请讲解上传的题目---24. As shown in the figure, it is known that the line y = kx - 6k passes through points A and B, and if S△OAB = 9.
(1) Find the value of k.
(2) If point C is a point on line segment OA, rotate line segment CB clockwise around point C by 90° to get CD, at this time point D happens to lie on line AB.
① Find the coordinates of points C and D.
② Let the line BE be the reflection of line AB across the y-axis, intersecting the x-axis at point E. If point P is on line BE, and Q is on line AB, does there exist a quadrilateral with vertices C, D, P, Q that is a parallelogram? If it exists, directly write down the coordinates of all points Q that satisfy the condition; if it does not exist, please explain the reason.
**Diagram 1 Description:**
* Type: Coordinate plane diagram.
* Main Elements:
* Coordinate Axes: X-axis (labeled 'x') and Y-axis (labeled 'y') intersecting at the origin O.
* Points: O (origin), A is on the positive x-axis, B is in the second quadrant.
* Line: A straight line segment connecting points B and A passes through the first quadrant.
**Diagram 2 Description:**
* Type: Coordinate plane diagram.
* Main Elements:
* Coordinate Axes: X-axis (labeled 'x') and Y-axis (labeled 'y') intersecting at the origin O.
* Points: O (origin), A is on the positive x-axis, E is on the negative x-axis, B is in the second quadrant.
* Lines: A straight line passing through points B, O, and A. Another straight line passing through points B and E. Both lines intersect at B and O. (Correction: The second diagram shows lines AB and BE. Line AB passes through A, B, and O. Line BE passes through B, E, and O).
* Labels: O, A, B, E, x (axis label), y (axis label).
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这是一道关于直线方程和三角形面积的综合题目。已知直线方程为 y = kx - 6k,可以写成 y = k(x - 6)的形式。当y等于0时,x等于6,所以直线必过点A(6,0)。点B在第二象限,三角形OAB的面积为9。我们需要利用面积公式来求出k的值。
现在我们来详细求解k的值。首先,直线方程y等于k乘以x减6k可以写成y等于k乘以括号x减6。当y等于0时,x等于6,所以点A的坐标是6逗号0。三角形OAB的面积等于二分之一乘以底边乘以高,底边OA长度为6,高为点B到x轴的距离。由面积等于9,可得高等于3。因为B在第二象限,所以y坐标为3。将B点坐标代入直线方程,经过计算可得k等于1减根号2。
现在解决第二问。点C在线段OA上,线段CB绕点C顺时针旋转90度得到CD,且点D恰好落在直线AB上。通过分析可知,当C为原点时满足条件。此时向量CB为括号3减3根号2逗号3,顺时针旋转90度后得到向量CD为括号3逗号负3加3根号2。因此C的坐标是括号0逗号0,D的坐标是括号3逗号负3加3根号2。
现在我们构造反射直线BE。直线BE是直线AB关于y轴的反射,其方程为y等于负的括号1减根号2乘以x减6乘以括号1减根号2。直线BE与x轴的交点E的坐标为负6逗号0。接下来我们需要寻找平行四边形CDPQ,其中P在直线BE上,Q在直线AB上。通过平行四边形的性质,对角线互相平分,我们可以找到满足条件的点。
最终我们得到了完整的答案。存在满足条件的平行四边形CDPQ,其中所有满足条件的点Q的坐标为:第一个是括号0逗号6根号2减6,第二个是括号6逗号0。这两个点对应着两个不同的平行四边形。我们通过平行四边形对角线互相平分的性质,分别考虑了三种可能的对角线组合,最终确定了这两个解。