Question

for exercises 6 - 9, find the coordinates of each point described in relation to line segment cd. 8. $\frac{2}{3}$ of the way from c to d 9. $\frac{1}{3}$ of the way from d to c

Explanation:

1. First, assume the coordinates of \(C=(0,6)\) and \(D=(5,1)\) (by observing the graph - the \(x\) - coordinate of \(C\) is \(0\), \(y\) - coordinate of \(C\) is \(6\); the \(x\) - coordinate of \(D\) is \(5\), \(y\) - coordinate of \(D\) is \(1\)).

• The formula for finding a point \(P=(x,y)\) that divides the line - segment joining \(C=(x_1,y_1)\) and \(D=(x_2,y_2)\) in the ratio \(m:n\) is given by:
• \(x=\frac{mx_2+nx_1}{m + n}\) and \(y=\frac{my_2+ny_1}{m + n}\).

1. For the point that is \(\frac{2}{3}\) of the way from \(C\) to \(D\) (Exercise 8):

• Here, \(m = 2\) and \(n = 1\), \(x_1 = 0\), \(y_1 = 6\), \(x_2 = 5\), \(y_2 = 1\).
• Calculate the \(x\) - coordinate:
• \(x=\frac{2\times5+1\times0}{2 + 1}=\frac{10}{3}\).
• Calculate the \(y\) - coordinate:
• \(y=\frac{2\times1+1\times6}{2 + 1}=\frac{2 + 6}{3}=\frac{8}{3}\).

1. For the point that is \(\frac{1}{3}\) of the way from \(D\) to \(C\) (Exercise 9):

• Here, \(m = 1\) and \(n = 2\), \(x_1 = 0\), \(y_1 = 6\), \(x_2 = 5\), \(y_2 = 1\).
• Calculate the \(x\) - coordinate:
• \(x=\frac{1\times0+2\times5}{1 + 2}=\frac{10}{3}\).
• Calculate the \(y\) - coordinate:
• \(y=\frac{1\times6+2\times1}{1 + 2}=\frac{6 + 2}{3}=\frac{8}{3}\).

Step1: Identify coordinates of endpoints

Assume \(C=(0,6)\), \(D=(5,1)\)

Step2: Use section - formula for Exercise 8

\(x=\frac{2\times5+1\times0}{3}=\frac{10}{3}\), \(y=\frac{2\times1 + 1\times6}{3}=\frac{8}{3}\)

Step3: Use section - formula for Exercise 9

\(x=\frac{1\times0+2\times5}{3}=\frac{10}{3}\), \(y=\frac{1\times6+2\times1}{3}=\frac{8}{3}\)

Answer:

1. \((\frac{10}{3},\frac{8}{3})\)
2. \((\frac{10}{3},\frac{8}{3})\)