Example

Verifying that 5x+4y=15x + 4y = 1 and 4x+5y=34x + 5y = 3 are Not Perpendicular

To determine whether the lines 5x+4y=15x + 4y = 1 and 4x+5y=34x + 5y = 3 are perpendicular, convert both equations to slope-intercept form to compare their slopes.

First equation: Solve 5x+4y=15x + 4y = 1 for yy. Subtract 5x5x from both sides: 4y=5x+14y = -5x + 1 Divide by 44: y=54x+14y = -\frac{5}{4}x + \frac{1}{4} The slope is m1=54m_1 = -\frac{5}{4}.

Second equation: Solve 4x+5y=34x + 5y = 3 for yy. Subtract 4x4x from both sides: 5y=4x+35y = -4x + 3 Divide by 55: y=45x+35y = -\frac{4}{5}x + \frac{3}{5} The slope is m2=45m_2 = -\frac{4}{5}.

Check for negative reciprocals: The slopes 54-\frac{5}{4} and 45-\frac{4}{5} are reciprocals, but they share the same negative sign. For lines to be perpendicular, their slopes must be negative reciprocals with opposite signs. Since their product is \left(-\frac{5}{4} ight)\left(-\frac{4}{5} ight) = 1 instead of 1-1, the lines are not perpendicular.

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Updated 2026-04-23

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