Question

grace has a segment with endpoints a (3, 2) and b (6, 11) that is partitioned by a point c such that ac and bc form a 2:3 ratio. she knows the distance between the x coordinates is 3 units. which of the following fractions will let her find the x coordinate for point c? (4 points) \\(\frac{3}{5}\\) \\(\frac{3}{2}\\) \\(\frac{2}{3}\\) \\(\frac{2}{5}\\)

Explanation:

Step1: Understand the Section Formula

When a point \( C \) divides a segment \( AB \) in the ratio \( m:n \) (where \( AC:CB = m:n \)), the \( x \)-coordinate of \( C \) can be found using the formula that involves the fraction \( \frac{m}{m + n} \) (or \( \frac{n}{m + n} \) depending on direction, but here we analyze the ratio given). Here, the ratio \( AC:BC = 2:3 \), so \( m = 2 \) and \( n = 3 \).

Step2: Determine the Fraction for x - coordinate

To find the \( x \)-coordinate of \( C \), we consider the movement from \( A \) to \( B \) (or vice - versa). The total number of parts in the ratio \( AC:BC=2:3 \) is \( m + n=2 + 3=5 \) parts.

If we are moving from \( A \) to \( B \), the fraction of the horizontal (x - coordinate) distance that we cover from \( A \) to \( C \) is determined by the ratio of \( AC \) to the total length \( AB \) (where \( AB=AC + BC \)). Since \( AC:BC = 2:3 \), the length of \( AC \) is 2 parts and the length of \( AB \) is \( 2 + 3=5 \) parts. But when finding the \( x \)-coordinate, we can also think in terms of the ratio of the part we take from the difference in \( x \)-coordinates. The difference in \( x \)-coordinates between \( A(3,2) \) and \( B(6,11) \) is \( 6 - 3 = 3 \) units (as given). To find the \( x \)-coordinate of \( C \), we start with the \( x \)-coordinate of \( A \) (or \( B \)) and add (or subtract) a fraction of this difference.

The fraction of the difference in \( x \)-coordinates that we need to take from \( A \) towards \( B \) is \( \frac{2}{5} \)? Wait, no. Wait, let's re - express. Let the ratio be \( AC:CB = 2:3 \). So, using the section formula for internal division, the \( x \)-coordinate of \( C \) is given by \( x_C=x_A+\frac{m}{m + n}(x_B - x_A) \), where \( m = 2 \) (the ratio of \( AC \)) and \( n = 3 \) (the ratio of \( BC \)). Wait, no, actually, if \( AC:BC=m:n \), then the formula is \( x_C=\frac{n\times x_A+m\times x_B}{m + n} \) (when the ratio is \( AC:BC = m:n \)). Wait, let's clarify the ratio notation. If \( AC:CB=m:n \), then the coordinates of \( C \) are given by \( x=\frac{n\times x_A+m\times x_B}{m + n} \) and \( y=\frac{n\times y_A+m\times y_B}{m + n} \). Here, \( AC:BC = 2:3 \), so \( m = 2 \), \( n = 3 \). Then \( x_C=\frac{3\times3+2\times6}{2 + 3}=\frac{9 + 12}{5}=\frac{21}{5}=4.2 \), and the \( x \)-coordinate of \( A \) is 3, the difference in \( x \)-coordinates is \( 6 - 3=3 \). So \( x_C=3+\frac{2}{5}\times3 \)? Wait, no, \( \frac{2}{5}\times3=\frac{6}{5} \), and \( 3+\frac{6}{5}=\frac{15 + 6}{5}=\frac{21}{5}=4.2 \), which matches. But wait, the question is about which fraction will let her find the \( x \)-coordinate. Let's think again.

