Question

graph c + d with its initial point at the initial point of c.

Explanation:

Step1: Find components of vector \( \boldsymbol{c} \)

Vector \( c \) has initial point \((-4, 5)\) and terminal point \((2, 1)\).
Change in \( x \): \( 2 - (-4) = 6 \)? Wait, no—wait, looking at the graph, vector \( c \): from its start (let's check coordinates). Wait, the terminal point of \( c \) is \((2,1)\)? Wait, no, the blue vector \( c \) starts at, say, \((-4, 5)\) (since it's on the grid: x=-4, y=5) and ends at \((2, 1)\)? Wait, no, the terminal point of \( c \) is (2,1)? Wait, the other vector \( d \): starts at (2,1)? No, wait, the problem says "initial point of \( c \)"—so first, find vector \( c \): initial point (let's see the grid: the start of \( c \) is at (-4, 5) (x=-4, y=5) and end at (2, 1) (x=2, y=1). So components of \( c \): \( \Delta x = 2 - (-4) = 6 \)? Wait, no, wait, maybe I misread. Wait, the terminal point of \( c \) is (2,1), and initial point: let's check the grid. The vector \( c \) is from, say, (-4,5) to (2,1)? Wait, no, the y-coordinate at start: the grid has y=5? Wait, the graph: y-axis has 8,6,4,2,0,-2,-4,-6,-8. So the start of \( c \) is at x=-4, y=5 (since it's on the line at y=5, x=-4) and ends at (2,1) (x=2, y=1). So vector \( c \): \( \langle 2 - (-4), 1 - 5 angle = \langle 6, -4 angle \)? Wait, no, wait, maybe the terminal point of \( c \) is (2,1), and initial point: let's check the other vector \( d \): starts at (2,1) (terminal point of \( c \))? Wait, no, the problem is to graph \( c + d \) with initial point at initial point of \( c \). So first, find vectors \( c \) and \( d \) in component form.

Wait, vector \( c \): initial point (let's find coordinates). Looking at the graph, the blue vector \( c \) starts at (-4, 5) (x=-4, y=5) and ends at (2, 1) (x=2, y=1). So \( c = \langle 2 - (-4), 1 - 5 angle = \langle 6, -4 angle \)? Wait, no, that can't be. Wait, maybe the terminal point of \( c \) is (2,1), and initial point: let's check the grid again. Wait, the y-coordinate at the start of \( c \) is 5? Wait, the grid lines: each square is 1 unit. So from x=-4, y=5 to x=2, y=1: that's a vector. Then vector \( d \): starts at (2,1) (terminal point of \( c \))? No, vector \( d \) is the other blue vector, starting at (2,1)? Wait, no, vector \( d \) has initial point (2,1)? Wait, no, the vector \( d \) is from (2,1) to (-4, -7)? Wait, no, looking at the graph, vector \( d \) goes from (2,1) down to (-4, -7)? Wait, no, the arrow of \( d \) is at x=-4, y=-7? Wait, no, the grid: x=-4, y=-7? Wait, the y-axis: 0, -2, -4, -6, -8. So x=-4, y=-7? No, maybe vector \( d \) has components: from (2,1) to (-4, -7)? Wait, no, let's do it properly.

Alternative approach: To add vectors \( c \) and \( d \) using the triangle law (or parallelogram law), we can find their components.

First, find vector \( c \):
- Initial point of \( c \): Let's say \( (x_1, y_1) = (-4, 5) \) (from the graph: x=-4, y=5 is the start of \( c \))
- Terminal point of \( c \): \( (x_2, y_2) = (2, 1) \) (end of \( c \))
So \( c = \langle x_2 - x_1, y_2 - y_1 angle = \langle 2 - (-4), 1 - 5 angle = \langle 6, -4 angle \)? Wait, no, that seems long. Wait, maybe I made a mistake. Wait, the terminal point of \( c \) is (2,1), and initial point: let's check the y-coordinate. The vector \( c \) is going from ( -4, 5 ) to (2, 1): so horizontal change is 2 - (-4) = 6, vertical change is 1 - 5 = -4.

