Respuesta :

[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ \frac{1}{8}}}\quad ,&{{-\frac{9}{5}}})\quad % (c,d) &({{ \frac{3}{8}}}\quad ,&{{ -\frac{4}{5}}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}[/tex]

[tex]\bf d=\sqrt{\left[ \frac{3}{8}-\frac{1}{8} \right]^2+\left[-\frac{4}{5}-\left( -\frac{9}{5} \right) \right]^2}\implies d=\sqrt{\left( \frac{3}{8}-\frac{1}{8} \right)^2+\left(-\frac{4}{5}+\frac{9}{5} \right)^2} \\\\\\ d=\sqrt{\left( \frac{2}{8}\right)^2+\left(\frac{5}{5} \right)^2}\implies d=\sqrt{\left( \frac{1}{4} \right)^2+\left( 1 \right)^2}\implies d=\sqrt{\frac{1^2}{4^2}+1}[/tex]

[tex]\bf d=\sqrt{\frac{1}{16}+1}\implies d=\sqrt{\cfrac{17}{16}}\implies d=\cfrac{\sqrt{17}}{\sqrt{16}}\implies d=\cfrac{\sqrt{17}}{4}[/tex]
mreijepatmat
If you have 2 pairs such that A(x₁ , y₁)     and B(x₂ , y₂). , the distance is:

AB = √[(x₂ - x₁)² + (y₂ - y₁)²]

A(1/8 , 9/5)   and B(3/8 , - 4/5)

Plug in the related value:

AB = √[(3/8 - 1/8)² + ( - 4/5 - 9/5)²] = √(2729/400) (ALREADY SIMPLIFIED)
AB = distance = 2.61 

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