Question

select the correct graph. which image shows triangle abc dilated by a scale factor of 3 centered at the point b?

Explanation:

Step1: Recall Dilation Rules

Dilation centered at \( B \) means \( B \) is the center, so \( B \) and its image \( B' \) are the same point. The scale factor is 3, so the distance from \( B \) to \( A' \) should be 3 times the distance from \( B \) to \( A \), and the distance from \( B \) to \( C' \) should be 3 times the distance from \( B \) to \( C \). Also, the lines \( BA \) and \( BA' \), \( BC \) and \( BC' \) should be colinear.

Step2: Analyze Each Image

- **Top - Left Image**: Check if \( B \) is the center. The line from \( B \) to \( A' \) and \( B \) to \( A \) are colinear, and the length from \( B \) to \( A' \) seems 3 times \( B \) to \( A \) (since \( BA \) has 1 segment, \( BA' \) has 3? Wait, no, original \( BA \): let's see, original triangle \( ABC \) (small triangle) has \( B \), \( A \), \( C \). In top - left, \( B \) is on the line \( BB' \), \( A \) is on \( AA' \), \( C \) is on \( CC' \), all passing through \( B \). The length from \( B \) to \( A \): if original \( BA \) is, say, 1 unit, \( BA' \) is 3 units? Wait, no, the small triangle \( ABC \): \( B \) to \( A \) is a side, \( B \) to \( C \) is vertical. In top - left, \( B \) is between \( C \) and \( B' \), \( A \) is between \( A \) (wait, no, labels: original \( A \), \( B \), \( C \); dilated \( A' \), \( B' \), \( C' \). Wait, the center is \( B \), so \( B' = B \). So in the image, \( B \) should be the same as the original \( B \). Let's check the bottom - right image: \( B \) is the same, but the distance from \( B \) to \( A \) and \( B \) to \( A' \): if original \( BA \) is length \( l \), \( BA' \) should be \( 3l \). In the top - left image, the line from \( B \) to \( A' \) is 3 times the line from \( B \) to \( A \) (since the segment from \( B \) to \( A \) is 1 part, and from \( B \) to \( A' \) is 3 parts? Wait, no, original \( ABC \): \( B \) to \( A \): let's count the segments. Original triangle \( ABC \) (small) has \( B \), \( A \), \( C \) with \( BC \) vertical, \( AC \) horizontal. In the top - left image, \( B \) is on the vertical line, \( A \) is on the horizontal line from \( C \), and \( A' \) is on the extension of \( BA \) beyond \( A \), with \( BA' = 3BA \) (since \( BA \) is 1 segment, \( BA' \) is 3 segments? Wait, no, the original \( BA \): if we consider the small triangle, \( BA \) is a side, and in the top - left, the line from \( B \) to \( A' \) passes through \( A \), and the length from \( B \) to \( A' \) is 3 times \( B \) to \( A \) (because \( A \) is between \( B \) and \( A' \), and the number of segments: from \( B \) to \( A \) is 1, from \( B \) to \( A' \) is 3? No, wait, dilation centered at \( B \): the vector from \( B \) to \( A \) is \( \vec{BA} \), so the vector from \( B \) to \( A' \) should be \( 3\vec{BA} \), so \( A' = B + 3(A - B)=3A - 2B \). So the line \( BA' \) is the same line as \( BA \), and \( |BA'| = 3|BA| \). Similarly for \( C' \).

