Question

fluency: sketch the graph of each line

1. $y = \frac{5}{3}x$
2. $x = 0$
3. $y = -\frac{1}{3}x + 3$
4. $y = \frac{1}{5}x - 4$

Explanation:

Problem 7: \( y = \frac{5}{3}x \)

Step 1: Identify the type of line

This is a linear equation in the form \( y = mx \) (where \( m = \frac{5}{3} \) is the slope and the y - intercept \( b = 0 \)). So, the line passes through the origin \((0,0)\).

Step 2: Find another point

We can use the slope to find another point. The slope \( m=\frac{5}{3}\) means that for a run (change in \( x \)) of \( 3 \), the rise (change in \( y \)) is \( 5 \). Starting from the origin \((0,0)\), if we move \( x = 3 \) units to the right (since the run is positive), then \( y=\frac{5}{3}\times3 = 5\). So, another point on the line is \((3,5)\).

Step 3: Sketch the line

Plot the points \((0,0)\) and \((3,5)\) on the coordinate plane and draw a straight line passing through them.

Problem 8: \( x = 0 \)

Step 1: Identify the type of line

The equation \( x = 0 \) represents a vertical line. In the coordinate system, the line \( x = 0 \) is the y - axis.

Step 2: Sketch the line

Draw a vertical line that coincides with the y - axis (passing through all points where \( x = 0 \), for example, \((0,0)\), \((0,1)\), \((0, - 1)\), etc.).

Problem 9: \( y=-\frac{1}{3}x + 3\)

Step 1: Identify the y - intercept

The equation is in the slope - intercept form \( y=mx + b\), where \( b = 3 \) is the y - intercept. So, the line passes through the point \((0,3)\).

Step 2: Find another point using the slope

The slope \( m=-\frac{1}{3}\). The slope \( m=\frac{\text{rise}}{\text{run}}\), so for a run (change in \( x \)) of \( 3 \), the rise (change in \( y \)) is \( - 1\) (negative because the slope is negative). Starting from \((0,3)\), if we move \( x = 3 \) units to the right, then \( y=-\frac{1}{3}\times3+3=- 1 + 3=2\)? Wait, no. Wait, \( y=mx + b\), when \( x = 3\), \( y=-\frac{1}{3}(3)+3=- 1 + 3 = 2\)? Wait, no, let's recalculate. \( y=-\frac{1}{3}x+3\), when \( x = 3\), \( y=-\frac{1}{3}\times3 + 3=-1 + 3=2\)? Wait, no, the slope is \( -\frac{1}{3}\), which means for a run of \( 3\) (increase in \( x \) by \( 3\)), the \( y \) - value decreases by \( 1\). So starting from \((0,3)\), when \( x=3\), \( y = 3-1=2\), so the point is \((3,2)\). Alternatively, if we take a run of \( - 3\) (decrease \( x \) by \( 3\)), then the rise is \( 1\) (since \( m =-\frac{1}{3}=\frac{1}{-3}\)). So when \( x=-3\), \( y=-\frac{1}{3}\times(-3)+3 = 1 + 3=4\), so the point \((-3,4)\) is also on the line.

Step 3: Sketch the line

Plot the points (e.g., \((0,3)\) and \((3,2)\) or \((0,3)\) and \((-3,4)\)) and draw a straight line passing through them.

Problem 10: \( y=\frac{1}{5}x-4\)

Answer:

Step 1: Identify the y - intercept

The equation is in the slope - intercept form \( y = mx + b\), where \( b=-4\). So, the line passes through the point \((0,-4)\).

Step 2: Find another point using the slope

The slope \( m=\frac{1}{5}\). This means that for a run (change in \( x \)) of \( 5 \), the rise (change in \( y \)) is \( 1\). Starting from \((0,-4)\), if we move \( x = 5\) units to the right, then \( y=\frac{1}{5}\times5-4=1 - 4=-3\). So another point on the line is \((5,-3)\).

Step 3: Sketch the line

Plot the points \((0,-4)\) and \((5,-3)\) on the coordinate plane and draw a straight line passing through them.