要求制作一个手写动画。不要漏步骤，讲慢一点。在白板上用红笔写，写字和声音输出同步，声音用男声大概30岁左右，普通话。---**Extracted Content:** **Question Stem:** 3. As shown in the figure, in the Cartesian coordinate system xOy, the vertices A and C of rectangle OABC have coordinates (0,6) and (10,0) respectively. Points P and Q are moving points on sides OA and OC respectively (not coinciding with endpoints). Point D, the reflection of point O across line PQ, happens to fall on side AB. Connect OD, which intersects PQ at point G. Draw a line through point D parallel to OA, intersecting PQ and OQ at points E and F respectively. Let the coordinates of point E be (m,n). **Sub-questions:** (1) Find the function expression for n in terms of m, and directly state the range of the independent variable m. (2) When the area of triangle OPG (S_△OPG) is equal to the area of triangle EFQ (S_△EFQ), find the function expression for line PQ. (3) Connect DQ, find the range of values for DQ · DA (dot product of vectors DQ and DA). **Chart/Diagram Description:** * **Type:** A geometric figure displayed on a 2D Cartesian coordinate system. * **Coordinate Axes:** * X-axis: Labeled 'x', extends horizontally to the right from the origin O. * Y-axis: Labeled 'y', extends vertically upwards from the origin O. * Origin: Labeled 'O'. * **Main Elements:** * **Rectangle OABC:** * Vertex O is the origin (0,0). * Vertex A is on the y-axis. * Vertex C is on the x-axis. * Vertex B is in the first quadrant. * **Points:** * O: Origin. * A: A point on the positive y-axis, forming a vertex of the rectangle. * B: A vertex of the rectangle, opposite to O. * C: A point on the positive x-axis, forming a vertex of the rectangle. * P: A point on the line segment OA (y-axis). * Q: A point on the line segment OC (x-axis). * D: A point on the line segment AB. It is also the reflection of O across the line segment PQ. * G: The intersection point of line segments OD and PQ. * E: A point on the line segment PQ. It is also an intersection point of the vertical line through D and PQ. Its coordinates are (m,n). * F: A point on the line segment OC (x-axis). It is also an intersection point of the vertical line through D and OC. * **Lines/Segments:** * Line segment PQ: Connects points P and Q. * Line segment OD: Connects points O and D. * Line segment DF: A vertical line segment from D to F on the x-axis (OC), parallel to OA. * Line segment AB: Part of the top side of the rectangle. * Line segment OA: Part of the left side of the rectangle (y-axis). * Line segment OC: Part of the bottom side of the rectangle (x-axis). * **Labels and Annotations:** All points (O, A, B, C, D, E, F, G, P, Q) and axes (x, y) are labeled. * **Relative Position and Direction:** * P is between O and A. * Q is between O and C. * D is on AB. * PQ bisects OD perpendicularly. * G is on OD and PQ. * The line segment DF is perpendicular to the x-axis and passes through D and F. E lies on this line segment and on PQ. The image contains handwritten mathematical problems and their solutions/derivations, presented in a structured format. --- **Chart/Diagram Description:** * **Type:** Geometric figure on a Cartesian coordinate system. * **Main Elements:** * **Coordinate Axes:** X-axis and Y-axis are shown, with the origin labeled 'O'. The positive X-axis extends to point 'C', and the positive Y-axis extends to point 'A'. * **Points:** * **O:** Origin (0,0). * **A:** A point on the Y-axis. Its y-coordinate is indicated as 6. (Implied A(0,6)). * **C:** A point on the X-axis. Its x-coordinate is indicated as 'm'. (Implied C(m,0)). * **D:** A point in the first quadrant, forming a rectangle OADC with O, A, and C. Its coordinates are given as D(m,6) in the text. * **B:** Another point in the first quadrant. Its coordinates are