The line y = 3x – 2 crosses the y-axis at G. Write down the coordinates of G.

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Here are the solutions to the problems.

Problem (b):
Given point $A = (4, 1)$ and vector $\vec{AB} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ . We need to find the coordinates of point B.

Understand the relationship between points and vectors.
If $A = (x_A, y_A)$ and $B = (x_B, y_B)$ , then the vector $\vec{AB}$ is given by $\begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}$ .
Set up equations for the coordinates of B.
We have $A = (4, 1)$ , so $x_A = 4$ and $y_A = 1$ .
We are given $\vec{AB} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ .
Therefore, we can write:
$\begin{pmatrix} x_B - 4 \\ y_B - 1 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$
This gives two separate equations:
$x_B - 4 = -3$
$y_B - 1 = 1$
Solve for $x_B$ and $y_B$ .
For the x-coordinate:
$x_B - 4 = -3$
$x_B = -3 + 4$
$x_B = 1$

For the y-coordinate:
$y_B - 1 = 1$
$y_B = 1 + 1$
$y_B = 2$

The coordinates of B are $\boxed{(1, 2)}$ .

Problem (c):
The line $y = 3x - 2$ crosses the y-axis at G. We need to write down the coordinates of G.
Recall the condition for a line crossing the y-axis.
A line crosses the y-axis when the x-coordinate is 0.
Substitute $x=0$ into the equation of the line.
Substitute $x=0$ into $y = 3x - 2$ :
$y = 3(0) - 2$
Calculate the corresponding y-value.
$y = 0 - 2$ $y = -2$
Write down the coordinates of G.
Since $x=0$ and $y=-2$ , the coordinates of G are $(0, -2)$ .

The coordinates of G are $\boxed{(0, -2)}$ .

Point	x	y
y-intercept	0	−2
x-intercept	0.67	0

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