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 \(AP:PB=2:3\).

- Create equal reference intervals on a ray from one endpoint. - Connect the last reference point to the other endpoint of the segment. - Use parallel lines to transfer the equal spacing to \(\overline{AB}\).

1. The app creates an auxiliary ray from \(A\) that is not collinear with \(\overline{AB}\). 2. Using one fixed length, the app marks five consecutive congruent segments on the ray, ending at points \(P_1,P_2,P_3,P_4,P_5\). 3. The app draws \(\overline{P_5B}\). 4. Through each of \(P_1,P_2,P_3,P_4\), the app creates a line parallel to \(\overline{P_5B}\). Let these lines meet \(\overline{AB}\) at \(Q_1,Q_2,Q_3,Q_4\). 5. Parallel lines cut transversals proportionally. Since the five segments on the auxiliary ray are congruent, the corresponding segments \(AQ_1\), \(Q_1Q_2\), \(Q_2Q_3\), \(Q_3Q_4\), and \(Q_4B\) are congruent.

The app marks five equal steps on an auxiliary ray from \(A\), connects the fifth point to \(B\), and creates parallels through the first four marked points. Their intersections divide \(\overline{AB}\) into five congruent parts by the proportional-segments theorem.

- 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 Priya'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. Priya'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, \(Q=(-2, 2)+\frac{3}{5}(10, 5)=(4, 5)\).

\(Q=(4, 5)\)