Question

what is the distance between (-2 1/2, -3) and (1, -3)​

Answer 1

Answer:

The distance between (-2 1/2, -3) and (1, -3)​ will be:

• $d=\frac{7}{2}$

Step-by-step explanation:

Given the points

• (-2 1/2, -3)

• (1, -3)​

$\mathrm{Convert\:mixed\:numbers\:to\:improper\:fraction:}\:a\frac{b}{c}=\frac{a\cdot \:c+b}{c}$

$-2\frac{1}{2}=-\frac{5}{2}$

so the point becomes (-5/2, -3)

Finding the distance between (-5/2, -3) and (1, -3):

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\mathrm{The\:distance\:between\:}\left(-\frac{5}{2},\:-3\right)\mathrm{\:and\:}\left(1,\:-3\right)\mathrm{\:is\:}$

$d=\sqrt{\left(1-\left(-\frac{5}{2}\right)\right)^2+\left(-3-\left(-3\right)\right)^2}$

  $=\sqrt{\left(\frac{5}{2}+1\right)^2+\left(3-3\right)^2}$

  $=\sqrt{\frac{7^2}{2^2}+0}$

  $=\sqrt{\frac{7^2}{2^2}}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

  $=\frac{\sqrt{7^2}}{\sqrt{2^2}}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

  $=\frac{7}{2}$

Thus, the distance between (-2 1/2, -3) and (1, -3)​ will be:

• $d=\frac{7}{2}$