Question

I NEED THIS PLEASE ONE OF MY LAST QUESTIONS

Line segment AB has endpoints A (-1.5, 0) and B (4.5, 8) Point C is on line segment AB and is located at (0,2)


What is the ratio of $\frac{AC}{CB\\}$

Answer 1

Answer:

Ratio of AC/CB = 1/3

Step-by-step explanation:

Distance between point A and point C

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

$\mathrm{The\:distance\:between\:}\left(-1.5,\:0\right)\mathrm{\:and\:}\left(0,\:2\right)\mathrm{\:is\:}$

$=\sqrt{\left(0-\left(-1.5\right)\right)^2+\left(2-0\right)^2}$

$=\sqrt{\left(0+1.5\right)^2+\left(2-0\right)^2}$

$=\sqrt{1.5^2+\left(2-0\right)^2}$

$=\sqrt{2.25+2^2}$

$=\sqrt{2.25+4}$

$=\sqrt{6.25}$

$=2.5$

Distance between point B and point C

$\mathrm{The\:distance\:between\:}\left(4.5,\:8\right)\mathrm{\:and\:}\left(0,\:2\right)\mathrm{\:is\:}$

$=\sqrt{\left(0-4.5\right)^2+\left(2-8\right)^2}$

$=\sqrt{4.5^2+6^2}$

$=\sqrt{20.25+6^2}$

$=\sqrt{20.25+36}$

$=\sqrt{56.25}$

$=7.5$

Ratio of AC/CB

= 2.5/7.5

=1/3