Example

Verifying that 7x+2y=37x + 2y = 3 and 2x+7y=52x + 7y = 5 are Not Perpendicular

To determine whether the lines 7x+2y=37x + 2y = 3 and 2x+7y=52x + 7y = 5 are perpendicular, convert both equations to slope-intercept form and check if their slopes are negative reciprocals.

First equation: Solve 7x+2y=37x + 2y = 3 for yy. Subtract 7x7x from both sides: 2y=7x+32y = -7x + 3 Divide both sides by 22: y=72x+32y = -\frac{7}{2}x + \frac{3}{2} The slope is m1=72m_1 = -\frac{7}{2}.

Second equation: Solve 2x+7y=52x + 7y = 5 for yy. Subtract 2x2x from both sides: 7y=2x+57y = -2x + 5 Divide both sides by 77: y=27x+57y = -\frac{2}{7}x + \frac{5}{7} The slope is m2=27m_2 = -\frac{2}{7}.

Check for negative reciprocals: The slopes 72-\frac{7}{2} and 27-\frac{2}{7} are reciprocals, but they share the same sign (both are negative). Therefore, they are not negative reciprocals. This can also be verified by computing their product:

ight)\left(-\frac{2}{7} ight) = 1$$ Because the product of the slopes is $$1$$ rather than $$-1$$, the lines are not perpendicular.

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Updated 2026-05-03

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