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.

determine the coordinates of point b and point d.

Explanation:

Step1: Find coordinates of point B

Use the section formula. If \(A(x_1,y_1)=(-6,10)\) and \(C(x_2,y_2)=(-3,4)\) and \(AB:BC = 2:1\), then for \(x\) - coordinate of \(B\):
\(x=\frac{2\times(-3)+1\times(-6)}{2 + 1}=\frac{-6-6}{3}=\frac{-12}{3}=-4\)
For \(y\) - coordinate of \(B\):
\(y=\frac{2\times4+1\times10}{2 + 1}=\frac{8 + 10}{3}=\frac{18}{3}=6\)

Step2: Find coordinates of point D

Since \(AC:CD=3:2\), let \(A(x_1,y_1)=(-6,10)\) and \(C(x_2,y_2)=(-3,4)\). Using the section formula for external division (as \(AC:CD = 3:2\) implies \(C\) divides \(AD\) in the ratio \(3:2\)).
For \(x\) - coordinate of \(D\):
\(x=\frac{2\times(-3)-3\times(-6)}{2-3}=\frac{-6 + 18}{-1}=\frac{12}{-1}=- 12\) (Wrong approach, let's use another way. If \(A(-6,10)\), \(C(-3,4)\) and \(AC:CD = 3:2\). The vector from \(A\) to \(C\) is \(\overrightarrow{AC}=(-3+6,4 - 10)=(3,-6)\). Let the coordinates of \(D=(x,y)\). Then \(\overrightarrow{CD}=(x + 3,y - 4)\). Since \(|\overrightarrow{AC}|:|\overrightarrow{CD}|=3:2\) and direction is same. \(\overrightarrow{CD}=\frac{2}{3}\overrightarrow{AC}\). \(x+3=\frac{2}{3}\times3\), \(x=-1\); \(y - 4=\frac{2}{3}\times(-6)\), \(y=-4\)

Answer:

Point \(B(-4,6)\) and Point \(D(-1,-4)\)