The pair of lines that are perpendicular is D) \(x + y = 10; x - y = 3\).
Lines are perpendicular if the product of their slopes is -1. The slope of a line in the form \(y = mx + b\) is \(m\).
Let's check each pair of lines:
A) Line 1: \(8x - 4y = 24\)
Rearrange to slope-intercept form: \(-4y = -8x + 24 \implies y = 2x - 6\)
Slope of Line 1: \(m_1 = 2\)
Line 2: \(y = -2x + 7\)
Slope of Line 2: \(m_2 = -2\)
\(m_1 \times m_2 = 2 \times (-2) = -4\)
The slopes are not negative reciprocals, so Lines A are not perpendicular.
B) Line 3: \(x - 3y = 15\)
Rearrange to slope-intercept form: \(-3y = -x + 15 \implies y = \frac{1}{3}x - 5\)
Slope of Line 3: \(m_3 = \frac{1}{3}\)
Line 4: \(y = 3x + 4\)
Slope of Line 4: \(m_4 = 3\)
\(m_3 \times m_4 = \frac{1}{3} \times 3 = 1\)
The slopes are not negative reciprocals, so Lines B are not perpendicular.
C) Line 5: \(2x - 5y = 5\)
Rearrange to slope-intercept form: \(-5y = -2x + 5 \implies y = \frac{2}{5}x - 1\)
Slope of Line 5: \(m_5 = \frac{2}{5}\)
Line 6: \(5x - 2y = 4\)
Rearrange to slope-intercept form: \(-2y = -5x + 4 \implies y = \frac{5}{2}x - 2\)
Slope of Line 6: \(m_6 = \frac{5}{2}\)
\(m_5 \times m_6 = \frac{2}{5} \times \frac{5}{2} = 1\)
The slopes are not negative reciprocals, so Lines C are not perpendicular.
D) Line 7: \(x + y = 10\)
Rearrange to slope-intercept form: \(y = -x + 10\)
Slope of Line 7: \(m_7 = -1\)
Line 8: \(x - y = 3\)
Rearrange to slope-intercept form: \(y = x - 3\)
Slope of Line 8: \(m_8 = 1\)
\(m_7 \times m_8 = -1 \times 1 = -1\)
The slopes are negative reciprocals, so Lines D are perpendicular.
E) Line 9: \(5x - y = 1\)
Rearrange to slope-intercept form: \(-y = -5x + 1 \implies y = 5x - 1\)
Slope of Line 9: \(m_9 = 5\)
Line 10: \(y = \frac{1}{5}x + 9\)
Slope of Line 10: \(m_{10} = \frac{1}{5}\)
\(m_9 \times m_{10} = 5 \times \frac{1}{5} = 1\)
The slopes are not negative reciprocals, so Lines E are not perpendicular.
The pair of lines that are perpendicular is D) \(x + y = 10; x - y = 3\).