Question

identify the lines which are perpendicular from the following.

\\(y = 3 + 5x\\) and \\(5y = x + 2\\)

\\(y = 3 - 5x\\) and \\(5y = x + 2\\)

Explanation:

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<reasoning>

Identify the slopes of the first pair of lines

We first analyze the first option containing the lines:
\[
y = 3 + 5x \quad \text{and} \quad 5y = x + 2
\]
Using the Slope of a Linear Function concept, we rewrite both equations in slope-intercept form \(y = mx + b\) to find their slopes:

• For the first line, \(y = 5x + 3\), so the slope is \(m_1 = 5\).
• For the second line, dividing both sides of \(5y = x + 2\) by \(5\) gives \(y = \frac{1}{5}x + \frac{2}{5}\), so the slope is \(m_2 = \frac{1}{5}\).

Check perpendicular condition for the first pair

Using the Parallel and Perpendicular Lines concept, two lines with non-vertical slopes \(m_1\) and \(m_2\) are perpendicular if and only if the product of their slopes is \(-1\):
\[
m_1 \cdot m_2 = -1
\]
Let's calculate the product for the first pair:
\[
m_1 \cdot m_2 = 5 \cdot \frac{1}{5} = 1
eq -1
\]
Thus, these lines are not perpendicular.

Identify the slopes of the second pair of lines

Next, we analyze the second option containing the lines:
\[
y = 3 - 5x \quad \text{and} \quad 5y = x + 2
\]
Using the Slope of a Linear Function concept, we find their slopes:

• For the first line, \(y = -5x + 3\), so the slope is \(m_1 = -5\).
• For the second line, \(5y = x + 2\) simplifies to \(y = \frac{1}{5}x + \frac{2}{5}\), so the slope is \(m_2 = \frac{1}{5}\).

Check perpendicular condition for the second pair

Using the Parallel and Perpendicular Lines concept, we calculate the product of their slopes:
\[
m_1 \cdot m_2 = -5 \cdot \frac{1}{5} = -1
\]
Since the product of their slopes is exactly \(-1\), these two lines are perpendicular.
</reasoning>

<answer>
<mcq-option>(A) \(y = 3 + 5x\) and \(5y = x + 2\)</mcq-option>
<mcq-correct>(B) \(y = 3 - 5x\) and \(5y = x + 2\)</mcq-correct>
</answer>

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Answer:

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<reasoning>

Identify the slopes of the first pair of lines

We first analyze the first option containing the lines:
\[
y = 3 + 5x \quad \text{and} \quad 5y = x + 2
\]
Using the Slope of a Linear Function concept, we rewrite both equations in slope-intercept form \(y = mx + b\) to find their slopes:

• For the first line, \(y = 5x + 3\), so the slope is \(m_1 = 5\).
• For the second line, dividing both sides of \(5y = x + 2\) by \(5\) gives \(y = \frac{1}{5}x + \frac{2}{5}\), so the slope is \(m_2 = \frac{1}{5}\).

Check perpendicular condition for the first pair

Using the Parallel and Perpendicular Lines concept, two lines with non-vertical slopes \(m_1\) and \(m_2\) are perpendicular if and only if the product of their slopes is \(-1\):
\[
m_1 \cdot m_2 = -1
\]
Let's calculate the product for the first pair:
\[
m_1 \cdot m_2 = 5 \cdot \frac{1}{5} = 1
eq -1
\]
Thus, these lines are not perpendicular.

Identify the slopes of the second pair of lines

Next, we analyze the second option containing the lines:
\[
y = 3 - 5x \quad \text{and} \quad 5y = x + 2
\]
Using the Slope of a Linear Function concept, we find their slopes:

• For the first line, \(y = -5x + 3\), so the slope is \(m_1 = -5\).
• For the second line, \(5y = x + 2\) simplifies to \(y = \frac{1}{5}x + \frac{2}{5}\), so the slope is \(m_2 = \frac{1}{5}\).

Check perpendicular condition for the second pair

Using the Parallel and Perpendicular Lines concept, we calculate the product of their slopes:
\[
m_1 \cdot m_2 = -5 \cdot \frac{1}{5} = -1
\]
Since the product of their slopes is exactly \(-1\), these two lines are perpendicular.
</reasoning>

<answer>
<mcq-option>(A) \(y = 3 + 5x\) and \(5y = x + 2\)</mcq-option>
<mcq-correct>(B) \(y = 3 - 5x\) and \(5y = x + 2\)</mcq-correct>
</answer>

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