Question

Draw an angle and label it as \[\angle BAC\] then draw its bisector ray AX and take a point P on it. From P draw line segments PM and PN such that \[PM\bot AB\] and \[PN\bot AC\] where M and N are points on AB and AC respectively. Measure the lengths PM and PN. Are the two lengths equal?

Answer 1

Hint: The diagram that represents the given data is shown below.

We solve this problem by assuming the ray AX as X – axis and point A as origin. Then we find the equation of lines AB and AC to find the perpendicular distance from point to line formula to find PM and PN.
For finding the equations of lines AB and AC we use the standard result that is if \[\theta \] is the angle between two lines with the slopes as \[{{m}_{1}},{{m}_{2}}\] then,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}.{{m}_{2}}} \right|\]
The formula for perpendicular distance from point \[A\left( h,k \right)\] to line \[ax+by+0\] is given as
\[D=\dfrac{\left| ah+bk+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]

Complete step by step answer:
Let us assume that the ray AX as X – axis and point A as origin.
We know that the equation of X – axis as
\[\Rightarrow y=0\]
Therefore the slope of AX is
\[\Rightarrow m=0\]
Let us assume that the slope of line AB as \[{{m}_{1}}\]
We know that the equation of line passing through origin and having slope \['m'\] is given as
\[\Rightarrow y=mx\]
By using this result we get the equation of AB as
\[\Rightarrow y={{m}_{1}}x\]
We know that if \[\theta \] is the angle between two lines with the slopes as \[{{m}_{1}},{{m}_{2}}\] then,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}.{{m}_{2}}} \right|\]
By applying the above formula to lines AB and AX we get
\[\Rightarrow \tan \theta =\left| \dfrac{{{m}_{1}}-m}{1+{{m}_{1}}.m} \right|\]
By substituting the value of \['m'\] in above equation we get
\[\begin{align}
  & \Rightarrow \tan \theta =\left| \dfrac{{{m}_{1}}-0}{1+{{m}_{1}}\left( 0 \right)} \right| \\
 & \Rightarrow \tan \theta =\left| {{m}_{1}} \right|.........equation(i) \\
\end{align}\]
Let us assume that the slope of line AC as \[{{m}_{2}}\]
We know that the equation of line passing through origin and having slope \['m'\] is given as
\[\Rightarrow y=mx\]
By using this result we get the equation of AC as
\[\Rightarrow y={{m}_{2}}x\]

We know that if \[\theta \] is the angle between two lines with the slopes as \[{{m}_{1}},{{m}_{2}}\] then,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}.{{m}_{2}}} \right|\]
By applying the above formula to lines AC and AX we get
\[\Rightarrow \tan \theta =\left| \dfrac{{{m}_{2}}-m}{1+{{m}_{2}}.m} \right|\]
By substituting the value of \['m'\] in above equation we get
\[\begin{align}
  & \Rightarrow \tan \theta =\left| \dfrac{{{m}_{2}}-0}{1+{{m}_{2}}\left( 0 \right)} \right| \\
 & \Rightarrow \tan \theta =\left| {{m}_{2}} \right|.........equation(i) \\
\end{align}\]
We know that the angular bisector divides the angle into two equal parts
By using the above condition, by equating both equation (i) and equation (ii) we get
\[\begin{align}
  & \Rightarrow \left| {{m}_{1}} \right|=\left| {{m}_{2}} \right| \\
 & \Rightarrow \left| \dfrac{{{m}_{1}}}{{{m}_{2}}} \right|=1......equation(iii) \\
\end{align}\]
Now, let us assume that the point P as \[P\left( h,0 \right)\]
We know that the formula for perpendicular distance from point \[A\left( h,k \right)\] to line \[ax+by+0\] is given as
\[D=\dfrac{\left| ah+bk+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
By using the above formula from point P to line AB we get
\[\begin{align}
  & \Rightarrow PM=\dfrac{\left| {{m}_{1}}\times h-0 \right|}{\sqrt{{{m}_{1}}^{2}+{{1}^{2}}}} \\
 & \Rightarrow PM=\dfrac{\left| {{m}_{1}} \right|\times \left| h \right|}{\sqrt{{{m}_{1}}^{2}+{{1}^{2}}}} \\
\end{align}\]
Now by using the distance formula from point P to line AC we get
\[\begin{align}
  & \Rightarrow PN=\dfrac{\left| {{m}_{2}}\times h \right|}{\sqrt{{{m}_{2}}^{2}+{{1}^{2}}}} \\
 & \Rightarrow PN=\dfrac{\left| {{m}_{2}} \right|\times \left| h \right|}{\sqrt{{{m}_{2}}^{2}+{{1}^{2}}}} \\
\end{align}\]
Now let us divide the length of PM with PN then we get
\[\Rightarrow \dfrac{PM}{PN}=\left| \dfrac{{{m}_{1}}}{{{m}_{2}}} \right|\]
By substituting the required value from equation (iii) we get
\[\begin{align}
  & \Rightarrow \dfrac{PM}{PN}=1 \\
 & \Rightarrow PM=PN \\
\end{align}\]

Therefore, the lengths PM and PN are equal.

Note: We can explain the above problem in another method.
We know that the definition of angular bisector is the line that divides the angle into two equal angles.
This is the general definition.
The mathematical definition of angular bisector is the locus of points that are equidistant from two intersecting lines is called the angular bisector.
Let us assume that the angular bisector as shown below

From the definition it says that point P is equidistant from both AB and AC
Therefore, we can conclude that \[PM=PN\]
Therefore, the lengths PM and PN are equal.