The difference in \( x \)-coordinates is \( 6 - 3 = 3 \) units. We need to find how much of this 3 - unit difference we add to the \( x \)-coordinate of \( A \) (or subtract from \( B \)) to get the \( x \)-coordinate of \( C \). Since \( AC:BC = 2:3 \), the length of \( AC \) is 2 parts and \( BC \) is 3 parts. So the fraction of the total length \( AB\) (where \( AB = AC+BC\)) that \( AC \) occupies is \( \frac{2}{5} \), but the fraction of the difference in \( x \)-coordinates (which is equal to the length of the horizontal component of \( AB \)) that we need to take from \( A \) towards \( B \) is \( \frac{2}{5} \)? Wait, no, maybe I mixed up. Wait, let's take an example. Suppose the ratio is \( AC:CB = 2:3 \). So starting at \( A(3,2) \), moving towards \( B(6,11) \), the total number of parts is \( 2 + 3=5 \) parts. The length from \( A \) to \( C \) is 2 parts and from \( C \) to \( B \) is 3 parts. So the horizontal distance from \( A \) to \( B \) is \( 6 - 3 = 3 \) units. So the horizontal distance from \( A \) to \( C \) is \( \frac{2}{5}\times3 \)? No, wait, no. Wait, if the ratio of \( AC \) to \( BC \) is \( 2:3 \), then the proportion of the \( x \)-distance (from \( A \) to \( B \)) that \( C \) is from \( A \) is \( \frac{2}{2 + 3}=\frac{2}{5} \)? No, that can't be. Wait, let's use the section formula correctly. The section formula for a point \( C \) dividing the line segment joining \( A(x_1,y_1) \) and \( B(x_2,y_2) \) in the ratio \( k:1 \) (where \( k=\frac{AC}{CB} \)) is \( x=\frac{x_1+kx_2}{1 + k} \), \( y=\frac{y_1+ky_2}{1 + k} \). Here, \( \frac{AC}{CB}=\frac{2}{3} \), so \( k = \frac{2}{3} \). Then \( x=\frac{3+ \frac{2}{3}\times6}{1+\frac{2}{3}}=\frac{3 + 4}{\frac{3 + 2}{3}}=\frac{7}{\frac{5}{3}}=\frac{21}{5}=4.2 \). Now, the difference in \( x \)-coordinates is \( 6 - 3 = 3 \). So \( 3\times\frac{2}{5}=\frac{6}{5} \), and \( 3+\frac{6}{5}=\frac{21}{5} \). Wait, but the question is which fraction. Wait, maybe I made a mistake in the ratio interpretation.

Wait, the problem says "AC and BC form a 2:3 ratio". So \( AC:BC = 2:3 \). So the total number of parts is \( 2+3 = 5 \) parts. If we are looking at the fraction of the distance between the \( x \)-coordinates (which is 3 units) that we need to use to find the \( x \)-coordinate of \( C \) relative to \( A \), we can think that from \( A \) to \( C \), we cover \( \frac{2}{5} \) of the total length \( AB \), but the distance between the \( x \)-coordinates is 3 units, which is equal to the horizontal component of \( AB \). Wait, no, the horizontal component of \( AB \) is 3 units (since \( 6 - 3=3 \)). So to find the \( x \)-coordinate of \( C \), we can start with \( A \)'s \( x \)-coordinate (3) and add a fraction of the 3 - unit difference. The fraction is \( \frac{2}{5} \)? But that's not one of the options. Wait, wait, maybe the ratio is \( AC:CB = 2:3 \), so when using the formula \( x_C=x_A+\frac{AC}{AC + CB}\times(x_B - x_A) \), \( \frac{AC}{AC + CB}=\frac{2}{2 + 3}=\frac{2}{5} \), but that's not an option. Wait, maybe I got the ratio reversed. If the ratio is \( BC:AC = 2:3 \), but the problem says \( AC \) and \( BC \) form a 2:3 ratio, so \( AC:BC = 2:3 \). Wait, the options are \( \frac{3}{5},\frac{3}{2},\frac{2}{3},\frac{2}{5} \). Wait, maybe I made a mistake in the direction. Let's consider the formula \( x_C=x_B-\frac{BC}{AC + BC}\times(x_B - x_A) \). Since \( BC \) is 3 parts and \( AC + BC \) is 5 parts, \( \frac{BC}{AC + BC}=\frac{3}{5} \), and \( x_B - x_A = 3 \). So \( x_C=6-\frac{3}{5}\times3=6-\frac{9}{5}=\frac{30 - 9}{5}=\frac{21}{5}=4.2 \), which is the same as before. Ah! So if we start from \( B \) and move towards \( A \), the fraction of the 3 - unit difference we subtract is \( \frac{3}{5} \). Wait, let's check:

\( x_A = 3 \), \( x_B=6 \), difference \( d = 6 - 3 = 3 \).

