Question

locating points on a coordinate plane - practice
exercise 1
find the coordinates of point x on the coordinate plane for each situation.

1. point x on (overline{ab}) is (\frac{1}{5}) the distance from a to b.
2. point x on (overline{rs}) is (\frac{1}{6}) the distance from r to s.
3. point x on (overline{jk}) is (\frac{1}{3}) the distance from j to k.

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:

1. For the problem of finding a point \(X\) on a line - segment \(AB\) that is \(\frac{1}{n}\) of the distance from \(A(x_1,y_1)\) to \(B(x_2,y_2)\):

• The formula for the coordinates of the point \(X=(x,y)\) is given by \(x = x_1+\frac{1}{n}(x_2 - x_1)\) and \(y=y_1+\frac{1}{n}(y_2 - y_1)\).
• For the problem of finding a point \(X\) on a line - segment \(AB\) such that the ratio of \(AX\) to \(XB\) is \(m:n\):
• The formula for the coordinates of the point \(X=(x,y)\) is \(x=\frac{mx_2+nx_1}{m + n}\) and \(y=\frac{my_2+ny_1}{m + n}\).

1. Let's take an example (for the first - type problem, finding a point that is \(\frac{1}{n}\) of the distance from one end - point to the other):

• Suppose \(A=(x_1,y_1)\) and \(B=(x_2,y_2)\) and we want to find a point \(X\) that is \(\frac{1}{5}\) of the distance from \(A\) to \(B\).
• Step 1: Calculate the \(x\) - coordinate of \(X\)
• The formula for the \(x\) - coordinate of \(X\) is \(x=x_1+\frac{1}{5}(x_2 - x_1)=\frac{5x_1+(x_2 - x_1)}{5}=\frac{4x_1 + x_2}{5}\).
• Step 2: Calculate the \(y\) - coordinate of \(X\)
• The formula for the \(y\) - coordinate of \(X\) is \(y=y_1+\frac{1}{5}(y_2 - y_1)=\frac{5y_1+(y_2 - y_1)}{5}=\frac{4y_1 + y_2}{5}\).

1. For the second - type problem (finding a point with a given ratio \(m:n\)):

• Suppose \(A=(x_1,y_1)\) and \(B=(x_2,y_2)\) and the ratio of \(AX\) to \(XB\) is \(1:3\).
• Step 1: Calculate the \(x\) - coordinate of \(X\)
• Using the formula \(x=\frac{1\times x_2+3\times x_1}{1 + 3}=\frac{x_2 + 3x_1}{4}\).
• Step 2: Calculate the \(y\) - coordinate of \(X\)
• Using the formula \(y=\frac{1\times y_2+3\times y_1}{1 + 3}=\frac{y_2 + 3y_1}{4}\).

Since the problem is not complete (no specific coordinates of the endpoints of the line - segments are given), we cannot provide specific numerical answers. But the general methods for solving such problems are as above. If we assume for a line - segment \(AB\) with \(A=(x_1,y_1)\) and \(B=(x_2,y_2)\) and we want to find a point \(X\) that is \(\frac{1}{3}\) of the distance from \(A\) to \(B\):

Step1: Calculate \(x\) - coordinate

Let \(A=(x_1,y_1)\) and \(B=(x_2,y_2)\). The \(x\) - coordinate of \(X\) is \(x=x_1+\frac{1}{3}(x_2 - x_1)=\frac{3x_1+(x_2 - x_1)}{3}=\frac{2x_1 + x_2}{3}\).

Step2: Calculate \(y\) - coordinate

The \(y\) - coordinate of \(X\) is \(y=y_1+\frac{1}{3}(y_2 - y_1)=\frac{3y_1+(y_2 - y_1)}{3}=\frac{2y_1 + y_2}{3}\).

Answer:

The coordinates of \(X\) are \((\frac{2x_1 + x_2}{3},\frac{2y_1 + y_2}{3})\)