Question

Segment upper A upper B is graphed on a coordinate plane. The x-axis ranges from 0 to 10 in increments of 1. The y-axis ranges from negative 2 to 2 in increments of 1.
Determine the coordinates of point G
, that divides the line AB
into two segments with lengths in a ratio of 3:5.

(1 point)
G
is located at (
,
).

Answer 1

A is (0,0) B is (0,8)

Answer 2

To find the coordinates of point G that divides the line segment AB in a ratio of 3:5, we can use the section formula. Let's denote the coordinates of points A and B as \(A(x_1, y_1)\) and \(B(x_2, y_2)\), respectively.

The section formula states that if a point G divides the line segment AB in the ratio \(m:n\), the coordinates \(G(x, y)\) can be found using the following formulas:

\[
x = \frac{mx_2 + nx_1}{m + n}
\]

\[
y = \frac{my_2 + ny_1}{m + n}
\]

In our case, the ratio is 3:5, so \(m = 3\) and \(n = 5\).

Let's denote:
- \(A = (x_1, y_1)\)
- \(B = (x_2, y_2)\)

Now, we would need the coordinates \(A\) and \(B\) to proceed. Since the problem did not provide specific coordinates, we need those values to calculate the coordinates of G.

Can you provide the coordinates of points A and B?

Answer 3

Given the coordinates of points A and B:

- \( A(0, 0) \)
- \( B(0, 8) \)

Now, we will use the section formula to find the coordinates of point G that divides AB in the ratio of 3:5.

Using the section formula:

\[
x = \frac{mx_2 + nx_1}{m + n}
\]
\[
y = \frac{my_2 + ny_1}{m + n}
\]

Substituting the coordinates and the ratio:

- \(m = 3\), \(n = 5\)
- \(x_1 = 0\), \(y_1 = 0\)
- \(x_2 = 0\), \(y_2 = 8\)

Now, calculate the x-coordinate of G:

\[
x = \frac{3 \cdot 0 + 5 \cdot 0}{3 + 5} = \frac{0 + 0}{8} = 0
\]

Next, calculate the y-coordinate of G:

\[
y = \frac{3 \cdot 8 + 5 \cdot 0}{3 + 5} = \frac{24 + 0}{8} = \frac{24}{8} = 3
\]

Thus, the coordinates of point G are:

\[
G(0, 3)
\]

So, point G is located at \( (0, 3) \).