Question

Without using the Pythagoras theorem, show that the points \[\left( 4,4 \right),\left( 3,5 \right),\left( -1,-1 \right)\] are the vertices of the right angled triangle.

Answer 1

Hint: Without using the Pythagoras theorem, in order to prove the triangle as a right-angled triangle we use slopes of lines formed by vertices of the triangle. We need to find the slopes of each line formed by using three points taken two at a time. Then we need to check whether the product of any two slopes is equal to ‘-1’ or not. Because if\[{{m}_{1}}\times {{m}_{2}}=-1\] where \[{{m}_{1}},{{m}_{2}}\] are slopes of two lines, then we can say that those two lines are perpendicular to each other.

Complete step-by-step solution
Let us assume that the given vertices of triangle as
\[P=\left( 4,4 \right),Q=\left( 3,5 \right),R=\left( -1,-1 \right)\]
We know that if \[A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\] are two points then we can write slope of AB as
\[\Rightarrow m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
By using this formulae let us find the slope of ‘PQ’ as
\[\begin{align}
  & \Rightarrow {{m}_{1}}=\dfrac{5-4}{3-4} \\
 & \Rightarrow {{m}_{1}}=-1 \\
\end{align}\]
Now, let us find the slope of ‘QR’
\[\begin{align}
  & \Rightarrow {{m}_{2}}=\dfrac{-1-5}{-1-3} \\
 & \Rightarrow {{m}_{2}}=\dfrac{3}{2} \\
\end{align}\]
Now, let us find the slope of ‘RP’
\[\begin{align}
  & \Rightarrow {{m}_{3}}=\dfrac{-1-4}{-1-4} \\
 & \Rightarrow {{m}_{3}}=1 \\
\end{align}\]
Here, we can see that the product of \[{{m}_{1}},{{m}_{3}}\] is ‘-1’.
So, we can say that the lines ‘PQ’ and ‘RP’ are perpendicular to each other.
So, if we consider the triangle \[\Delta PQR\], we can say that \[PQ\bot RP\].
Here, as the common vertex of two perpendicular lines is ‘P’ we can say that the right angle is formed at vertex ‘P’.

Note: Some students will make mistakes in taking the formula of the slope. That is, instead of taking \[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\], some students will consider \[m=\dfrac{{{x}_{2}}-{{x}_{1}}}{{{y}_{2}}-{{y}_{1}}}\]. This will be very wrong when we get a different line equation. Also, some students will find the equation of lines of the triangle which are not at all necessary. Finding an equation of lines is good, but here it is unnecessary as it will take more time.