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for exercises 9 - 16, refer to the number line. p o r s t u -5 - 4 - 3 …

Question

for exercises 9 - 16, refer to the number line.
p o r s t u
-5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7 8 9

  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: Recall the section - formula on a number - line

If a point \(G\) divides the line - segment joining \(x_1\) and \(x_2\) in the ratio \(m:n\), the coordinate of \(G\) is given by \(x=\frac{mx_2+nx_1}{m + n}\).

Step2: Solve problem 9

The coordinate of \(R=-1\) and the coordinate of \(U = 9\). Given the ratio \(RG:GU = 3:2\), using the section - formula \(x=\frac{3\times9+2\times(-1)}{3 + 2}=\frac{27-2}{5}=\frac{25}{5}=5\).

Step3: Solve problem 10

The coordinate of \(R=-1\) and the coordinate of \(T = 8\). The distance from \(R\) to \(T\) is \(8-(-1)=9\). Point \(E\) is \(\frac{1}{3}\) of the distance from \(R\) to \(T\). The coordinate of \(E=-1+\frac{1}{3}(8 - (-1))=-1 + 3=2\).

Step4: Solve problem 11

The coordinate of \(Q=-2\) and the coordinate of \(S = 5\). The distance from \(Q\) to \(S\) is \(5-(-2)=7\). Point \(N\) is \(\frac{1}{7}\) of the distance from \(Q\) to \(S\). The coordinate of \(N=-2+\frac{1}{7}(5 - (-2))=-2 + 1=-1\).

Step5: Solve problem 12

The coordinate of \(Q=-2\) and the coordinate of \(T = 8\). Given the ratio \(QO:OT = 1:4\), using the section - formula \(x=\frac{1\times8+4\times(-2)}{1 + 4}=\frac{8-8}{5}=0\).

Step6: Solve problem 13

The coordinate of \(S = 5\) and the coordinate of \(U = 9\). Given the ratio \(SM:MU = 2:3\), using the section - formula \(x=\frac{2\times9+3\times5}{2 + 3}=\frac{18 + 15}{5}=\frac{33}{5}=6.6\).

Step7: Solve problem 14

The coordinate of \(P=-5\) and the coordinate of \(R=-1\). The distance from \(P\) to \(R\) is \(-1-(-5)=4\). Point \(E\) is \(\frac{5}{4}\) of the distance from \(P\) to \(R\). The coordinate of \(E=-5+\frac{5}{4}(-1-(-5))=-5 + 5=0\).

Step8: Solve problem 15

The coordinate of \(S = 5\) and the coordinate of \(P=-5\). The distance from \(S\) to \(P\) is \(-5 - 5=-10\). Point \(J\) is \(\frac{1}{4}\) of the distance from \(S\) to \(P\). The coordinate of \(J=5+\frac{1}{4}(-5 - 5)=5-2.5 = 2.5\).

Step9: Solve problem 16

The coordinate of \(R=-1\), the coordinate of \(S = 5\), and the coordinate of \(T = 8\). The distance from \(T\) to \(R\) is \(8-(-1)=9\), and the distance from \(T\) to \(S\) is \(8 - 5=3\). The fractional distance of \(S\) from \(T\) to \(R\) is \(\frac{8 - 5}{8-(-1)}=\frac{3}{9}=\frac{1}{3}\).

Answer:

  1. 5
  2. 2
  3. -1
  4. 0
  5. 6.6
  6. 0
  7. 2.5
  8. \(\frac{1}{3}\)