Question

number line
for exercises 9 - 16, refer to the number line.

1. find the coordinate of point g such that the ratio of rg to gu is 3:2.
2. find the coordinate of point e that is 1/3 of the distance from r to t.
3. find the coordinate of point n that is 1/7 of the distance from q to s.
4. find the coordinate of point o such that the ratio of qo to ot is 1:4.
5. find the coordinate of point m such that the ratio of sm to mu is 2:3.
6. find the coordinate of point e that is 5/4 of the distance from p to r.
7. find the coordinate of point j that is 1/4 of the distance from s to p.
8. point s is what fractional distance from t to r?

Explanation:

Step1: Identify the formula for finding a point on a number - line given a ratio

If we have two points \(x_1\) and \(x_2\) on a number - line and we want to find a point \(x\) that divides the line segment from \(x_1\) to \(x_2\) in the ratio \(m:n\), the formula is \(x=\frac{nx_1 + mx_2}{m + n}\). If we want to find a point that is \(k\) fraction of the distance from \(x_1\) to \(x_2\), the formula is \(x=(1 - k)x_1+kx_2\).

Step2: Solve problem 9

The coordinate of \(R=-1\) and the coordinate of \(U = 8\). The ratio of \(RG\) to \(GU\) is \(3:2\), so \(m = 3\) and \(n = 2\). Using the formula \(x=\frac{nx_1+mx_2}{m + n}\), we substitute \(x_1=-1\) and \(x_2 = 8\). Then \(x=\frac{2\times(-1)+3\times8}{3 + 2}=\frac{-2 + 24}{5}=\frac{22}{5}=4.4\).

Step3: Solve problem 10

The coordinate of \(R=-1\) and the coordinate of \(T = 7\). The point \(E\) is \(\frac{1}{3}\) of the distance from \(R\) to \(T\). Using the formula \(x=(1 - k)x_1+kx_2\) with \(k=\frac{1}{3}\), \(x_1=-1\) and \(x_2 = 7\), we get \(x=(1-\frac{1}{3})\times(-1)+\frac{1}{3}\times7=\frac{2}{3}\times(-1)+\frac{7}{3}=\frac{-2 + 7}{3}=\frac{5}{3}\approx1.67\).

Step4: Solve problem 11

The coordinate of \(Q=-2\) and the coordinate of \(S = 5\). The point \(N\) is \(\frac{1}{7}\) of the distance from \(Q\) to \(S\). Using the formula \(x=(1 - k)x_1+kx_2\) with \(k = \frac{1}{7}\), \(x_1=-2\) and \(x_2 = 5\), we have \(x=(1-\frac{1}{7})\times(-2)+\frac{1}{7}\times5=\frac{6}{7}\times(-2)+\frac{5}{7}=\frac{-12 + 5}{7}=-1\).

Step5: Solve problem 12

The coordinate of \(Q=-2\) and the coordinate of \(T = 7\). The ratio of \(QO\) to \(OT\) is \(1:4\), so \(m = 1\) and \(n = 4\). Using the formula \(x=\frac{nx_1+mx_2}{m + n}\), we substitute \(x_1=-2\) and \(x_2 = 7\). Then \(x=\frac{4\times(-2)+1\times7}{1 + 4}=\frac{-8 + 7}{5}=-\frac{1}{5}=-0.2\).

Step6: Solve problem 13

The coordinate of \(S = 5\) and the coordinate of \(U = 8\). The ratio of \(SM\) to \(MU\) is \(2:3\), so \(m = 2\) and \(n = 3\). Using the formula \(x=\frac{nx_1+mx_2}{m + n}\), we substitute \(x_1 = 5\) and \(x_2 = 8\). Then \(x=\frac{3\times5+2\times8}{2 + 3}=\frac{15+16}{5}=\frac{31}{5}=6.2\).

Step7: Solve problem 14

The coordinate of \(P=-5\) and the coordinate of \(R=-1\). The point \(E\) is \(\frac{5}{4}\) of the distance from \(P\) to \(R\). Using the formula \(x=(1 - k)x_1+kx_2\) with \(k=\frac{5}{4}\), we note that \(\frac{5}{4}>1\) which means the point is beyond \(R\). First, find the distance between \(P\) and \(R\) which is \(|-1-(-5)| = 4\). The point \(E\) is \(\frac{5}{4}\times4 = 5\) units from \(P\) towards \(R\). So \(x=-5+5 = 0\).

Step8: Solve problem 15

The coordinate of \(S = 5\) and the coordinate of \(P=-5\). The point \(J\) is \(\frac{1}{4}\) of the distance from \(S\) to \(P\). Using the formula \(x=(1 - k)x_1+kx_2\) with \(k=\frac{1}{4}\), \(x_1 = 5\) and \(x_2=-5\), we get \(x=(1-\frac{1}{4})\times5+\frac{1}{4}\times(-5)=\frac{3}{4}\times5-\frac{5}{4}=\frac{15 - 5}{4}=\frac{10}{4}=2.5\).

Step9: Solve problem 16

The coordinate of \(T = 7\) and the coordinate of \(R=-1\). The distance between \(T\) and \(R\) is \(7-(-1)=8\). The distance between \(S = 5\) and \(R\) is \(5-(-1)=6\). The fractional distance of \(S\) from \(T\) to \(R\) is \(\frac{7 - 5}{7-(-1)}=\frac{2}{8}=\frac{1}{4}\).

Answer:

1. \(4.4\)
2. \(\frac{5}{3}\approx1.67\)
3. \(-1\)
4. \(-0.2\)
5. \(6.2\)
6. \(0\)
7. \(2.5\)
8. \(\frac{1}{4}\)