Question

quadrilateral abcd is the result of dilating quadrilateral abcd about point a by a scale factor of \\(\frac{1}{2}\\).
determine whether each claim about the properties of abcd and abcd is true or false.
the lengths of \\(\overline{ab}\\) and \\(\overline{ab}\\) are equal. true/false
\\(\overline{ab}\\) and \\(\overline{ab}\\) are on distinct parallel lines. true/false

Explanation:

Step1: Analyze the first claim

Dilation with a scale factor of $\frac{1}{2}$ about point $A$ means the length of $A'B'$ is $\frac{1}{2}$ the length of $AB$. So $AB$ and $A'B'$ are not equal. Thus, the first claim is false.

Step2: Analyze the second claim

Dilation preserves the slope of lines. So the line containing $AB$ and the line containing $A'B'$ have the same slope, meaning they are parallel. Also, since it's a dilation about $A$, $A$ is on both lines (or the lines are the same? Wait, no, dilation about $A$: the image of $A$ is $A$ itself (since scale factor about $A$: $AA' = 0$). Wait, maybe I made a mistake. Wait, the original quadrilateral $ABCD$, dilated about $A$ by scale factor $\frac{1}{2}$. So the coordinates: let's find coordinates. From the graph, $A$ is at $(4,0)$, $B$ is at $(-2, -4)$. The vector from $A$ to $B$ is $(-2 - 4, -4 - 0)=(-6, -4)$. Dilation about $A$ by $\frac{1}{2}$: the vector from $A$ to $A'$ is $(0,0)$ (since $A' = A$), and the vector from $A$ to $B'$ is $\frac{1}{2}(-6, -4)=(-3, -2)$. So $B'$ is at $A + (-3, -2)=(4 - 3, 0 - 2)=(1, -2)$. Now, the line $AB$: slope is $\frac{-4 - 0}{-2 - 4}=\frac{-4}{-6}=\frac{2}{3}$. The line $A'B'$: $A'$ is $A=(4,0)$, $B'=(1, -2)$. Slope is $\frac{-2 - 0}{1 - 4}=\frac{-2}{-3}=\frac{2}{3}$. So same slope, so they are parallel. Also, are they distinct? $A$ is on both lines, but the lines are the same? Wait, no, because $A$ is a common point. Wait, maybe the problem says "distinct parallel lines" – but if they share a point (A), they are the same line. Wait, maybe I misread. Wait, the original line $AB$ and the image line $A'B'$: since dilation about $A$, the line $AB$ and $A'B'$ are the same line? Wait, no, $A' = A$, so the line through $A$ and $B'$ is the same as the line through $A$ and $B$? Wait, yes, because $B'$ is on the line $AB$ (since it's a dilation along the line from $A$ to $B$). So actually, the lines are the same, not distinct parallel lines. Wait, that contradicts my earlier thought. Wait, maybe the problem has a typo, or I misinterpret the graph. Wait, maybe the original quadrilateral: let's check other points. $C$ is at $(1,0)$, $D$ is at $(4,3)$. Dilation about $A(4,0)$ by $\frac{1}{2}$: vector from $A$ to $C$ is $(1 - 4, 0 - 0)=(-3, 0)$, so $C'$ is at $A + \frac{1}{2}(-3,0)=(4 - 1.5, 0)=(2.5, 0)$. Vector from $A$ to $D$ is $(4 - 4, 3 - 0)=(0, 3)$, so $D'$ is at $A + \frac{1}{2}(0,3)=(4, 1.5)$. Now, line $AB$: from $(4,0)$ to $(-2, -4)$, line $A'B'$: from $(4,0)$ to $(1, -2)$. These are the same line, because $(1, -2)$ is on line $AB$ (let's check: equation of $AB$: $y - 0=\frac{2}{3}(x - 4)$. Plug in $x=1$: $y=\frac{2}{3}(-3)= -2$, which matches $B'$'s y-coordinate. So line $AB$ and line $A'B'$ are the same line, not distinct parallel lines. Wait, but that would make the second claim false? But that contradicts my earlier thought. Wait, maybe the problem has a mistake, or I misread the graph. Alternatively, maybe the original quadrilateral is not with $A$ at $(4,0)$? Wait, the graph: $A$ is at (4,0) (the orange dot), $B$ is at (-2, -4) (the black dot), $C$ is at (1,0), $D$ is at (4,3). So dilation about $A$ by $\frac{1}{2}$: $A' = A$, $B'$ is midpoint of $A$ and $B$? Wait, no, dilation about $A$: the distance from $A$ to $B'$ is half the distance from $A$ to $B$. So vector from $A$ to $B$ is $(-6, -4)$, so vector from $A$ to $B'$ is $(-3, -2)$, so $B'$ is at (4 - 3, 0 - 2)=(1, -2), which is on line $AB$ (as we saw). So line $AB$ and line $A'B'$ are the same line, so they are not distinct parallel lines. Wait, but that would mean the second claim is false? But I'm confused. Wait, maybe the problem meant "the lines containing $AB$ and $A'B'$ are parallel" – but if they are the same line, they are parallel (since a line is parallel to itself). But "distinct parallel lines" – if they share a point, they are the same line, so not distinct. So the second claim: "AB and A'B' are on distinct parallel lines" – is that true? Let's check the definition. Two lines are parallel if they never meet (or same slope). If they share a point, they are coincident (same line). So coincident lines are parallel (by some definitions, parallel lines include coincident lines; by others, they are distinct). This is a bit ambiguous. But in the context of dilation, when you dilate a line about a point on the line, the image is the same line. So the lines are coincident, not distinct. So the second claim would be false? But I'm not sure. Wait, maybe I made a mistake in coordinates. Let's recalculate:

