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question 8 in the figure, point f is common to all the lines, rays, and…

Question

question 8
in the figure, point f is common to all the lines, rays, and line segments shown, and points b, e, and f lie on the same line.
which of the following statements about the figure is true?
a (overleftrightarrow{ef}) is the same object as (overleftrightarrow{fh})
b (overline{fc}) is the only line segment in the figure.

Explanation:

Step1: Analyze Option A

A line segment is defined by its two endpoints. $\overline{EF}$ and $\overline{FH}$: Let's check their endpoints. $\overline{EF}$ has endpoints E and F, $\overline{FH}$ has endpoints F and H. But wait, in a line, a line segment and another segment on the same line—wait, no, actually, a line segment is the same as another if they have the same endpoints (or are part of the same line with same length? Wait, no, the key is that a line segment is uniquely defined by its two endpoints, but also, if two segments are on the same line and share the same endpoints (or are the same set of points). Wait, actually, $\overleftrightarrow{EF}$ is a line, and $\overline{FH}$—wait, no, the option A says $\overleftrightarrow{EF}$ is the same object as $\overline{FH}$. No, a line ($\overleftrightarrow{EF}$) is infinite in both directions, while a line segment ($\overline{FH}$) is finite. Wait, maybe I misread. Wait, the figure: point F is common to all lines, rays, and segments. E, F, H are on the same line (since F is common, and E, F, H are colinear). So $\overleftrightarrow{EF}$ is the line containing E, F, H. $\overline{FH}$ is a segment on that line. But wait, no—wait, the option A: is $\overleftrightarrow{EF}$ (line) the same as $\overline{FH}$ (segment)? No, that can't be. Wait, maybe the other option: option B says $\overline{FC}$ is the only line segment? No, there are other segments. Wait, maybe I made a mistake. Wait, let's re-express.

Wait, the problem: "In the figure, point F is common to all the lines, rays, and line segments shown, and points E, F, and H lie on the same line."

Option A: $\overleftrightarrow{EF}$ is the same object as $\overline{FH}$. Wait, $\overleftrightarrow{EF}$ is a line (infinite, through E, F, H), and $\overline{FH}$ is a segment (finite, from F to H). But actually, a line segment is part of a line. But the "same object"—no, a line and a segment are different. Wait, maybe the correct reasoning is: since E, F, H are colinear, $\overleftrightarrow{EF}$ is the line that includes E, F, H, and $\overline{FH}$ is a segment on that line. But the option A says they are the same object? No, that doesn't make sense. Wait, maybe the other option is wrong. Wait, option B: $\overline{FC}$ is the only line segment? No, there are other segments like $\overline{FH}$, $\overline{EF}$, etc. Wait, maybe I misread the options. Wait, the user's image: the options are A: $\overleftrightarrow{EF}$ is the same as $\overline{FH}$, B: $\overline{FC}$ is the only segment. Wait, no, maybe the correct answer is A? Wait, no—wait, a line and a segment can't be the same. Wait, maybe the notation: $\overleftrightarrow{EF}$ is the line, and $\overline{FH}$ is a segment on that line, but the line $\overleftrightarrow{EF}$ contains the segment $\overline{FH}$, but they are not the same object. Wait, maybe the problem has a typo, or I'm missing something. Wait, alternatively, maybe $\overleftrightarrow{EF}$ and $\overleftrightarrow{FH}$ are the same line (since E, F, H are colinear), so $\overleftrightarrow{EF} = \overleftrightarrow{FH}$, but the option A says $\overleftrightarrow{EF}$ (line) is the same as $\overline{FH}$ (segment). No, that's not. Wait, maybe the option A is correct because $\overleftrightarrow{EF}$ is the line through E, F, H, and $\overline{FH}$ is a segment on that line, but the question says "same object"—no, that's incorrect. Wait, maybe I made a mistake. Let's think again.

Wait, the key: a line is defined by two points and extends infinitely. A line segment is defined by two endpoints and is finit…

Answer:

A. $\overleftrightarrow{EF}$ is the same object as $\overline{FH}$