Question

Select the pair of equations that represent perpendicular lines



a
2y+6x=10

4y+8x=12

b
2x+4y=12

2x=y+6

c
4y=10x+14

y=(5/2)x+3

Answer 1

To determine which pair of equations represents perpendicular lines, we need to find the slopes of the lines in each pair. Lines are perpendicular if the product of their slopes is -1.

Let's convert each equation to slope-intercept form (y = mx + b) where m is the slope.

### Pair a:
1. \(2y + 6x = 10\)
\[
2y = -6x + 10 \\
y = -3x + 5 \quad \text{(slope = -3)}
\]

2. \(4y + 8x = 12\)
\[
4y = -8x + 12 \\
y = -2x + 3 \quad \text{(slope = -2)}
\]

Slope product: \(-3 \times -2 = 6\) (not perpendicular)

### Pair b:
1. \(2x + 4y = 12\)
\[
4y = -2x + 12 \\
y = -\frac{1}{2}x + 3 \quad \text{(slope = -\frac{1}{2})}
\]

2. \(2x = y + 6\)
\[
y = 2x - 6 \quad \text{(slope = 2)}
\]

Slope product: \(-\frac{1}{2} \times 2 = -1\) (perpendicular)

### Pair c:
1. \(4y = 10x + 14\)
\[
y = \frac{10}{4}x + \frac{14}{4} \\
y = \frac{5}{2}x + \frac{7}{2} \quad \text{(slope = \frac{5}{2})}
\]

2. \(y = \frac{5}{2}x + 3\)
\[
\text{(slope = \frac{5}{2})}
\]

Slope product: \(\frac{5}{2} \times \frac{5}{2} = \frac{25}{4}\) (not perpendicular)

### Conclusion:
The pair of equations that represent perpendicular lines is **pair b**:

1. \(2x + 4y = 12\) (slope = -\frac{1}{2})
2. \(2x = y + 6\) (slope = 2)