Question

The triangle with vertices \[\left( 4,3 \right),\left( -3,2 \right),\left( 1,-6 \right)\] is
\[A)\] An obtuse angled triangle
\[B)\] An acute angled triangle
\[C)\] right angled triangle
\[D)\] right angled isosceles

Answer 1

Hint: We know that if \[A\left( {{x}_{1}},{{y}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}} \right)\] are two points on a line equation AB is \[y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)\]. We know that if the a line equation has slope \[{{m}_{1}}\] and other line equation has slope \[{{m}_{2}}\], then the angle between these two lines is equal to \[\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}\]. We know that the slope of \[ax+by+c=0\] is equal to \[\dfrac{-a}{b}\]. By using this concept, we can solve the problem.

Complete step by step answer:
From the question, we were given that a triangle of the vertices \[\left( 4,3 \right),\left( -3,2 \right),\left( 1,-6 \right)\]. Let us assume the vertices of the triangle as \[A\left( 4,3 \right),B\left( -3,2 \right),C\left( 1,-6 \right)\].
We know that if \[A\left( {{x}_{1}},{{y}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}} \right)\] are two points on a line equation AB is \[y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)\].
Now let us find the equation of AB whose vertices are \[A\left( 4,3 \right)\] and \[B\left( -3,2 \right)\]. So, the equation of line AB is
\[\begin{align}
  & \Rightarrow y-3=\left( \dfrac{2-3}{-3-4} \right)\left( x-4 \right) \\
 & \Rightarrow y-3=\left( \dfrac{-1}{-7} \right)\left( x-4 \right) \\
 & \Rightarrow y-3=\left( \dfrac{1}{7} \right)\left( x-4 \right) \\
 & \Rightarrow y-3=\dfrac{x-4}{7} \\
\end{align}\]
Now by using cross multiplication, we get
\[\begin{align}
  & \Rightarrow x-4=7\left( y-3 \right) \\
 & \Rightarrow x-4=7y-21 \\
 & \Rightarrow x-7y+17=0.....(1) \\
\end{align}\]
So, it is clear that the line equation of AB is \[x-7y+8=0\].
Now let us find the equation of BC whose vertices are \[B\left( -3,2 \right)\] and \[C\left( 1,-6 \right)\] . So, the equation of line BC is
\[\begin{align}
  & \Rightarrow y-2=\left( \dfrac{-6-2}{1-(-3)} \right)\left( x-(-3) \right) \\
 & \Rightarrow y-2=\left( \dfrac{-8}{4} \right)\left( x+3 \right) \\
 & \Rightarrow y-2=\left( -2 \right)\left( x+3 \right) \\
 & \Rightarrow y-2=-2x-6 \\
 & \Rightarrow 2x+y+4=0....(2) \\
\end{align}\]
So, it is clear that the line equation of AB is \[2x+y+4=0\].
Now let us find the equation of AC whose vertices are \[A\left( 4,3 \right)\] and \[C\left( 1,-6 \right)\]. So, the equation of line AB is
\[\begin{align}
  & \Rightarrow y-3=\left( \dfrac{-6-3}{1-4} \right)\left( x-4 \right) \\
 & \Rightarrow y-3=\left( \dfrac{-9}{-3} \right)\left( x-4 \right) \\
 & \Rightarrow y-3=\left( 3 \right)\left( x-4 \right) \\
 & \Rightarrow y-3=3x-12 \\
 & \Rightarrow 3x-y-9=0...(3) \\
\end{align}\]
So, it is clear that the line equation of AC is \[3x-y-9=0\].
We know that the slope of \[ax+by+c=0\] is equal to \[\dfrac{-a}{b}\].
Now we should find the slope of AB, BC and CA.
