Question

Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than to B.
Given points A(3, -5) and B(19, -1), find the coordinates of point C such that CB/AC = 1/7.

Answer 1

1) C (9,-3.5)

2) C (17,-1.5)

Step-by-step explanation:

1)

To solve this problem, we must divide the segment AB into 8 equal intervals, and then find the point sitting at 3/8 of the whole segment.

The end points of the segment in this problem are:

$A(3,-5)$

and

$B(19,-1)$

This means that the x- and y-coordinates of point C are given by the equations:

$x_c=x_a + 3\frac{x_b-x_a}{8}\\y_c=y_a+3\frac{y_b-y_a}{8}$

And substituting the values of the coordinates of A and B, we find:

$x_c = x_a + 3 \frac{19-3}{8}=3+3\cdot 2 =9\\y_x = y_a + 3 \frac{-1-(-5)}{8}=-5+3\cdot 0.5 =-3.5$

2)

In this problem, we want to find the coordinates of point C such that:

$\frac{CB}{AC}=\frac{1}{7}$ (1)

As before, the coordinates of the endpoints of the segment AB are:

$A(3,-5)$

and

$B(19,-1)$

We can call the coordinates of point C as follows:

$C(x_c,y_c)$

To satisfy eq.(1) for the x-coordinate, we have:

$\frac{x_b-x_c}{x_c-x_a}=\frac{1}{7}$

Therefore, by substitution we find:

$\frac{19-x_c}{x_c-3}=\frac{1}{7}\\7(19-x_c)=x_c-3\\8x_c=136 \rightarrow x_c = 17$

Similarly on the y-coordinate we find:

$\frac{y_b-y_c}{y_c-y_a}=\frac{1}{7}$

And solving we get:

$\frac{-1-y_c}{y_c-(-5)}=\frac{1}{7}\\7(-1-y_c)=y_c+5\\8y_c=-12 \rightarrow y_c = -1.5$