Now vector \( d \):
- Initial point of \( d \): (2, 1) (since it's the terminal point of \( c \), but wait, no—vector \( d \) is drawn from (2,1) to (-4, -7)? Wait, no, the vector \( d \) has an arrow at x=-4, y=-7? Wait, no, looking at the graph, vector \( d \) starts at (2,1) and goes to (-4, -7)? Wait, no, the grid: from (2,1) to (-4, -7): horizontal change is -4 - 2 = -6, vertical change is -7 - 1 = -8? That can't be. Wait, maybe the vector \( d \) is from (2,1) to (-4, -7)? No, that's too long. Wait, maybe the initial point of \( d \) is (2,1), and terminal point is (-4, -7)? Wait, no, let's check the coordinates again.

Wait, the vector \( d \): starts at (2,1) (x=2, y=1) and ends at (x=-4, y=-7)? Wait, no, the y-coordinate at the end of \( d \) is -7? The grid has y=-8 at the bottom. Wait, maybe the vector \( d \) is from (2,1) to (-4, -7)? No, that's 6 units left and 8 units down. But maybe I misread the vectors.

Wait, another way: To graph \( c + d \) with initial point at initial point of \( c \), we can use the parallelogram method or the triangle method (place the initial point of \( d \) at the terminal point of \( c \), then \( c + d \) is from initial point of \( c \) to terminal point of \( d \)).

But maybe the components are simpler. Let's re-express:

Vector \( c \): from (-4, 5) to (2, 1). So \( \Delta x = 2 - (-4) = 6 \), \( \Delta y = 1 - 5 = -4 \). So \( c = \langle 6, -4 angle \).

Vector \( d \): from (2, 1) to (-4, -7)? Wait, no, the vector \( d \) is drawn from (2,1) down to (-4, -7)? Wait, no, the arrow of \( d \) is at x=-4, y=-7? Wait, the grid lines: each square is 1 unit. So from (2,1) to (-4, -7): that's 6 left (x from 2 to -4: 2 - 6 = -4) and 8 down (y from 1 to -7: 1 - 8 = -7). So \( d = \langle -6, -8 angle \).

Now, \( c + d = \langle 6 + (-6), -4 + (-8) angle = \langle 0, -12 angle \)? That can't be right. Wait, maybe I messed up the vectors.

Wait, maybe the initial point of \( c \) is (-4, 5), terminal point (2, 1). Then vector \( c \) is \( (2 - (-4), 1 - 5) = (6, -4) \).

Vector \( d \): initial point (2, 1), terminal point (-4, -7)? No, that's not matching. Wait, maybe the vector \( d \) is from (2,1) to (-4, -7)? Wait, no, the graph shows vector \( d \) going from (2,1) down to (-4, -7)? Wait, the arrow of \( d \) is at x=-4, y=-7? Let's count the grid squares: from (2,1) to (-4, -7): x decreases by 6 (2 to -4: 6 units left), y decreases by 8 (1 to -7: 8 units down). So \( d = (-6, -8) \).

Then \( c + d = (6 - 6, -4 - 8) = (0, -12) \). So the resultant vector \( c + d \) has initial point at (-4, 5) (initial point of \( c \)) and terminal point at (-4 + 0, 5 + (-12)) = (-4, -7). Wait, that's the terminal point of \( d \)! So that makes sense: using the triangle law, \( c + d \) is from initial point of \( c \) to terminal point of \( d \) when \( d \) is placed at terminal point of \( c \).

So to graph \( c + d \):
- Start at the initial point of \( c \): (-4, 5)
- End at the terminal point of \( d \): (-4, -7) (since \( c + d \) has components (0, -12), so from (-4,5), move 0 in x, -12 in y: (-4, 5 - 12) = (-4, -7)).

Answer:

To graph \( \boldsymbol{c + d} \):

1. Identify the initial point of \( \boldsymbol{c} \): \( (-4, 5) \) (from the graph).
2. The resultant vector \( \boldsymbol{c + d} \) has the same initial point as \( c \) (\( (-4, 5) \)) and its terminal point is the terminal point of \( \boldsymbol{d} \) (since \( c + d \) is formed by placing \( d \) at the terminal point of \( c \)). From the graph, the terminal point of \( d \) is \( (-4, -7) \).
3. Draw a vector from \( (-4, 5) \) to \( (-4, -7) \) (a vertical vector downward along \( x = -4 \), 12 units long).

(Note: The key is using the triangle law of vector addition: place the initial point of \( d \) at the terminal point of \( c \); then \( c + d \) is the vector from the initial point of \( c \) to the terminal point of \( d \).)