Respuesta :
the so-called pythagorean triple will just be the distance between both points,
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 5}} &,&{{ 4}}~) % (c,d) &&(~{{ 1}} &,&{{ \frac{1}{2}}}~) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}[/tex]
[tex]\bf d=\sqrt{(1-5)^2+\left( \frac{1}{2}-4 \right)^2}\implies d=\sqrt{(-4)^2+\left( -\frac{7}{2} \right)^2} \\\\\\ d=\sqrt{16+\frac{7^2}{2^2}}\implies d=\sqrt{16+\frac{49}{4}}\implies d=\sqrt{\frac{64+49}{4}} \\\\\\ d=\sqrt{\frac{113}{4}}\implies d=\cfrac{\sqrt{113}}{\sqrt{4}}\implies d=\cfrac{\sqrt{113}}{2}[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 5}} &,&{{ 4}}~) % (c,d) &&(~{{ 1}} &,&{{ \frac{1}{2}}}~) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}[/tex]
[tex]\bf d=\sqrt{(1-5)^2+\left( \frac{1}{2}-4 \right)^2}\implies d=\sqrt{(-4)^2+\left( -\frac{7}{2} \right)^2} \\\\\\ d=\sqrt{16+\frac{7^2}{2^2}}\implies d=\sqrt{16+\frac{49}{4}}\implies d=\sqrt{\frac{64+49}{4}} \\\\\\ d=\sqrt{\frac{113}{4}}\implies d=\cfrac{\sqrt{113}}{\sqrt{4}}\implies d=\cfrac{\sqrt{113}}{2}[/tex]
We are giving two pairs of values for x and y. To calculate the Pythagorean triples for each pair, we need to focus on each pair individually.
a. x1 = 5, y1 = 4
Answer: sq(5^2 + 4^2) = sq(41)
b. x2 = 1, y2 = ?
Answer: r = sq(1^2 + ?^2)
