Question

Both of these tables represent linear graphs:


-3
8
-2
10

12
0
14
1
16
2
18
3
20
4
22
-3
-2
0
-2
0
3
4
8
10
12
Does the line through the first set of points intersect with the line through the second set of points?
(1 point)
• Yes...these lines are parallel
• Yes...these lines are perpendicular
• You cannot tell without a graph
• Yes...these lines are not parallel nor perpendicular
• No... these lines are perpendicular
• No...these lines are parallel

Answer 1

To determine if the lines through the two sets of points intersect, we need to analyze their slope and y-intercept.

**First Set of Points:**
- Points: (-3, 8), (-2, 10)
- The slope (m) can be calculated using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
- Using the points:
\[ m = \frac{10 - 8}{-2 + 3} = \frac{2}{1} = 2 \]
- Now we take one point, say (-3, 8), and plug it into the slope-intercept form \( y = mx + b \) to find the y-intercept (b):
\[ 8 = 2(-3) + b \]
\[ 8 = -6 + b \]
\[ b = 14 \]
- The equation of the first line is \( y = 2x + 14 \).

**Second Set of Points:**
- Points: (12, 0), (14, 1), (16, 2), etc.
- Using the points (12, 0) and (14, 1):
\[ m = \frac{1 - 0}{14 - 12} = \frac{1}{2} \]
- Using the point (12, 0) to find the y-intercept:
\[ 0 = \frac{1}{2}(12) + b \]
\[ 0 = 6 + b \]
\[ b = -6 \]
- The equation of the second line is \( y = \frac{1}{2}x - 6 \).

**Determine Intersection**:
1. Set the equations equal to find the x-coordinate of the intersection:
\[ 2x + 14 = \frac{1}{2}x - 6 \]

2. Multiply through by 2 to eliminate the fraction:
\[ 4x + 28 = x - 12 \]
\[ 4x - x = -12 - 28 \]
\[ 3x = -40 \]
\[ x = -\frac{40}{3} \]

3. Substitute \( x = -\frac{40}{3} \) into either equation to find the y-coordinate:
For the first line:
\[ y = 2(-\frac{40}{3}) + 14 = -\frac{80}{3} + 14 = -\frac{80}{3} + \frac{42}{3} = -\frac{38}{3} \]
So the intersection point is \((- \frac{40}{3}, - \frac{38}{3})\).

Since we determined two lines with distinct slopes (2 and \( \frac{1}{2} \)), they are not parallel. Thus, they are intersecting lines.

Therefore, the answer is:

**Yes...these lines are not parallel nor perpendicular.**