Question

y = -x + 3\
x - 3y = 3\
plot two lines by clicking

Explanation:

Step1: Analyze the first equation \( y = -x + 3 \)

This is in slope - intercept form \( y=mx + b \), where the slope \( m=- 1 \) and the y - intercept \( b = 3 \). To find two points on this line:

• When \( x = 0 \), \( y=-0 + 3=3 \), so we have the point \( (0,3) \).
• When \( y = 0 \), \( 0=-x + 3\), then \( x = 3 \), so we have the point \( (3,0) \).

Step2: Analyze the second equation \( x-3y=3 \)

We can rewrite it in slope - intercept form. Solve for \( y \):
\[

$$\begin{align*} x-3y&=3\\ - 3y&=-x + 3\\ y&=\frac{1}{3}x-1 \end{align*}$$

\]
Here, the slope \( m = \frac{1}{3} \) and the y - intercept \( b=-1 \). To find two points on this line:

• When \( x = 0 \), \( y=\frac{1}{3}(0)-1=-1 \), so we have the point \( (0, - 1) \).
• When \( y = 0 \), \( 0=\frac{1}{3}x-1\), then \( \frac{1}{3}x=1\), and \( x = 3 \), so we have the point \( (3,0) \).

(If we want to find the intersection point of the two lines, we can substitute \( y=-x + 3 \) into \( x - 3y=3 \)):

Step3: Find the intersection point (optional, if we want to solve the system)

Substitute \( y=-x + 3 \) into \( x-3y = 3 \):
\[

$$\begin{align*} x-3(-x + 3)&=3\\ x + 3x-9&=3\\ 4x&=3 + 9\\ 4x&=12\\ x&=3 \end{align*}$$

\]
When \( x = 3 \), \( y=-3 + 3=0 \). So the intersection point of the two lines is \( (3,0) \).

To plot the lines:

• For \( y=-x + 3 \), plot the points \( (0,3) \) and \( (3,0) \) and draw a line through them.
• For \( y=\frac{1}{3}x-1 \), plot the points \( (0,-1) \) and \( (3,0) \) and draw a line through them.

Answer:

To plot the lines:

• For \( y=-x + 3 \): Use points \( (0,3) \) and \( (3,0) \).
• For \( x - 3y=3 \) (or \( y=\frac{1}{3}x-1 \)): Use points \( (0,-1) \) and \( (3,0) \). The intersection point of the two lines is \( (3,0) \).