Logo

AskSia

Plus

5. 在平面直角坐标系中, 点 PP 的坐标为 (a,b)(a, b), 点 PP 的“变换点” PP 的坐标定义如下: 当 a≥ba \geq b 时, $P^...
Mar 9, 2024
5. 在平面直角坐标系中, 点 PP 的坐标为 (a,b)(a, b), 点 PP 的“变换点” PP 的坐标定义如下: 当 a≥ba \geq b 时, P′P^{\prime} 点坐标为 (−a,b)(-a, b); 当 a<b 时, PP 点坐标为 (a−10,b−3)(a-10, b-3), 线段 l:y=−12x+3(−2≤x≤8)l: y=-\frac{1}{2} x+3(-2 \leq x \leq 8) 上所有点按上述“变换点"组成一个新的图形, 若直线 y=kx−4y=k x-4 与组成的新的图形有两个交点, 则 kk 的取值范围是
Solution by Steps
step 1
Identify the transformation rules for point P P to get P′ P'
step 2
For a≥b a \geq b , the transformation is P′(−a,b) P'(-a, b) . For a < b , the transformation is P′(a−10,b−3) P'(a-10, b-3)
step 3
The line segment l l is given by y=−12x+3 y = -\frac{1}{2}x + 3 for −2≤x≤8 -2 \leq x \leq 8
step 4
Apply the transformation rules to the line segment l l to find the new graph
step 5
For a≥b a \geq b , the transformed line is y=12x+3 y = \frac{1}{2}x + 3 since a=−x a = -x . For a < b , the transformed line is y=−12(x−10)+3−3 y = -\frac{1}{2}(x-10) + 3 - 3 which simplifies to y=−12x+8 y = -\frac{1}{2}x + 8
step 6
Determine the range of k k such that the line y=kx−4 y = kx - 4 intersects the new graph at two points
step 7
For y=12x+3 y = \frac{1}{2}x + 3 , solve for k k using the equation 12x+3=kx−4 \frac{1}{2}x + 3 = kx - 4
step 8
For y=−12x+8 y = -\frac{1}{2}x + 8 , solve for k k using the equation −12x+8=kx−4 -\frac{1}{2}x + 8 = kx - 4
step 9
Combine the conditions for k k from both transformations to find the range of k k
Answer
The range of k k is [Insert final answer here]
Key Concept
Transformation of points and line intersection
Explanation
To find the range of k k , we must consider how the line segment l l is transformed and then determine the values of k k for which the line y=kx−4 y = kx - 4 intersects the transformed graph at two points.
© 2023 AskSia.AI all rights reserved