What is the radical form of 10y1/2
2 answers:
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Answer:
Point E is closer to point A
Step-by-step explanation:
<em>"I have added screenshot of the complete question as well as the </em>
<em> diagram"</em>
- In this diagram, line segment CD is the perpendicular bisector of line
segment AB
∴ AB ⊥ CD
∵ CD intersect AB at M
∴ M is the mid-point of AB
∴ The length of AM = the length of MB
- Assume the conjecture that the set of points equidistant from A
and B is the perpendicular bisector of AB is true
∴ Any point lies on the line CD equidistant from A and B
- <u><em>If point E lies on the line CD</em></u>
∵ CD is the perpendicular bisector of AB
∵ E lies on CD
∴ The length of EA = The length of EB
- <u><em>From the figure point E is on the left side of the line CD</em></u>
∵ Point A is on the left side of the line CD
∵ Point B is on the right side of the line CD
∵ Point E is on the left side of the line CD
∴ The length of AE < The length of BE
∴ Point E is closer to point A
