Free math worksheets

- Read the three positions from left to right on the graph. - Compare the length from \(A\) to \(P\) with the length from \(P\) to \(B\). - Keep the ratio in the order requested, then simplify it.

1. From the graph, \(A=(0, 0)\), \(P=(2, 0)\), and \(B=(8, 0)\). 2. Therefore, \(AP=2\) units and \(PB=6\) units. 3. The ratio is \(2:6=1:3\).

\(1:3\)

- Count the equal grid intervals between each pair of labeled points. - For \(AP:PB\), compare the length from \(A\) to \(P\) with the length from \(P\) to \(B\). - For \(AP:AB\), compare \(AP\) with the entire segment.

1. Count the grid intervals from \(A\) to \(P\): \(AP=2\) units. 2. Count the grid intervals from \(P\) to \(B\): \(PB=3\) units. Therefore, \(AP:PB=2:3\). 3. The whole segment has \(AB=5\) units, so \(AP:AB=2:5\).

\(AP:PB=2:3\) \(AP:AB=2:5\)

- Add the two ratio parts to determine what fraction of the whole segment lies from \(A\) to \(P\). - Find the total horizontal and vertical changes from \(A\) to \(B\). - Apply the same fraction to both coordinate changes before starting from \(A\).

1. The ratio \(2:3\) has \(5\) total parts, so \(P\) is \(\frac{2}{5}\) of the way from \(A\) to \(B\). 2. The coordinate change from \(A\) to \(B\) is \((10, 10)\). Taking \(\frac{2}{5}\) of that change gives \((4, 4)\). 3. Starting at \(A\), \(P=(-3+4, 2+4)=(1, 6)\).

\(P=(1, 6)\)

- Convert the part-to-part ratio into the fraction of the entire segment measured from \(A\). - Find the displacement vector from \(A\) to \(B\). - Take the required fraction of that displacement and add it to the starting endpoint.

1. The ratio \(2:3\) means that \(P\) is \(\frac{2}{5}\) of the way from \(A\) to \(B\). 2. The displacement from \(A\) to \(B\) is \((7, 5)\), so \(\frac{2}{5}(7, 5)=(\frac{14}{5}, 2)\). 3. Add this displacement to \(A\): \((-1, 3)+(\frac{14}{5}, 2)=(\frac{9}{5}, 5)\).

\(P=(\frac{9}{5}, 5)\)

- Translate the ratio into a fraction of the whole segment measured from \(A\). - Compare each candidate point with the corresponding fraction of the displacement from \(A\) to \(B\). - Be careful about the order of the two parts in the ratio.

1. From the graph, \(A=(0, 0)\), \(B=(10, 5)\), \(P=(4, 2)\), and \(Q=(6, 3)\). 2. A \(2:3\) division places the point \(\frac{2}{5}\) of the way from \(A\) to \(B\). 3. \(\frac{2}{5}(10, 5)=(4, 2)\), which is the position of \(P\). 4. Therefore, \(P\) gives \(AP:PB=2:3\). Point \(Q\) is \(\frac{3}{5}\) of the way from \(A\), so it gives the reversed ratio \(3:2\).

\(P\). It is \(\frac{2}{5}\) of the way from \(A\) to \(B\), so \(AP:PB=2:3\).

- Read the coordinates of all three collinear points from the graph. - Compare the coordinate changes from \(A\) to \(P\) and from \(P\) to \(B\). - The common direction factor cancels when the two segment lengths are put in a ratio.

1. From the graph, \(A=(1, -2)\), \(P=(7, 4)\), and \(B=(11, 8)\). 2. From \(A\) to \(P\), the coordinate change is \((6, 6)\), so \(AP=6\sqrt{2}\). 3. From \(P\) to \(B\), the coordinate change is \((4, 4)\), so \(PB=4\sqrt{2}\). 4. Therefore, \(AP:PB=6\sqrt{2}:4\sqrt{2}=3:2\).

\(AP:PB=3:2\)

- Convert each part-to-part ratio into a fraction of the whole segment measured from \(A\). - Apply each fraction to the same coordinate change from \(A\) to \(B\). - After finding \(P\) and \(Q\), use the midpoint formula twice and compare the results.

a) From the graph, \(A=(-6, 0)\) and \(B=(6, 6)\). Point \(P\) is \(\frac{1}{3}\) of the way from \(A\) to \(B\), so \(P=(-2, 2)\). Point \(Q\) is \(\frac{2}{3}\) of the way from \(A\) to \(B\), so \(Q=(2, 4)\). b) The midpoint of \(\overline{PQ}\) is \(\left(\frac{-2+2}{2},\frac{2+4}{2}\right)=(0, 3)\). The midpoint of \(\overline{AB}\) is \(\left(\frac{-6+6}{2},\frac{0+6}{2}\right)=(0, 3)\). The two midpoints coincide.

a) \(P=(-2, 2)\), \(Q=(2, 4)\) b) Both midpoints are \((0, 3)\).

- Translate \(AP:PB=2:3\) into the fraction of the whole segment traveled from \(A\). - Compare that fraction with the endpoint weights in the student's expression. - Check the proposed point by locating what fraction of the displacement from \(A\) to \(B\) it represents.

1. The ratio \(AP:PB=2:3\) places \(P\) \(\frac{2}{5}\) of the way from \(A\) to \(B\), so the endpoint weights are reversed: \(P=\frac{3A+2B}{5}\). 2. Equivalently, \(B-A=(15, 5)\), and \(A+\frac{2}{5}(B-A)=(-4, 2)+(6, 2)=(2, 4)\). 3. The student's formula gives \(\frac{2A+3B}{5}=(5, 5)\), which is \(\frac{3}{5}\) of the way from \(A\) to \(B\) and therefore corresponds to the reversed ratio \(3:2\).

The weighting is reversed. The correct point is \(P=(2, 4)\), using \(P=\frac{3A+2B}{5}\).

- Translate the part-to-part ratio into the fraction of the whole segment from \(A\) to \(P\). - Work backward from \(\overrightarrow{AP}\) to the full vector \(\overrightarrow{AB}\). - Add the full coordinate change to endpoint \(A\) and check that \(P\) has the required ratio.

1. The ratio \(3:2\) means \(P\) is \(\frac{3}{5}\) of the way from \(A\) to \(B\). Thus, \(\overrightarrow{AP}=\frac{3}{5}\overrightarrow{AB}\). 2. Therefore, \(\overrightarrow{AB}=\frac{5}{3}\overrightarrow{AP}\). 3. Since \(\overrightarrow{AP}=(6, -6)\), \(\overrightarrow{AB}=\frac{5}{3}(6, -6)=(10, -10)\). 4. Add this vector to \(A\): \(B=(-3+10, 4-10)=(7, -6)\).

\(B=(7, -6)\)

- Treat the two partition conditions one at a time. - First locate \(P\) as a fraction of the way from \(A\) to \(B\). - Then use \(P\) as the new starting endpoint when applying the second ratio.

1. Since \(AP:PB=2:5\), point \(P\) is \(\frac{2}{7}\) of the way from \(A\) to \(B\). This gives \(P=(-2, 2)\). 2. The displacement from \(P\) to \(B\) is \((10, 5)\). 3. Since \(PQ:QB=3:2\), point \(Q\) is \(\frac{3}{5}\) of the way from \(P\) to \(B\). 4. Thus, \(Q=(-2, 2)+\frac{3}{5}(10, 5)=(4, 5)\).

\(Q=(4, 5)\)