Example

Verifying that y=βˆ’3x+2y = -3x + 2 and xβˆ’3y=4x - 3y = 4 are Perpendicular

To determine whether the lines y=βˆ’3x+2y = -3x + 2 and xβˆ’3y=4x - 3y = 4 are perpendicular, rewrite each equation in slope-intercept form and check if their slopes are negative reciprocals.

First equation: y=βˆ’3x+2y = -3x + 2 is already in slope-intercept form. Its slope is m1=βˆ’3m_1 = -3.

Second equation: Solve xβˆ’3y=4x - 3y = 4 for yy. Subtract xx from both sides: βˆ’3y=βˆ’x+4-3y = -x + 4 Divide every term by βˆ’3-3: y=13xβˆ’43y = \frac{1}{3}x - \frac{4}{3} The slope is m2=13m_2 = \frac{1}{3}.

Check for negative reciprocals: The slopes βˆ’3-3 and 13\frac{1}{3} have opposite signs and are reciprocals of each other. Verify by computing their product:

ight) = -1$$ Because the product of the slopes is $$-1$$, the lines are perpendicular.

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Updated 2026-04-23

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