Show that the points $\left( -2,3 \right),\left( 8,3 \right)$ and $\left( 6,7 \right)$ are the vertices of a right triangle.
Answer
Verified
Hint: Use the basic rule of Pythagoras theorem to prove it. Let us roughly draw a triangle with the given vertices:
Now, let us calculate distances between the points of triangle: As we have the distance formula for two vertices in $2-D$ as If two points $X\left( {{x}_{1}},{{y}_{1}} \right)\And Y\left( {{x}_{2}}{{y}_{2}} \right)$ are given then
Now, $\begin{align} & AC=\sqrt{{{\left( -2-8 \right)}^{2}}+{{\left( 3-3 \right)}^{2}}} \\ & AC=\sqrt{100}=10...........\left( 3 \right) \\ \end{align}$ Hence we can observe that $AC$ has the highest length in $AB,BC\And AC$ . Therefore if $ABC$ will represent a right angle triangle then it will show or follow Pythagoras property and $AC$ will be the Hypotenuse length. As Pythagoras property can be expressed as following:
$x{{y}^{2}}+y{{z}^{2}}=x{{z}^{2}}$ Where $xz$ is Hypotenuse and biggest in length among the three sides. Hence, if $ABC$ will represent right angle triangle then:
It will follow \[\begin{align} & A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} \\ & {{\left( 4\sqrt{5} \right)}^{2}}+{{\left( 2\sqrt{5} \right)}^{2}}={{\left( 10 \right)}^{2}}=100 \\ \end{align}\] Let LHS (Left Hand Side): $\begin{align} & {{\left( 4\sqrt{5} \right)}^{2}}+{{\left( 2\sqrt{5} \right)}^{2}}=80+20 \\ & =100=RHS \\ \end{align}$
Hence, It is proved that $ABC$ is a right angled triangle at $B$ with $AC$ as Hypotenuse.
Note: In straight line we learn concept of calculating slope of a line and property of perpendicular lines as well which is “If two lines are perpendicular then; $Slope\left( Line1 \right)\times Slope\left(Line2 \right)=-1.........\left( 1 \right)$ As right angle triangle will have 3 lines and all have slope, and if two of them will follow equation $\left( 1 \right)$ then the triangle will be a right angled triangle. This proving is more advanced than the provided solution. We have formula of slope as $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ with a line of two points given as $\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}}{{y}_{2}} \right)$ .
& {{m}_{BC}}=\dfrac{7-3}{6-8}=\dfrac{-4}{2}=-2 \\ & \text{As }{{\text{m}}_{AB}}\times {{m}_{BC}}=-1 \\ \end{align}\] Hence, $ABC$ have $B$ angle as $90{}^\circ $. Therefore $ABC$is a right angle triangle. We can calculate angle between the lines by using formula $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$ where ${{m}_{1}}\And {{m}_{2}}$ are slopes of two lines between which we need to find angle and slopes can be calculated by formula $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ . Hence; this can be the angel approach as well but it may be lengthy than the above two discussed problems.
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