Question

1. which of the given pairs of lines is perpendicular?

$3x + y = 8$
$y = 3x - 8$
$y = \frac{1}{4}x - 3$
$4x - y = 7$
$y = 10$
$x = -1$
$y = 2x + 5$
$y = 2x + 7$

Explanation:

Step1: Recall perpendicular lines rule

Two lines are perpendicular if the product of their slopes is $-1$ (or one is horizontal, the other vertical).

Step2: Analyze first pair

For $3x + y = 8$, rewrite as $y = -3x + 8$, slope $m_1 = -3$. For $y = 3x - 8$, slope $m_2 = 3$. Product: $-3\times3 = -9
eq -1$. Not perpendicular.

Step3: Analyze second pair

For $y = \frac{1}{4}x - 3$, slope $m_1 = \frac{1}{4}$. For $4x - y = 7$, rewrite as $y = 4x - 7$, slope $m_2 = 4$. Product: $\frac{1}{4}\times4 = 1
eq -1$. Not perpendicular.

Step4: Analyze third pair

$y = 10$ is horizontal (slope $0$), $x = -1$ is vertical (undefined slope). Horizontal and vertical lines are perpendicular.

Step5: Analyze fourth pair

Both $y = 2x + 5$ and $y = 2x + 7$ have slope $2$. Parallel, not perpendicular.

Answer:

The pair $y = 10$ and $x = -1$ (the third option: $y = 10$; $x = -1$)