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**Question 21:**
(8 分) (2024 七年级·湖南长沙·期末) 在平面直角坐标系中, 已知点 A(a,0), B(b,3), C(4,0), 且满足 $|a+b| + (a-b+6)^2 = 0$. 线段 AB 交 y 轴于点 F, 点 D 是 y 轴正半轴上的一点.
**Diagram Description:**
* **Figure 1:**
* Type: Geometric figure on a coordinate plane.
* Elements: X-axis and Y-axis intersecting at origin O. Points A, F, B, C are plotted. Line segment AB intersects the Y-axis at F. Line segment AC is on the X-axis. BC is a line segment forming a triangle ABC. Point A is on the negative X-axis, O is the origin, C is on the positive X-axis. Point B is in the first quadrant. Point F is on the positive Y-axis.
* Labels: x, y, O, A, F, B, C, 图 1.
* **Figure 2:**
* Type: Geometric figure on a coordinate plane, similar to Figure 1 but with additional elements.
* Elements: X-axis and Y-axis intersecting at origin O. Points A, F, B, C are plotted. D is a point on the positive Y-axis. M is a point inside triangle ABD. Lines/segments: AB, AC, BC, AF, BF, AD, BD, AM, DM, MN (dashed line). N is a point on the negative X-axis. There are angle bisector markings on ∠CAB and ∠ODB.
* Labels: x, y, O, A, F, D, M, B, C, N, 图 2, α is marked as ∠CAB.
* **Figure 3:**
* Type: Geometric figure on a coordinate plane, similar to Figure 1.
* Elements: X-axis and Y-axis intersecting at origin O. Points A, F, B, C are plotted. Line segment AB intersects the Y-axis at F. Line segment AC is on the X-axis. BC is a line segment forming a triangle ABC. Point A is on the negative X-axis, O is the origin, C is on the positive X-axis. Point B is in the first quadrant. Point F is on the positive Y-axis.
* Labels: x, y, O, A, F, B, C, 图 3.
**Question 21 Sub-questions:**
(1) 求出点 A, B 的坐标;
(2) 如图 2, 若 DB∥AC, ∠BAC = α, AM, DM 分别平分∠CAB, ∠ODB, 求∠AMD (用含 α 的代数式表示);
(3) 如图 3, 坐标轴上是否存在一点 P, 使得 △ABP 的面积和 △ABC 的面积相等? 若存在请求出 P 点坐标; 若不存在, 请说明理由.
**Question 22:**
(8 分) (2024 七年级·四川凉山·期末) 某体育用品店准备购进甲、乙两种品牌跳绳, 若购买甲种跳绳 10 根, 乙种跳绳 5 根, 需要 100 元, 若购买甲种跳绳 5 根, 乙种跳绳 3 根, 需要 55 元.
**Other Relevant Text:**
* 第 5 页 共 6 页 (Page 5 of 6)
**Handwritten Notes (Likely scratch work):**
* |a+b| + (a-b+6)^2 = 0
* { a+b = 0, a-b+6 = 0 }
* { a = -b, -b-b+6 = 0 }
* { a = -b, -2b = -6 }
* { a = -b, b = 3 }
* { a = -3, b = 3 }
* ∴ A (-3, 0)
* B (3, 3)
* C (4, 0)
* a+b=0 (a+b)2=0
* ∴ a+b=0 a+b=0
* (2)
* ∠BAM = ∠CAM = 1/2∠CAB
* BD || AC
* ∴ ∠BDO + ∠DOA = 180 (likely a mistake in angle notation)
* ∴ BD || AC ∴ ∠BDP + ∠DPC = 180 (likely a mistake in angle notation)
* ∠ODB
* DM 平分 ∠ODB
* ∠ODM = 1/2 ∠ODB
* ∠BDP = 90 (possibly BD ⊥ y-axis, but BD || AC || x-axis implies BD is horizontal, not vertical, maybe related to some other line)
* ∠BDF + ∠FDO = ∠BDO
* ∠BDM = ∠FMD ? ± ∠BDF ? 45° (unclear)
* ∠BAM = ∠CAM = 1/2 α
* ∠AMD = ?
* ∠ADM = ?
* ∠LDMA = ∠DMNT?N?? (unclear)
* = 45° + 1/2 α B (likely an expression for ∠AMD)
* 45° + 1/2 α (repetition)
* BP⊥y轴 (BP perpendicular to y-axis)
* (3)
* 存在点 P (Point P exists)
* △ABP 的面积 = △ABC 的面积
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Let's solve this coordinate geometry problem step by step. We have points A at coordinates (a,0), B at (b,3), and C at (4,0). The key constraint is that the absolute value of a plus b, plus the square of (a minus b plus 6), equals zero. Since both terms are non-negative, each must equal zero individually. This gives us a plus b equals zero, and a minus b plus 6 equals zero. Solving this system: from the first equation, a equals negative b. Substituting into the second equation: negative b minus b plus 6 equals zero, so negative 2b equals negative 6, giving us b equals 3. Therefore a equals negative 3. So point A is at (-3,0) and point B is at (3,3).
Now let's solve part 2. We have DB parallel to AC, angle BAC equals alpha, AM bisects angle CAB, and DM bisects angle ODB. We need to find angle AMD in terms of alpha. Since DB is parallel to AC, we can use properties of parallel lines. Since AM bisects angle CAB, angle CAM equals angle BAM, which is alpha over 2. Since DB is parallel to AC, angle BDO equals 90 degrees. Since DM bisects angle ODB, angle ODM equals angle BDM, which is 45 degrees. Using angle relationships in triangle AMD, we can show that angle AMD equals 45 degrees plus alpha over 2.
For part 3, we need to find points P on the coordinate axes such that triangle ABP has the same area as triangle ABC. First, let's calculate the area of triangle ABC. Using the coordinates A(-3,0), B(3,3), and C(4,0), the area is 21 over 2. For triangle ABP to have the same area, point P must be positioned so that the distance from P to line AB creates the same area. There are four possible positions: P1 at (0,6) and P2 at (0,-6) on the y-axis, and P3 at (7,0) and P4 at (-11,0) on the x-axis. Each of these points creates a triangle with area equal to 21 over 2.
Now let's solve problem 22, which is about jump rope pricing. We have a system of linear equations: 10 type A ropes plus 5 type B ropes cost 100 yuan, and 5 type A ropes plus 3 type B ropes cost 55 yuan. Let a be the price of type A and b be the price of type B. First, we simplify the first equation by dividing by 5 to get 2a plus b equals 20. From this, we can express b as 20 minus 2a. Substituting into the second equation: 5a plus 3 times (20 minus 2a) equals 55. Expanding: 5a plus 60 minus 6a equals 55. Simplifying: negative a equals negative 5, so a equals 5 yuan. Substituting back: b equals 20 minus 2 times 5, which equals 10 yuan. Let's verify: 10 times 5 plus 5 times 10 equals 100, and 5 times 5 plus 3 times 10 equals 55. Both equations check out.
Let's summarize our solutions to both problems. For problem 21, we found that point A is at (-3,0) and point B is at (3,3) by solving the constraint equation. For the angle bisector problem, we determined that angle AMD equals 45 degrees plus alpha over 2. For the equal area problem, we found four possible positions for point P: (0,6), (0,-6), (7,0), and (-11,0). For problem 22, the jump rope pricing problem, we solved the system of linear equations to find that type A costs 5 yuan and type B costs 10 yuan. These problems showcase important mathematical concepts including absolute value equations, parallel line properties, triangle area calculations, and systems of linear equations. Coordinate geometry provides powerful tools for solving both theoretical and practical mathematical problems.