Question

example 2
refer to the coordinate grid.

1. find point x on (overline{ab}) such that the ratio of ax to xb is 1:3.
2. find point y on (overline{cd}) such that the ratio of dy to yc is 2:1.
3. find point z on (overline{ef}) such that the ratio of ez to zf is 2:3.

examples 1 and 2
refer to the coordinate grid.

1. find point c on (overline{ab}) that is (\frac{1}{5}) of the distance from a to b.
2. find point q on (overline{rs}) that is (\frac{5}{8}) of the distance from r to s.
3. find point w on (overline{uv}) that is (\frac{1}{7}) of the distance from u to v.
4. find point d on (overline{ab}) that is (\frac{3}{4}) of the distance from a to b.
5. find point z on (overline{rs}) such that the ratio of rz to zs is 1:3.
6. find point g on (overline{ab}) such that the ratio of ag to gb is 3:2.
7. find point e on (overline{uv}) such that the ratio of ue to ev is 3:4.

Explanation:

Step1: Recall the section - formula

If a point \(P(x,y)\) divides the line - segment joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in the ratio \(m:n\), then \(x=\frac{mx_2+nx_1}{m + n}\) and \(y=\frac{my_2+ny_1}{m + n}\).

Step2: Solve for problem 4

Let \(A(x_1,y_1)\) and \(B(x_2,y_2)\) be the endpoints of \(\overline{AB}\). Since the ratio of \(AX\) to \(XB\) is \(1:3\), \(m = 1\) and \(n = 3\). Then the \(x\) - coordinate of \(X\) is \(x=\frac{1\times x_2+3\times x_1}{1 + 3}=\frac{x_2 + 3x_1}{4}\), and the \(y\) - coordinate of \(X\) is \(y=\frac{1\times y_2+3\times y_1}{1 + 3}=\frac{y_2 + 3y_1}{4}\).

Step3: Solve for problem 5

For the line - segment \(\overline{CD}\) with the ratio of \(DY\) to \(YC\) is \(2:1\), \(m = 2\) and \(n = 1\). The \(x\) - coordinate of \(Y\) is \(x=\frac{2\times x_2+1\times x_1}{2 + 1}=\frac{2x_2+x_1}{3}\), and the \(y\) - coordinate of \(Y\) is \(y=\frac{2\times y_2+1\times y_1}{2 + 1}=\frac{2y_2+y_1}{3}\).

Step4: Solve for problem 6

For the line - segment \(\overline{EF}\) with the ratio of \(EZ\) to \(ZF\) is \(2:3\), \(m = 2\) and \(n = 3\). The \(x\) - coordinate of \(Z\) is \(x=\frac{2\times x_2+3\times x_1}{2+3}=\frac{2x_2 + 3x_1}{5}\), and the \(y\) - coordinate of \(Z\) is \(y=\frac{2\times y_2+3\times y_1}{2 + 3}=\frac{2y_2+3y_1}{5}\).

Step5: Solve for problem 7

If a point \(C\) on \(\overline{AB}\) is \(\frac{1}{5}\) of the distance from \(A\) to \(B\), then the ratio of \(AC\) to \(CB\) is \(1:4\) (\(m = 1\), \(n = 4\)). The \(x\) - coordinate of \(C\) is \(x=\frac{1\times x_2+4\times x_1}{1 + 4}=\frac{x_2 + 4x_1}{5}\), and the \(y\) - coordinate of \(C\) is \(y=\frac{1\times y_2+4\times y_1}{1 + 4}=\frac{y_2 + 4y_1}{5}\).

Step6: Solve for problem 8

If a point \(Q\) on \(\overline{RS}\) is \(\frac{5}{8}\) of the distance from \(R\) to \(S\), then the ratio of \(RQ\) to \(QS\) is \(5:3\) (\(m = 5\), \(n = 3\)). The \(x\) - coordinate of \(Q\) is \(x=\frac{5\times x_2+3\times x_1}{5 + 3}=\frac{5x_2+3x_1}{8}\), and the \(y\) - coordinate of \(Q\) is \(y=\frac{5\times y_2+3\times y_1}{5 + 3}=\frac{5y_2+3y_1}{8}\).

Step7: Solve for problem 9

If a point \(W\) on \(\overline{UV}\) is \(\frac{1}{7}\) of the distance from \(U\) to \(V\), then the ratio of \(UW\) to \(WV\) is \(1:6\) (\(m = 1\), \(n = 6\)). The \(x\) - coordinate of \(W\) is \(x=\frac{1\times x_2+6\times x_1}{1 + 6}=\frac{x_2+6x_1}{7}\), and the \(y\) - coordinate of \(W\) is \(y=\frac{1\times y_2+6\times y_1}{1 + 6}=\frac{y_2+6y_1}{7}\).