given as B(m,n) in the text. Visually, it appears vertically aligned with D and C, and horizontally aligned with some 'n' on the Y-axis. * **P:** A point on the line segment OA (Y-axis). * **Q:** A point on the line segment OC (X-axis). * **F:** An intersection point, visually located on the line segment OD. * **E:** An intersection point, visually located on the line segment AD. * **G:** A point G(m/2, 3) is referenced in the calculations, though not explicitly labeled on the diagram. * **Lines/Shapes:** * A rectangle OADC with vertices O(0,0), A(0,6), D(m,6), C(m,0). * A line segment PQ connecting point P on OA and point Q on OC. * Other visible line segments include OA, OC, AD, and OD. * **Labels and Annotations:** Points O, A, B, C, D, P, Q, F, E are labeled. The x-coordinate 'm' is indicated on the X-axis (near C), and the y-coordinate '6' is indicated on the Y-axis (near A). 'n' is labeled near point B. --- **Extracted Content:** **Problem/Derivation <1>** * **Given Points:** * B(m,n) * D(m,6) * G(m/2, 3) * **Conditions/Derivations:** * PQ ⊥ OD (Line PQ is perpendicular to line OD) * kOD = 6/m * kPQ * kPA = -1 (This seems like a typo, should be kPQ * kOD = -1 if PQ ⊥ OD. Given the next line, it probably means kPQ * kOD = -1, which makes kPQ = -m/6) * ∴ kPA = -m/6 (This is kPQ, derived from kOD * kPQ = -1) * E and G are on PQ (E和G在PQ上) * Therefore, kPQ = (n-3) / (m - m/2) = (n-3) / (m/2) = 2(n-3)/m * Equating the slopes: 2(n-3)/m = -m/6 * ∴ 2(n-3) = -m^2/6 * And n = 3 - m^2/12 * **Conditions on P and Q:** * P is on OA (P在OA上) * Let P(0, p), then 0 < p < 6 * By symmetry property, PD = PB (由对称性质 PD=PB) * ∴ p = sqrt(m^2 + (6-p)^2) (This is actually the distance formula for PD=PB, simplified to find p) * The equation simplifies to p = (m^2+36)/12 * Since p < 6, we have (m^2+36)/12 < 6 * m^2+36 < 72 * m^2 < 36 * ∴ 0 < m < 6 (Because m is a length, it must be positive) * **Q is on OC (Q在OC上):** * Let Q(q, 0), then 0 < q < 10 * QO = QD * Therefore, q = sqrt((q-m)^2 + 6^2) * The equation simplifies to q = (m^2+36)/(2m) * Since q < 10, we have (m^2+36)/(2m) < 10 * m^2+36 < 20m * m^2 - 20m + 36 < 0 * Factoring the quadratic: (m-2)(m-18) < 0 * ∴ 2 < m < 18 * **Combined Conclusion:** * 综上 2 < m < 6. (Combining 0 < m < 6 and 2 < m < 18, the valid range for m is 2 < m < 6.) **Problem/Derivation <2>** * **Given Points (from previous context):** * O(0,0) * P(0,p) * G(m/2, 3) * **Area Calculations:** * SΔOPG = 1/2 * base * height = 1/2 * p * (m/2) = pm/4 (Area of triangle OPG) * SΔEFQ = 1/2 * (q-m) * n (Area of triangle EFQ) - assuming EF is parallel to y-axis and F has x-coordinate m. 'n' is the y-coordinate for a point on EF. * **System of Equations (relating variables from Problem <1>):** * Equating areas: pm/4 = 1/2 * (q-m) * n * q = (m^2+36)/(2m) * p = (m^2+36)/12 * n = 3 - m^2/12 * **Solution for m, n, p, q:** * Substituting p, q, n into the area equation and solving leads to: * => m = 2√3 * At this point: (此时) * n = 3 - (2√3)^2/12 = 3 - 12/12 = 3 - 1 = 2 * p = ((2√3)^2+36)/12 = (12+36)/12 = 48/12 = 4 * q = ((2√3)^2+36)/(2 * 2√3) = (12+36)/(4√3) = 48/(4√3) = 12/√3 = 4√3 * **Equation of Line PQ:** * ∴ PQ: y = (-√3/3)x + 4 (Derived using P(0,4) and Q(4√3, 0)) **Problem/Derivation <3>** * **Given Points/Lengths:** * D(m,6) * A(0,6) * DA = m (Length of segment DA, which is the difference in x-coordinates for A and D) * QD = QO = q = (m^2+36)/(2m) (From Problem <1>) * **Calculation:** * ∴ DQ * DA = m * (m^2+36)/(2m) = (m^2+36)/2 * **Range of Values:** * The previous result for m: 2 < m < 6 * Squaring the inequality: 4 < m^2 < 36 * Substitute this into the expression for DQ * DA: * (4+36)/2 < (m^2+36)/2 < (36+36)/2 * 40/2 < (m^2+36)/2 < 72/2 * 20 < (m^2+36)/2 < 36 * Therefore: 即 20 < DQ * DA < 36