Coordinates:

$A: (4, 0)$

$B: (-2, -4)$

Vector $\overrightarrow{AB} = B - A = (-2 - 4, -4 - 0) = (-6, -4)$

Dilation about $A$ with scale factor $\frac{1}{2}$: $\overrightarrow{AB'} = \frac{1}{2}\overrightarrow{AB} = (-3, -2)$

So $B' = A + \overrightarrow{AB'} = (4 - 3, 0 - 2) = (1, -2)$

Now, line $AB$: passes through $(4,0)$ and $(-2, -4)$. Slope $m_1 = \frac{-4 - 0}{-2 - 4} = \frac{2}{3}$

Line $A'B'$: passes through $A'(4,0)$ and $B'(1, -2)$. Slope $m_2 = \frac{-2 - 0}{1 - 4} = \frac{2}{3}$

Since $m_1 = m_2$, the lines are parallel. Now, are they distinct? A line is determined by two points. Line $AB$ has points $A(4,0)$ and $B(-2, -4)$. Line $A'B'$ has points $A'(4,0)$ and $B'(1, -2)$. Since $A(4,0)$ is common, and $B'(1, -2)$ is on line $AB$ (as shown by the equation), the two lines are the same. Therefore, they are not distinct parallel lines. Wait, but that would mean the second claim is false. But I'm not sure if the problem considers a line parallel to itself. Some definitions say parallel lines are distinct, some say they can be coincident. This is a bit confusing. But in most high school math, parallel lines are distinct (non-intersecting, and not coincident). So if they are coincident, they are not distinct parallel lines. Therefore, the second claim is false? But I'm not sure. Wait, maybe I made a mistake in the dilation. Wait, dilation about a point: the image of the point is itself. So $A' = A$, $B'$ is the image of $B$ under dilation about $A$ by $\frac{1}{2}$, so $A$, $B$, $B'$ are colinear, with $AB' = \frac{1}{2}AB$. So line $AB$ and line $A'B'$ are the same line, so they are not distinct. Therefore, the second claim is false. But let's check the first claim again: length of $AB$ and $A'B'$. Distance $AB$: $\sqrt{(-2 - 4)^2 + (-4 - 0)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$. Distance $A'B'$: $\sqrt{(1 - 4)^2 + (-2 - 0)^2} = \sqrt{9 + 4} = \sqrt{13}$, which is half of $AB$. So $AB$ and $A'B'$ are not equal, so first claim is false. Second claim: are they on distinct parallel lines? Since they are the same line, no, so false? But I'm confused. Wait, maybe the problem meant "the segments $AB$ and $A'B'$ are on parallel lines" – but segments: the lines containing the segments. Since the lines have the same slope, they are parallel. And since $A$ is on both lines, but the segments: $AB$ is from $A$ to $B$, $A'B'$ is from $A'$ (which is $A$) to $B'$. So the lines containing the segments are the same, so not distinct. So both claims are false? But that seems odd. Wait, maybe I misread the graph. Let me check coordinates again. $A$: (4,0), $B$: (-2, -4), $C$: (1,0), $D$: (4,3). Dilation about $A$ by $\frac{1}{2}$: $A' = A$, $B'$: vector from $A$ to $B$ is (-6, -4), so $\frac{1}{2}$ of that is (-3, -2), so $B' = (4 - 3, 0 - 2) = (1, -2)$. $C'$: vector from $A$ to $C$ is (1 - 4, 0 - 0) = (-3, 0), so $\frac{1}{2}$ of that is (-1.5, 0), so $C' = (4 - 1.5, 0 + 0) = (2.5, 0)$. $D'$: vector from $A$ to $D$ is (0, 3), so $\frac{1}{2}$ of that is (0, 1.5), so $D' = (4 + 0, 0 + 1.5) = (4, 1.5)$. Now, line $AB$: through (4,0) and (-2, -4), slope $\frac{2}{3}$. Line $A'B'$: through (4,0) and (1, -2), slope $\frac{2}{3}$. So same slope, so parallel. Are they distinct? No, because they share the point (4,0) and have the same slope, so they are the same line. Therefore, the lines are not distinct, so the second claim is false. The first claim: lengths are not equal, so false. But maybe the problem has a different interpretation. Alternatively, maybe the dilation is not about $A$ as the center, but about the origin? Wait, the problem says "dilating quadrilateral $ABCD$ about point $A$ by a scale factor of $\frac{1}{2}$". So center is $A$. So my calculation is correct. So first claim: false, second claim: false. But I'm not sure. Maybe the second claim is true because the lines have the same slope, so they are parallel, and even though they share a point, in some definitions, parallel lines can be coincident. But in most cases, distinct parallel lines are non-coincident. So I think both are false.

Answer:

The first claim "The lengths of $\overline{AB}$ and $\overline{A'B'}$ are equal" is False.
The second claim "$\overline{AB}$ and $\overline{A'B'}$ are on distinct parallel lines" is False.

(If the options are to mark each as True/False, then for the first dropdown: False; for the second dropdown: False)