Free math worksheets

- 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

- 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

- Match x- and y-coordinates to complete the axis-aligned rectangle. - Find the two side lengths from coordinate differences before using either measurement formula. - Keep linear units for perimeter and square units for area.

1. Point \(S\) has the same x-coordinate as \(P\) and the same y-coordinate as \(R\), so \(S=(1, 5)\). 2. The side lengths are \(9-1=8\,\text{cm}\) and \(5-1=4\,\text{cm}\). The perimeter is \(2(8+4)=24\,\text{cm}\). 3. The area is \(8\,\text{cm}\cdot4\,\text{cm}=32\,\text{cm}^2\).

1. \(S=(1, 5)\) 2. \(24\,\text{cm}\) 3. \(32\,\text{cm}^2\)

- 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.

- 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.

- 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 \(QA=\sqrt{17}\ne\sqrt{41}=QB\).

- 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 \(QA=\sqrt{17}\ne\sqrt{41}=QB\).