Question

The vertices of the triangle are\[A\left( {1,{\rm{ }}1} \right),{\rm{ }}B\left( {4,{\rm{ }} - 2} \right)\]and \[C\left( {5,{\rm{ }}5} \right)\]. Find the equation of the perpendicular dropped from \[C\] to the interior bisector of the angle \[A\].
A. \[{\rm{y - 5 = 0}}\]
B. \[{\rm{ x - 5 = 0}}\]
C. \[{\rm{ y + 5 = 0}}\]

Answer 1

Hint: We are given the vertices of the triangle and we have to find the equation of the perpendicular. We will use the concept of slope in the given question. So firstly we will find the value of the slope and put it into the slope point form.

Formula used:
The following formula will be useful for this question
\[\begin{array}{l}{m_1}\; = {\rm{ }}\dfrac{{({y_2}\;-{\rm{ }}{y_1})}}{{({x_2}\;-{\rm{ }}{x_1})}}\\tan{\rm{ }}\alpha {\rm{ }} = {\rm{ }}\left| {\dfrac{{({m_1}\;-{\rm{ }}{m_2})}}{{(1{\rm{ }} + {\rm{ }}{m_1}{m_2})}}} \right|\\{m_1}{m_2}\; = {\rm{ }} - 1\end{array}\]

Complete step-by-step solution:


Image: The triangle having vertices\[A\left( {1,{\rm{ }}1} \right),{\rm{ }}B\left( {4,{\rm{ }} - 2} \right)\]and\[C\left( {5,{\rm{ }}5} \right)\]

Firstly we will find the slope of\[AC\]
\[\begin{array}{l}{m_1}\; = \dfrac{{\left( { - 2 - 1} \right)}}{{\left( {4 - 1} \right)}}\\{m_1} = \dfrac{{ - 3}}{3}\\{m_1} = - 1\end{array}\]
Now, we will find the slope of\[AC\]
\[\begin{array}{l}{m_2}\; = {\rm{ }}\dfrac{{\left( {5 - 1} \right)}}{{\left( {5 - 1} \right)}}\\{m_2}\; = 1\end{array}\]
Here, \[{m_1}{m_2}\; = {\rm{ }} - 1\]
So \[AC\] is perpendicular to\[AC\]
Let slope of bisector line \[AD = m\]
Since\[\angle A = {90^0}\]the angle between \[\;AD\] and\[AC\]is \[{45^0}\]
\[\begin{array}{l}tan{\rm{ }}\alpha {\rm{ }} = {\rm{ }}\left| {\dfrac{{({m_1}\;-{\rm{ }}{m_2})}}{{(1{\rm{ }} + {\rm{ }}{m_1}{m_2})}}} \right|\\\begin{array}{*{20}{l}}{tan{\rm{ }}45{\rm{ }} = {\rm{ }}\left| {\dfrac{{\left( {m{\rm{ }}-{\rm{ }}1} \right)}}{{\left( {1{\rm{ }} + {\rm{ }}m} \right)}}} \right|}\\{\dfrac{{\left( {m{\rm{ }}-{\rm{ }}1} \right)}}{{(1 + m)}} = {\rm{ }} \pm 1}\\{\dfrac{{\left( {m{\rm{ }}-{\rm{ }}1} \right)}}{{\left( {1 + {\rm{ }}m} \right)}} = 1{\rm{ }}or{\rm{ }}\dfrac{{\left( {m{\rm{ }}-{\rm{ }}1} \right)}}{{\left( {1 + {\rm{ }}m} \right)}} = - 1}\\{m{\rm{ }} = {\rm{ }}0}\end{array}\end{array}\]
This line will be parallel to the x-axis.
Then the slope of the perpendicular on \[AD,{\rm{ }}m = \infty \]
Equation of line is
\[\begin{array}{l}y{\rm{ }}-{\rm{ }}5{\rm{ }} = {\rm{ }}\left( {\dfrac{1}{0}} \right)\left( {x{\rm{ }}-{\rm{ }}5} \right)\\0 = x - 5\\x - 5 = 0\end{array}\]

Therefore, the option (B) is correct.

Additional Information: The slope, or rate of change along the regression line, is calculated by dividing the vertical distance between any two points on the line by the horizontal distance between those same two locations. When the product of the slope of two different lines is \[{\rm{ - 1}}\] then the two lines are perpendicular to each other and if the slope of two-line is the same then they are parallel to each other.

Note: Students generally make the mistake of finding the slope of the line. To avoid this, the slope and slope point form formula should be clear.