Slope of Perpendicular Lines Theorem

If you have already learned about systems of linear equations, then you have probably discussed that the product of the slope of perpendicular lines is -1. The proof of this theorem comes from the fact that any point (x,y) rotated 90 degrees at about the origin becomes (-y,x). One example of this is shown below. The point (3, 4), when rotated 90 degrees counterclockwise becomes (-4,3).

With this fact, we prove this theorem.

Slope of Perpendicular Lines Theorem

If two lines with slopes m_1 and m_2 are perpendicular, then m_1 m_2 = -1


Let P(x_1, y_1) and Q(x_2, y_2) be points on line PQ passing through the origin.



slope of perpendicular linesIf we rotate the points on the origin, then, the new coordinates of the points will be

P^\prime= (-y_1, x_1) and Q^\prime(-y_2, x_2).

If we let m_1 be the slope of PQ and m_2 be the slope of P^\prime Q^\prime , then

m_1 = \displaystyle \frac{y_2 - y_1}{x_2 - x_2}

m_2 = \displaystyle \frac{x_2 - x_1}{-y_2 - (-y_1)} = \frac{x_2 - x_1}{-(y_2 - y_1)} = -\frac{x_2 - x_1}{y_2 - y_1}

Multiplying m_1 and m_2, we have

m_1 m_2 = \left ( \displaystyle \frac{y_2 - y_1}{x_2 - x_1} \right ) \left ( \displaystyle -\frac{x_2 - x_1}{y_2 - y_1} \right ).

= - 1

That proves the theorem above.

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