3680.31 – Finding a Slope

Consider the line y = x and the line y = 0, both of which go through the origin. They form a 45 degree angle in the first quadrant. The bisector of this angle is therefore a line with equation y = mx. What's the value of m? It's tempting to think that m = 1/2, but, heh heh, this is not the case.

Solution

We recollect: ab=xy\frac{a}{b}=\frac{x}{y}

So, here's snazzy diagram no. 1 (below)

Using our recollection, we have 21=1−yy→y2=1−y→y2+y=1→y(1+2)=1\frac{\sqrt2}{1}=\frac{1-y}{y}\rightarrow y\sqrt2 = 1 - y \rightarrow y\sqrt2 + y = 1 \rightarrow y(1+\sqrt2) = 1

→y=11+2=1(1−2(1+2)(1−2)=1−21−2=2−1\rightarrow y = \frac{1}{1+\sqrt2} = \frac{1(1-\sqrt2}{(1+\sqrt2)(1-\sqrt2)} = \frac{1-\sqrt2}{1-2} = \sqrt2 - 1.

So y1\frac{y}{1} = m, and m = 2−1\sqrt2 - 1. That's snazzy geometry.

We can also use trig. We recollect: tanx2=1−cosxsinxtan\frac{x}{2}= \frac{1-cosx}{sinx}.

And here's snazzy diagram no. 2, also below.

m = m/1 =tan22.5 = tan(45/2) = 1−cos45sin45\frac{1 - cos45}{sin45}.

We know that cos45 = sin45 = 1/\sqrt 2.

So, continuing, m = 1−1212\frac{1 - \frac{1}{\sqrt2}} {\frac{1}{\sqrt2}} = 2−1212=2−1\frac{\frac{\sqrt2 - 1}{\sqrt2}} {\frac{1}{\sqrt2}} =\sqrt2 - 1.

So m = 2−1\sqrt2 - 1, by snazzy trig.

It's all just snazzy.