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) (9,-3.5)

2) (17,-1.5)

Step-by-step explanation:

1)

In order to solve this problem, we have to divide the segment into 8 equal parts, and find the point that sits at 3/8 of the whole segment.

The end points of the segment in this problem are:

$A(3,-5)$

and

$B(19,-1)$

First of all, we find the distance between the x-coordinates and between the y-coordinates:

$d_x = |19-3|=16\\d_y=|-1-(-5)|=4$

Then we divide the distances by 8 parts:

$\frac{d_x}{8}=\frac{16}{8}=2\\\frac{d_y}{8}=\frac{4}{8}=0.5$

Now we find the coordinates of point C, which sits 3/8 of the way along the segment, by using the equations:

$x_c = x_a + 3 \frac{d_x}{8}=3+3\cdot 2 =9\\y_x = y_a + 3 \frac{d_y}{8}=-5+3\cdot 0.5 =-3.5$

2)

Here instead we want to find the coordinates of point C such that

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

The coordinates of the endpoints of the segment AB are:

$A(3,-5)$

and

$B(19,-1)$

We call the coordinates of point C as:

$C(x_c,y_c)$

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

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

Substittuing the values of the x-coordinates of A and B we find:

$\frac{19-x_c}{x_c-3}=\frac{1}{7}\\7(19-x_c)=x_c-3\\133-7x_c=x_c-3\\8x_c=136 \rightarrow x_c = 17$

And similarly for the y-coordinate we have:

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