Blog: Prove that if two non-vertical lines are perpendicular, then the product of their slopes is -1.

Saturday, January 31, 2015

Prove that if two non-vertical lines are perpendicular, then the product of their slopes is -1.

One way to prove this is by using Pythagorean theorem.

Let's label the coordinates of point of the intersection of the two perpendicular lines as $\displaystyle{\left({x}_{{0}},{y}_{{0}}\right)}$ . Then, since the lines are perpendicular, if we pick any point on the first line with the coordinates $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}$ , and a point on the second line with the coordinates $\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}$ , the line segments connecting these three points will form a right triangle, with the right angle at the point $\displaystyle{\left({x}_{{0}},{y}_{{0}}\right)}$ .

The lengths of the sides of a right triangle must obey Pythagorean theorem. So, for our right triangle, the hypotenuse is the segment connecting points 1 and 2 an it has the length

$\displaystyle{{d}_{{{12}}}^{{2}}}={{\left({x}_{{2}}-{x}_{{1}}\right)}}^{{2}}+{{\left({y}_{{2}}-{y}_{{1}}\right)}}^{{2}}$

The other sides' lengths are the distances between the point of the intersection of the two lines and points 1 and 2, respectively. They are equal

$\displaystyle{{d}_{{{10}}}^{{2}}}={{\left({x}_{{1}}-{x}_{{0}}\right)}}^{{2}}+{{\left({y}_{{1}}-{y}_{{0}}\right)}}^{{2}}$ and

$\displaystyle{{d}_{{{20}}}^{{2}}}={{\left({x}_{{2}}-{x}_{{0}}\right)}}^{{2}}+{{\left({y}_{{2}}-{y}_{{0}}\right)}}^{{2}}$

According to Pythagorean Theorem, $\displaystyle{{d}_{{{10}}}^{{2}}}+{{d}_{{{20}}}^{{2}}}={{d}_{{{12}}}^{{2}}}$ . If we now plug in the expressions for the d's in terms of the coordinates x and y, then by doing a lot of algebra, we can show that the resultant equation will lead to the required relationship between the slopes.

First, rewrite the equation above as

$\displaystyle{{d}_{{{10}}}^{{2}}}-{{d}_{{{12}}}^{{2}}}+{{d}_{{{20}}}^{{2}}}={0}$

For now, I will work with the part of the equation that involves x-coordinates only, because the part involving y-coordinates will be identical.

The part involving x-coordinates is

$\displaystyle{{\left({x}_{{1}}-{x}_{{0}}\right)}}^{{2}}-{{\left({x}_{{2}}-{x}_{{1}}\right)}}^{{2}}+{{\left({x}_{{2}}-{x}_{{0}}\right)}}^{{2}}$

The first two terms make up a difference of two squares and can be factored as such:

$\displaystyle{\left[{\left({x}_{{1}}-{x}_{{0}}\right)}-{\left({x}_{{2}}-{x}_{{1}}\right)}\right]}\cdot{\left[{\left({x}_{{1}}-{x}_{{0}}\right)}+{\left({x}_{{2}}-{x}_{{1}}\right)}\right]}+{{\left({x}_{{2}}-{x}_{{0}}\right)}}^{{2}}$

Opening parenthesis and combining like terms in square brackets, we get:

$\displaystyle{\left({2}{x}_{{1}}-{x}_{{0}}-{x}_{{2}}\right)}{\left({x}_{{2}}-{x}_{{0}}\right)}+{{\left({x}_{{2}}-{x}_{{0}}\right)}}^{{2}}$

Notice that these two terms have a common binomial factor, which can be factored out:

$\displaystyle{\left({x}_{{2}}-{x}_{{0}}\right)}{\left({2}{x}_{{1}}-{x}_{{0}}-{x}_{{2}}+{x}_{{2}}-{x}_{{0}}\right)}={\left({x}_{{2}}-{x}_{{0}}\right)}{\left({2}{x}_{{1}}-{2}{x}_{{0}}\right)}$ , or

$\displaystyle{2}{\left({x}_{{2}}-{x}_{{0}}\right)}{\left({x}_{{1}}-{x}_{{0}}\right)}$

Again, working with the y-coordinates will yield identical result, so the equation of Pythagorean Theorem will become

$\displaystyle{2}{\left({x}_{{2}}-{x}_{{0}}\right)}{\left({x}_{{1}}-{x}_{{0}}\right)}+{2}{\left({y}_{{2}}-{y}_{{0}}\right)}{\left({y}_{{1}}-{y}_{{0}}\right)}={0}$

The factor of 2 can be canceled, and now we can separate the coordinates:

$\displaystyle{\left({y}_{{2}}-{y}_{{0}}\right)}{\left({y}_{{1}}-{y}_{{0}}\right)}=-{\left({x}_{{2}}-{x}_{{0}}\right)}{\left({x}_{{1}}-{x}_{{0}}\right)}$

Divide both sides by $\displaystyle{\left({x}_{{2}}-{x}_{{0}}\right)}{\left({y}_{{1}}-{y}_{{0}}\right)}$ :

$\displaystyle\frac{{{y}_{{2}}-{y}_{{0}}}}{{{x}_{{2}}-{x}_{{0}}}}=-\frac{{{x}_{{1}}-{x}_{{0}}}}{{{y}_{{1}}-{y}_{{0}}}}$

We can recognize that the left side is the slope of the line passing through the points with the coordinates $\displaystyle{\left({x}_{{0}},{y}_{{0}}\right)}$ and $\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}$ (let's call it $\displaystyle{m}_{{2}}$ ), and the right side is the negative reciprocal of the slope of the line passing through the points with the coordinates $\displaystyle{\left({x}_{{0}},{y}_{{0}}\right)}$ and $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}$ (call it $\displaystyle{m}_{{1}}$ ). So what we get is

$\displaystyle{m}_{{2}}=-\frac{{1}}{{m}_{{1}}}$ , which can also be written as $\displaystyle{m}_{{1}}\cdot{m}_{{2}}=-{1}$

The product of the slopes of the two perpendicular lines is -1.

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Saturday, January 31, 2015

Prove that if two non-vertical lines are perpendicular, then the product of their slopes is -1.

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Saturday, January 31, 2015

Prove that if two non-vertical lines are perpendicular, then the product of their slopes is -1.

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What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?