Let us assume the slope of AB is equal to \[{{m}_{AB}}\].
Then we get
\[\begin{align}
  & \Rightarrow {{m}_{AB}}=\dfrac{-1}{-7} \\
 & \Rightarrow {{m}_{AB}}=\dfrac{1}{7}.....(4) \\
\end{align}\]
Let us assume the slope of BC is \[{{m}_{BC}}\].
Then we get
\[\begin{align}
  & \Rightarrow {{m}_{BC}}=\dfrac{-2}{1} \\
 & \Rightarrow {{m}_{BC}}=-2.....(5) \\
\end{align}\]
Let us assume the slope of CA is \[{{m}_{CA}}\].
\[\begin{align}
  & \Rightarrow {{m}_{CA}}=\dfrac{3}{1} \\
 & \Rightarrow {{m}_{CA}}=3.....(6) \\
\end{align}\]
We know if the a line equation has slope \[{{m}_{1}}\] and other line equation has slope \[{{m}_{2}}\], then the angle between these two lines is equal to \[\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}\].
Let us assume the angle between AB and BC is equal to \[\angle B\]. So, now let us find the angle between AB and BC.
\[\begin{align}
  & \Rightarrow \tan \angle B=\dfrac{\dfrac{1}{7}-\left( -2 \right)}{1+\left( \dfrac{1}{7} \right)\left( -2 \right)} \\
 & \Rightarrow \tan \angle B=\dfrac{\dfrac{1}{7}+2}{1-\dfrac{2}{7}} \\
 & \Rightarrow \tan \angle B=\dfrac{\dfrac{15}{7}}{\dfrac{5}{7}} \\
 & \Rightarrow \tan \angle B=\dfrac{15}{5} \\
 & \Rightarrow \tan \angle B=3....(7) \\
\end{align}\]
Let us assume the angle between AC and BC is equal to \[\angle C\]. So, now let us find the angle between AC and BC.
\[\begin{align}
  & \Rightarrow \tan \angle C=\dfrac{-2-3}{1+\left( 3 \right)\left( -2 \right)} \\
 & \Rightarrow \tan \angle C=\dfrac{-5}{1+\left( -6 \right)} \\
 & \Rightarrow \tan \angle C=\dfrac{-5}{-5} \\
 & \Rightarrow \tan \angle C=1....(8) \\
\end{align}\]
Let us assume the angle between AC and CA is equal to \[\angle A\]. So, now let us find the angle between AC and CA.
\[\begin{align}
  & \Rightarrow \tan \angle A=\dfrac{3-\dfrac{1}{7}}{1+\left( 3 \right)\left( \dfrac{1}{7} \right)} \\
 & \Rightarrow \tan \angle A=\dfrac{\dfrac{20}{7}}{1+\dfrac{3}{7}} \\
 & \Rightarrow \tan \angle A=\dfrac{\dfrac{20}{7}}{\dfrac{10}{7}} \\
 & \Rightarrow \tan \angle A=2....(8) \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \tan \angle A=\dfrac{3-\dfrac{1}{7}}{1+\left( 3 \right)\left( \dfrac{1}{7} \right)} \\
 & \Rightarrow \tan \angle A=\dfrac{\dfrac{20}{7}}{1+\dfrac{3}{7}} \\
 & \Rightarrow \tan \angle A=\dfrac{\dfrac{20}{7}}{\dfrac{10}{7}} \\ \
 & \Rightarrow \tan \angle A=2....(9) \\
\end{align}\]
So, from equation (7), equation (8) and equation (9) it is clear that the triangle ABC is acute angle triangle.

Note:
Students may have a misconception that if the a line equation has slope \[{{m}_{1}}\] and other line equation has slope \[{{m}_{2}}\], then the angle between these two lines is equal to \[\tan \theta =\dfrac{{{m}_{2}}+{{m}_{1}}}{1-{{m}_{1}}{{m}_{2}}}\]. But we know that if the a line equation has slope \[{{m}_{1}}\] and other line equation has slope \[{{m}_{2}}\], then the angle between these two lines is equal to \[\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}\]. So, this misconception should be avoided.