Step8: Solve for problem 10

If a point \(D\) on \(\overline{AB}\) is \(\frac{3}{4}\) of the distance from \(A\) to \(B\), then the ratio of \(AD\) to \(DB\) is \(3:1\) (\(m = 3\), \(n = 1\)). The \(x\) - coordinate of \(D\) is \(x=\frac{3\times x_2+1\times x_1}{3 + 1}=\frac{3x_2+x_1}{4}\), and the \(y\) - coordinate of \(D\) is \(y=\frac{3\times y_2+1\times y_1}{3 + 1}=\frac{3y_2+y_1}{4}\).

Step9: Solve for problem 11

For the line - segment \(\overline{RS}\) with the ratio of \(RZ\) to \(ZS\) is \(1:3\), \(m = 1\) and \(n = 3\). The \(x\) - coordinate of \(Z\) is \(x=\frac{1\times x_2+3\times x_1}{1 + 3}=\frac{x_2 + 3x_1}{4}\), and the \(y\) - coordinate of \(Z\) is \(y=\frac{1\times y_2+3\times y_1}{1 + 3}=\frac{y_2 + 3y_1}{4}\).

Step10: Solve for problem 12

For the line - segment \(\overline{AB}\) with the ratio of \(AG\) to \(GB\) is \(3:2\), \(m = 3\) and \(n = 2\). The \(x\) - coordinate of \(G\) is \(x=\frac{3\times x_2+2\times x_1}{3 + 2}=\frac{3x_2+2x_1}{5}\), and the \(y\) - coordinate of \(G\) is \(y=\frac{3\times y_2+2\times y_1}{3 + 2}=\frac{3y_2+2y_1}{5}\).

Step11: Solve for problem 13

For the line - segment \(\overline{UV}\) with the ratio of \(UE\) to \(EV\) is \(3:4\), \(m = 3\) and \(n = 4\). The \(x\) - coordinate of \(E\) is \(x=\frac{3\times x_2+4\times x_1}{3 + 4}=\frac{3x_2+4x_1}{7}\), and the \(y\) - coordinate of \(E\) is \(y=\frac{3\times y_2+4\times y_1}{3 + 4}=\frac{3y_2+4y_1}{7}\).

Since the coordinates of the endpoints of the line - segments are not given, the general formulas for finding the coordinates of the points that divide the line - segments in the given ratios are provided above. To get the actual numerical answers, we need to know the coordinates of the endpoints of each line - segment.

Answer:

Step1: Recall the section - formula

If a point \(P(x,y)\) divides the line - segment joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in the ratio \(m:n\), then \(x=\frac{mx_2+nx_1}{m + n}\) and \(y=\frac{my_2+ny_1}{m + n}\).

Step2: Solve for problem 4

Let \(A(x_1,y_1)\) and \(B(x_2,y_2)\) be the endpoints of \(\overline{AB}\). Since the ratio of \(AX\) to \(XB\) is \(1:3\), \(m = 1\) and \(n = 3\). Then the \(x\) - coordinate of \(X\) is \(x=\frac{1\times x_2+3\times x_1}{1 + 3}=\frac{x_2 + 3x_1}{4}\), and the \(y\) - coordinate of \(X\) is \(y=\frac{1\times y_2+3\times y_1}{1 + 3}=\frac{y_2 + 3y_1}{4}\).

Step3: Solve for problem 5

For the line - segment \(\overline{CD}\) with the ratio of \(DY\) to \(YC\) is \(2:1\), \(m = 2\) and \(n = 1\). The \(x\) - coordinate of \(Y\) is \(x=\frac{2\times x_2+1\times x_1}{2 + 1}=\frac{2x_2+x_1}{3}\), and the \(y\) - coordinate of \(Y\) is \(y=\frac{2\times y_2+1\times y_1}{2 + 1}=\frac{2y_2+y_1}{3}\).

Step4: Solve for problem 6

For the line - segment \(\overline{EF}\) with the ratio of \(EZ\) to \(ZF\) is \(2:3\), \(m = 2\) and \(n = 3\). The \(x\) - coordinate of \(Z\) is \(x=\frac{2\times x_2+3\times x_1}{2+3}=\frac{2x_2 + 3x_1}{5}\), and the \(y\) - coordinate of \(Z\) is \(y=\frac{2\times y_2+3\times y_1}{2 + 3}=\frac{2y_2+3y_1}{5}\).

