(10, 5)\) is \(Q=\left(\frac{5+10}{2},\frac{1+5}{2}\right)=(7.5, 3)\).

a) \(D=(4, 5)\) b) \(P=(5, 1)\) c) \(Q=(7.5, 3)\)

- Read the four vertex coordinates from the graph first. - Apply the midpoint formula separately to each requested segment. - Compare the two final midpoint coordinate pairs rather than judging by appearance.

1. From the graph, \(A=(0, 0)\), \(B=(2, 6)\), \(C=(8, 6)\), and \(D=(10, 0)\). 2. \(P=\left(\frac{0+2}{2},\frac{0+6}{2}\right)=(1, 3)\), and \(Q=\left(\frac{8+10}{2},\frac{6+0}{2}\right)=(9, 3)\). 3. \(M_1=\left(\frac{0+10}{2},\frac{0+0}{2}\right)=(5, 0)\), and \(M_2=\left(\frac{2+8}{2},\frac{6+6}{2}\right)=(5, 6)\). 4. The midpoint of \(\overline{PQ}\) is \(\left(\frac{1+9}{2},\frac{3+3}{2}\right)=(5, 3)\). The midpoint of \(\overline{M_1M_2}\) is \(\left(\frac{5+5}{2},\frac{0+6}{2}\right)=(5, 3)\). The two midpoints coincide.

1. \(P=(1, 3)\), \(Q=(9, 3)\) 2. \(M_1=(5, 0)\), \(M_2=(5, 6)\) 3. Both final midpoints are \((5, 3)\).

- Read the endpoint and midpoint coordinates from the graph. - Each midpoint coordinate is the average of the corresponding endpoint coordinates. - Treat the two unknown coordinates of \(B\) separately, then check that \(M\) is halfway between the endpoints.

1. From the graph, \(A=(-2, 3)\) and \(M=(4, -1)\). 2. If \(B=(x, y)\), the midpoint equations are \(\frac{-2+x}{2}=4\) and \(\frac{3+y}{2}=-1\). 3. Solving gives \(x=10\) and \(y=-5\), so \(B=(10, -5)\).

\(B=(10, -5)\)

- Use each midpoint coordinate to form an equation for the corresponding unknown endpoint coordinate. - After recovering the endpoint, compare both coordinate changes from \(A\) to \(B\). - Keep the distance exact and simplify the radical if possible.

1. Let \(B=(x,y)\). From the midpoint formula, \(\frac{-6+x}{2}=1\) and \(\frac{1+y}{2}=5\). 2. Solving gives \(x=8\) and \(y=9\), so \(B=(8,9)\). 3. Now \(AB=\sqrt{(8-(-6))^2+(9-1)^2}=\sqrt{14^2+8^2}=\sqrt{260}=2\sqrt{65}\) units.

a) \(B=(8, 9)\) b) \(2\sqrt{65}\) units

- Read all four coordinates from the graph before comparing the candidates. - For each candidate, compute its distance to both endpoints of \(\overline{AB}\). - The relevant criterion is equality of the two endpoint distances, not visual closeness in the drawing.

1. From the graph, \(A=(-4,1)\), \(B=(4,5)\), \(P=(1,1)\), and \(Q=(0,0)\). 2. For \(P\), \(PA=\sqrt{(1-(-4))^2+(1-1)^2}=5\) and \(PB=\sqrt{(1-4)^2+(1-5)^2}=5\). Thus, \(P\) is equidistant from \(A\) and \(B\). 3. For \(Q\), \(QA=\sqrt{(0-(-4))^2+(0-1)^2}=\sqrt{17}\), while \(QB=\sqrt{(0-4)^2+(0-5)^2}=\sqrt{41}\). These are not equal. 4. Therefore, \(P\), and not \(Q\), lies on the perpendicular bisector of \(\overline{AB}\).

\(P\) lies on the perpendicular bisector because \(PA=PB=5\). Point \(Q\) does not because \(QA=\sqrt{17}\ne\sqrt{41}=QB\).