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Analytic Geometry:

AG 004 Rectangular coordinates:

By William E. Steinman:

October 1, 2007:

In our last notes on rectangular coordinates, we found we could calculate the angle of inclination or slope of a line in our coordinate system.

Okay, it should be clear that if two lines are parallel, their slopes will be equal.

What is not so clear is that if two lines are perpendicular, the slope of one is the negative reciprocal of the other.

It would be easy to say that is intuitively obvious. Some math books do that, but it is not obvious to me. And I will not say it.

What are we really saying?

Take two lines, l₁ and l₂ that are perpendicular.

Let the slope of l₁ be m₁ and the slope of l₂ be m₂.

What the above statement asserts is that m₁ is equal to –1/m₂, or m₁m₂ = -1.

Can we prove that?

If this is mathematics, we ought to be able to prove it.

Let’s fetch out our old coordinate sketch and modify it a bit.

We can begin with a square that has two sides parallel to each axis.

Then we can connect the corners. The two connecting lines are perpendicular, and they create 4 right triangles with the corners of the square at points a, b, c, and d.

Let us select angles a and d as the angles of interest.

Then, by our assertion, the slope of line ac is equal to the negative reciprocal of line db.

The slope of the line ac is tan a = (c_x – d_x)/(d_y – a_y).

The slope of the line db is tan d = (b_x – a_x)/(d_y – a_y).

Since the line db is such that it really belongs to the second quadrant, the slope will be negative. So tan d = -(b_x – a_x)/(d_y – a_y).

If our assertion is true, then (c_x – d_x)/(d_y – a_y) = - 1/-(b_x – a_x)/(d_y – a_y).

Or, (c_x – d_x)/(d_y – a_y) . -(d_y – a_y)/ (b_x – a_x) = -1

Or (c_x – d_x)/ -(b_x – a_x) = -1.

However, a look at our diagram reveals that b_x is the same as c_x and a_x is the same as d_x.

So we get -1 = -1

There is probably an easier way to prove this, but this worked okay.

Intersecting lines:

The angle measured in a positive direction between line l₁ with slope = m₁ to line l₂ with slope m₂ is the angle whose tangent is:

(m₂ – m₁)/(1 + m₂m₁)

Use this sketch to prove that statement.

< c = < a + < b

or < a = < c - < b

tan < a = tan (< c - < b) = (tan < c – tan < b)/(1 + tan < c . tan < b)

= (m₂ – m₁)/(1 + m₂m₁)

Okay?

We can also find the area of any polygon whose vertices are given.

For sure, it is easier if we choose our polygon to be a triangle.

Let’s choose the easier way this time.

Look at this diagram.

The area of this triangle is given as

A = ½(x₁y₂ + x₂y₃ + x₃y₁ – x₃y₂ – x₂y₁ – x₁y₃)

You should be able to prove this to yourself by combining the areas of three trapezoids like this.

m₁p₁p₃m₃, + m₃p₃p₂m₂ – m₁p₁p₂m₂

Remember from study notes 022

The area of a trapezoid is equal to ½ the product of the length of the altitude and the sum of the lengths of its bases.

A trapezoid is a quadrilateral having two parallel sides.

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