Question

drag each coordinate pair to the correct location on the figure. not all coordinate pairs will be used
the ratios of the line segments are given below.
$ab : bc = 2 : 1$
$ac : cd = 3 : 2$
determine the coordinates of point b and point d.
$(-1,0)$ $(1,2)$ $(-2,-6)$ $(-1,-4)$ $(5,0)$ $(-4,6)$
(figure with point a at $(-6,10)$, point c at $(-3,4)$, and points b, d on the line segment)

Explanation:

Step1: Find Coordinates of B (Section Formula)

Given \( A(-6,10) \), \( C(-3,4) \), and \( AB:BC = 2:1 \). Let \( B(x,y) \). Using section formula \( x=\frac{2(-3)+1(-6)}{2 + 1}=\frac{-6-6}{3}=-4 \), \( y=\frac{2(4)+1(10)}{2 + 1}=\frac{8 + 10}{3}=6 \)? Wait, no—wait, \( AB:BC = 2:1 \), so \( B \) divides \( AC \) in ratio \( AB:BC = 2:1 \), so \( m:n = 2:1 \), \( A(x_1,y_1)=(-6,10) \), \( C(x_2,y_2)=(-3,4) \). Then \( x=\frac{2(-3)+1(-6)}{2 + 1}=\frac{-6-6}{3}=-4 \)? No, wait, section formula is \( x=\frac{m x_2 + n x_1}{m + n} \) when dividing internally in ratio \( m:n \). Wait, \( AB:BC = 2:1 \), so from \( A \) to \( B \) to \( C \), so \( B \) is between \( A \) and \( C \), ratio \( AB:BC = 2:1 \), so \( m = 2 \), \( n = 1 \), so \( x=\frac{2(-3)+1(-6)}{3}=\frac{-6 - 6}{3}=-4 \), \( y=\frac{2(4)+1(10)}{3}=\frac{8 + 10}{3}=6 \). Wait, but the options have \( (-4,6) \). Wait, but let's check \( AC \): distance from \( A(-6,10) \) to \( C(-3,4) \): \( \Delta x = 3 \), \( \Delta y = -6 \). \( AB:BC = 2:1 \), so \( AB \) is \( \frac{2}{3} \) of \( AC \). So \( B = A + \frac{2}{3}(C - A)=(-6,10)+\frac{2}{3}(3,-6)=(-6 + 2,10 - 4)=(-4,6) \). So \( B(-4,6) \).

Step2: Find Coordinates of D (Section Formula)

Given \( AC:CD = 3:2 \), so \( AC:CD = 3:2 \), so \( C \) divides \( AD \) in ratio \( AC:CD = 3:2 \), so \( m = 3 \), \( n = 2 \), \( A(-6,10) \), \( C(-3,4) \), let \( D(x,y) \). Using section formula: \( -3=\frac{3x + 2(-6)}{3 + 2} \), \( 4=\frac{3y + 2(10)}{3 + 2} \). Solve for \( x \): \( -15 = 3x - 12 \Rightarrow 3x = -3 \Rightarrow x = -1 \)? No, wait, ratio \( AC:CD = 3:2 \), so from \( A \) to \( C \) to \( D \), so \( C \) is between \( A \) and \( D \), ratio \( AC:CD = 3:2 \), so \( m = 3 \), \( n = 2 \), so \( x=\frac{3x_D + 2(-6)}{5}=-3 \Rightarrow 3x_D - 12 = -15 \Rightarrow 3x_D = -3 \Rightarrow x_D = -1 \). \( y=\frac{3y_D + 2(10)}{5}=4 \Rightarrow 3y_D + 20 = 20 \Rightarrow 3y_D = 0 \Rightarrow y_D = 0 \). Wait, but options have \( (-1,0) \)? Wait, no—wait, maybe \( D \) is beyond \( C \), ratio \( AC:CD = 3:2 \), so external division? Wait, no, the figure shows \( D \) below \( C \). Wait, let's check the vector from \( A \) to \( C \): \( (3, -6) \). \( AC:CD = 3:2 \), so \( CD = \frac{2}{3}AC \), so vector from \( C \) to \( D \) is \( \frac{2}{3}(3, -6)=(2, -4) \). So \( D = C + (2, -4)=(-3 + 2,4 - 4)=(-1,0) \). Yes! So \( D(-1,0) \).

Wait, let's verify \( B \): \( A(-6,10) \) to \( B(-4,6) \): \( \Delta x = 2 \), \( \Delta y = -4 \). \( B \) to \( C(-3,4) \): \( \Delta x = 1 \), \( \Delta y = -2 \). Ratio \( 2:1 \) (since \( 2/1 = (2)/1 \) in \( \Delta x \) and \( \Delta y \)). Correct. \( AC \) vector: \( (3, -6) \), \( CD \) vector: \( (2, -4) \), ratio \( 3:2 \) (since \( 3/2 = 3/2 \) in \( \Delta x \) and \( \Delta y \)). Correct.

Step3: Match with Options

For \( B \): \( (-4,6) \) is an option. For \( D \): \( (-1,0) \) is an option.

Answer:

Point B: (-4, 6), Point D: (-1, 0)