Step5: Solve for problem 7

If a point \(C\) on \(\overline{AB}\) is \(\frac{1}{5}\) of the distance from \(A\) to \(B\), then the ratio of \(AC\) to \(CB\) is \(1:4\) (\(m = 1\), \(n = 4\)). The \(x\) - coordinate of \(C\) is \(x=\frac{1\times x_2+4\times x_1}{1 + 4}=\frac{x_2 + 4x_1}{5}\), and the \(y\) - coordinate of \(C\) is \(y=\frac{1\times y_2+4\times y_1}{1 + 4}=\frac{y_2 + 4y_1}{5}\).

Step6: Solve for problem 8

If a point \(Q\) on \(\overline{RS}\) is \(\frac{5}{8}\) of the distance from \(R\) to \(S\), then the ratio of \(RQ\) to \(QS\) is \(5:3\) (\(m = 5\), \(n = 3\)). The \(x\) - coordinate of \(Q\) is \(x=\frac{5\times x_2+3\times x_1}{5 + 3}=\frac{5x_2+3x_1}{8}\), and the \(y\) - coordinate of \(Q\) is \(y=\frac{5\times y_2+3\times y_1}{5 + 3}=\frac{5y_2+3y_1}{8}\).

Step7: Solve for problem 9

If a point \(W\) on \(\overline{UV}\) is \(\frac{1}{7}\) of the distance from \(U\) to \(V\), then the ratio of \(UW\) to \(WV\) is \(1:6\) (\(m = 1\), \(n = 6\)). The \(x\) - coordinate of \(W\) is \(x=\frac{1\times x_2+6\times x_1}{1 + 6}=\frac{x_2+6x_1}{7}\), and the \(y\) - coordinate of \(W\) is \(y=\frac{1\times y_2+6\times y_1}{1 + 6}=\frac{y_2+6y_1}{7}\).

Step8: Solve for problem 10

If a point \(D\) on \(\overline{AB}\) is \(\frac{3}{4}\) of the distance from \(A\) to \(B\), then the ratio of \(AD\) to \(DB\) is \(3:1\) (\(m = 3\), \(n = 1\)). The \(x\) - coordinate of \(D\) is \(x=\frac{3\times x_2+1\times x_1}{3 + 1}=\frac{3x_2+x_1}{4}\), and the \(y\) - coordinate of \(D\) is \(y=\frac{3\times y_2+1\times y_1}{3 + 1}=\frac{3y_2+y_1}{4}\).

Step9: Solve for problem 11

For the line - segment \(\overline{RS}\) with the ratio of \(RZ\) to \(ZS\) is \(1:3\), \(m = 1\) and \(n = 3\). The \(x\) - coordinate of \(Z\) is \(x=\frac{1\times x_2+3\times x_1}{1 + 3}=\frac{x_2 + 3x_1}{4}\), and the \(y\) - coordinate of \(Z\) is \(y=\frac{1\times y_2+3\times y_1}{1 + 3}=\frac{y_2 + 3y_1}{4}\).

Step10: Solve for problem 12

For the line - segment \(\overline{AB}\) with the ratio of \(AG\) to \(GB\) is \(3:2\), \(m = 3\) and \(n = 2\). The \(x\) - coordinate of \(G\) is \(x=\frac{3\times x_2+2\times x_1}{3 + 2}=\frac{3x_2+2x_1}{5}\), and the \(y\) - coordinate of \(G\) is \(y=\frac{3\times y_2+2\times y_1}{3 + 2}=\frac{3y_2+2y_1}{5}\).

Step11: Solve for problem 13

For the line - segment \(\overline{UV}\) with the ratio of \(UE\) to \(EV\) is \(3:4\), \(m = 3\) and \(n = 4\). The \(x\) - coordinate of \(E\) is \(x=\frac{3\times x_2+4\times x_1}{3 + 4}=\frac{3x_2+4x_1}{7}\), and the \(y\) - coordinate of \(E\) is \(y=\frac{3\times y_2+4\times y_1}{3 + 4}=\frac{3y_2+4y_1}{7}\).

Since the coordinates of the endpoints of the line - segments are not given, the general formulas for finding the coordinates of the points that divide the line - segments in the given ratios are provided above. To get the actual numerical answers, we need to know the coordinates of the endpoints of each line - segment.