If we use the fraction \( \frac{2}{5} \) with respect to \( A \): \( 3+\frac{2}{5}\times3=3+\frac{6}{5}=\frac{21}{5}=4.2 \)

If we use the fraction \( \frac{3}{5} \) with respect to \( B \): \( 6-\frac{3}{5}\times3=6 - \frac{9}{5}=\frac{21}{5}=4.2 \)

Now, let's re - examine the problem statement: "Which of the following fractions will let her find the x coordinate for point C?".

We know that the distance between the \( x \)-coordinates is 3 units. Let's assume we start with the \( x \)-coordinate of \( A \) (3) and we want to find how much to add. The amount to add is \( \frac{2}{5}\times3 \), but that's not an option. Wait, maybe the ratio is interpreted as \( AC:BC = 2:3 \), so the weight for \( B \)'s \( x \)-coordinate is \( \frac{2}{5} \) and for \( A \)'s \( x \)-coordinate is \( \frac{3}{5} \) in the section formula \( x_C=\frac{3\times3 + 2\times6}{2+3}=\frac{9 + 12}{5}=\frac{21}{5} \). Wait, here, the formula is \( x_C=\frac{n\times x_A+m\times x_B}{m + n} \), where \( m = 2 \) (the ratio of \( AC \)) and \( n = 3 \) (the ratio of \( BC \)). So \( n = 3 \), \( m = 2 \), so the coefficient of \( x_A \) is \( \frac{n}{m + n}=\frac{3}{5} \) and the coefficient of \( x_B \) is \( \frac{m}{m + n}=\frac{2}{5} \). But the question is about the fraction of the distance between the \( x \)-coordinates (which is 3 units) that we use. If we think of the distance between \( x \)-coordinates as 3 units, and we want to find how much of this 3 - unit distance we need to move from \( A \) towards \( B \), we can also think in terms of the ratio of \( AC \) to \( AB \), but \( AB=AC + BC \), so \( AC:AB = 2:5 \), but that's \( \frac{2}{5} \). Wait, I'm confused. Wait, let's look at the options. The options are \( \frac{3}{5},\frac{3}{2},\frac{2}{3},\frac{2}{5} \).

Wait, maybe the problem is about the fraction of the 3 - unit distance. Let's calculate the \( x \)-coordinate of \( C \) in two ways.

Method 1: Using the section formula with ratio \( AC:BC = 2:3 \).

\( x_C=\frac{3\times3+2\times6}{2 + 3}=\frac{9 + 12}{5}=\frac{21}{5}=4.2 \)

The \( x \)-coordinate of \( A \) is 3, the difference between \( x \)-coordinates is \( 6 - 3 = 3 \). So \( 4.2-3 = 1.2 \), and \( 1.2=\frac{6}{5} \), and \( \frac{6}{5}=\frac{2}{5}\times3 \). But that's \( \frac{2}{5} \). Wait, but \( \frac{2}{5} \) is an option. Wait, the options are:

1. \( \frac{3}{5} \)

2. \( \frac{3}{2} \)

3. \( \frac{2}{3} \)

4. \( \frac{2}{5} \)

Ah! I think I made a mistake earlier. Let's re - express the ratio. If \( AC:BC = 2:3 \), then when finding the \( x \)-coordinate, we can use the formula \( x_C=x_A+\frac{AC}{AC + BC}\times(x_B - x_A) \). Here, \( AC = 2 \) parts, \( AC + BC=5 \) parts, so \( \frac{AC}{AC + BC}=\frac{2}{5} \). The difference in \( x \)-coordinates is \( x_B - x_A=6 - 3 = 3 \). So \( x_C=3+\frac{2}{5}\times3=3+\frac{6}{5}=\frac{21}{5}=4.2 \), which is correct. So the fraction of the 3 - unit distance (between \( x \)-coordinates) that we use is \( \frac{2}{5} \).

Answer:

\(\boldsymbol{\frac{2}{5}}\) (the option with \(\frac{2}{5}\))