Question

Show that the points (5,3) ;(-2,0) and (3,-2) are vertices of a right angled triangle and also show that mid point of the hypotenuse is equidistant from the vertices of the triangle .

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\mathrm{A}\left(5,3\right),\mathrm{B}\left(-2,0\right)\mathrm{and}\mathrm{C}\left(3,-2\right)\mathrm{be}\mathrm{the}\mathrm{vertices}\mathrm{of}\mathrm{triangle}. \\ \mathrm{Then}, \\ \mathrm{AB}=\sqrt{{\left(-2-5\right)}^2+{\left(0-3\right)}^2} \\ =\sqrt{7^2+3^2}=\sqrt{49+9}=\sqrt{58} \\ \mathrm{BC}=\sqrt{{\left(3+2\right)}^2+{\left(-2-0\right)}^2}=\sqrt{25+4}=\sqrt{29} \\ \mathrm{CA}=\sqrt{{\left(3-5\right)}^2+{\left(-2-3\right)}^2}=\sqrt{4+25}=\sqrt{29} \\ \mathrm{So},{\mathrm{BC}}^2+{\mathrm{CA}}^2 \\ =29+29 \\ =58 \\ ={\mathrm{AB}}^2 \\ \mathrm{So},△\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle}. \\ \mathrm{Let}\mathrm{P}\mathrm{be}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{the}\mathrm{hypotenuse}\mathrm{AB}. \\ \mathrm{Coordinates}\mathrm{of}\mathrm{midpoint}\mathrm{of}\mathrm{P}=\left(\frac{5+2}{2},\frac{3+0}{2}\right)=\left(\frac{7}{2},\frac{3}{2}\right) \\ \mathrm{CP}=\sqrt{{\left(\frac{7}{2}-3\right)}^2+{\left(\frac{3}{2}+2\right)}^2} \\ =\sqrt{\frac{1}{4}+\frac{49}{4}}=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2} \\ \mathrm{Similarly},\mathrm{find}\mathrm{AP}\mathrm{and}\mathrm{BP} \\ \mathrm{which}\mathrm{implies}\mathrm{AP}=\mathrm{CP}=\mathrm{BP} \end{gathered}$$
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