Looking at the top - left image: \( B \) is the center (since all lines from \( A \) to \( A' \), \( C \) to \( C' \), \( B \) to \( B' \) pass through \( B \), and \( B' = B \) (since \( B \) is on the line and the length from \( B \) to \( B' \) is 0, which is correct as center). The length from \( B \) to \( A \): if original \( BA \) is, say, 1 unit, \( BA' \) is 3 units (since the segment from \( B \) to \( A \) is 1, and from \( B \) to \( A' \) is 3 times that? Wait, no, the small triangle \( ABC \): \( BA \) is a side, \( BC \) is vertical. In the top - left image, \( B \) is between \( C \) and \( B' \), \( A \) is between \( A \) (wait, no, labels: original \( A \), \( B \), \( C \); dilated \( A' \), \( B' \), \( C' \). So \( B' = B \), \( A' \) is along \( BA \) extended, \( C' \) is along \( BC \) extended, with \( BA' = 3BA \) and \( BC' = 3BC \). This matches the dilation by scale factor 3 centered at \( B \).

The other images: bottom - left has a smaller triangle inside, which would be a scale factor less than 1. Bottom - right has \( B \) as center but the length from \( B \) to \( A \) and \( B \) to \( A' \) is not 3 times (maybe 1/3 or something). The top - right is the original triangle. So the top - left image (the first image on the top - left) shows the dilation correctly. Wait, but the user's image: let's re - check. The problem says "which image shows triangle \( ABC \) dilated by a scale factor of 3 centered at the point \( B \)". So center at \( B \), so \( B \) is fixed. So the dilated triangle \( A'B'C' \) has \( B' = B \), and \( \vec{BA'} = 3\vec{BA} \), \( \vec{BC'} = 3\vec{BC} \). So the lines \( BA \) and \( BA' \) are colinear, \( BC \) and \( BC' \) are colinear, and \( |BA'| = 3|BA| \), \( |BC'| = 3|BC| \).

Looking at the images:

- **Top - Left Image**: \( B \) is on the vertical line (same as original \( B \)), \( A \) is on the horizontal line from \( C \), and \( A' \) is on the extension of \( BA \) beyond \( A \), with \( BA' = 3BA \) (if \( BA \) is length \( l \), \( BA' = 3l \)), and \( BC' = 3BC \) (since \( BC \) is from \( B \) to \( C \), \( BC' \) is from \( B \) to \( C' \) with \( C' \) above \( C \)? Wait, no, original \( BC \) is vertical, from \( B \) up to \( C \). Wait, maybe I got the direction wrong. If \( B \) is the center, and we dilate by 3, then if \( A \) is to the right of \( B \) and \( C \) is above \( B \) (in the small triangle), then \( A' \) should be 3 times as far to the right of \( B \) as \( A \) is, and \( C' \) should be 3 times as far above \( B \) as \( C \) is. Wait, the small triangle \( ABC \): \( B \) is the bottom vertex, \( C \) is above \( B \), \( A \) is to the right of \( C \). So \( BA \) is a line from \( B \) (bottom) to \( A \) (right - top), \( BC \) is from \( B \) (bottom) to \( C \) (top). So dilating centered at \( B \), \( A' \) should be along \( BA \) extended beyond \( A \), and \( C' \) along \( BC \) extended beyond \( C \), with \( BA' = 3BA \) and \( BC' = 3BC \).

In the top - left image, \( B \) is the same as the original \( B \), \( A \) is between \( B \) and \( A' \), \( C \) is between \( B \) and \( C' \)? No, wait, the small triangle \( ABC \): \( B \), \( C \), \( A \) with \( BC \) vertical, \( AC \) horizontal. Then dilated triangle \( A'B'C' \) with \( B' = B \), \( A' \) such that \( B \) to \( A' \) is 3 times \( B \) to \( A \), and \( C' \) such that \( B \) to \( C' \) is 3 times \( B \) to \( C \). So the line \( BA \) is from \( B \) to \( A \), so \( A' \) is on the line \( BA \) (same line) and \( |BA'| = 3|BA| \). Similarly for \( BC \) and \( C' \).

The top - left image has \( B \) as the common point, \( A \) on \( BA \), \( A' \) on the same line, \( C \) on \( BC \), \( C' \) on the same line, with lengths \( BA' = 3BA \) and \( BC' = 3BC \). So this image is the correct one.

Answer:

The top - left image (the first image in the top - left corner)