Lesson Finding the slope of the bisector to the angle formed by two given lines in a coordinate plane

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<H2>Finding the slope of the bisector to the angle formed by two given lines in a coordinate plane</H2>

<H3>Problem 1</H3>An acute angle is formed by two lines of slopes &nbsp;(1/2) &nbsp;and &nbsp;(2/11). 
What is the slope of the line which bisects the angle?

<B>Solution</B>

<pre>
The given slopes are the tangent values of corresponding angles.

Thus we have  tan(a) = {{{1/2}}}  for one angle, "a", and  tan(b) = {{{2/11}}}  for the other angle, "b".

They ask you about  {{{tan((a+b)/2)}}}.

Using well known formulas of Trigonometry,

{{{tan(a+b)}}} = {{{(tan(a) + tan(b))/(1 - tan(a)*tan(b))}}} = {{{((1/2 + 2/11))/(1 - (1/2)*(2/11))}}} = {{{((11/22 + 4/22))/((1 - 1/11))}}} = {{{((15/22))/(10/11))}}} = {{{(15*11)/(10*22)}}} = {{{15/20}}} = {{{3/4}}}.

The last step is to use the formula for  tan(c/2) via  tan(c)

{{{tan(c/2)}}} = {{{(-1 +- sqrt(1+tan^2(c)))/tan(c)}}}.

When you apply it, you will get the ANSWER

{{{tan((a+b)/2)}}} = {{{((-1 + sqrt(1 + (3/4)^2)))/(3/4))}}} = {{{(-1 + sqrt(25/16))/(3/4)}}} = {{{(-1 + 5/4)/(3/4)}}} = {{{((1/4))/((3/4))}}} = {{{1/3}}}.

According to the condition, you may use the sign " + " at sqrt instead of " +/- ".

<U>ANSWER</U>.  The slope is  {{{1/3}}}.
</pre>

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