Question

find the values ot ac, ab, and bc from the points a(3-2), b(6-2) c(3-4) ​

Answer 1

Answer:

AC=2 UNITS     AB=3 UNITS    BC=√13 UNITS

Step-by-step explanation:

AC=$\sqrt{(3-3)^2+(-4-(-2)^2}$

AC=$\sqrt{-2^2}$      

AC=2 UNITS

AB=$\sqrt{(6-3)^2+(-2-(-2)^2}$

AB=$\sqrt{3^2}$      

AB=3 UNITS

BC=$\sqrt{(3-6)^2+(-4-(-2)^2}$

BC=$\sqrt{(-3)^2+(-2)^2 }$  

BC=$\sqrt{9+4} = \sqrt{13}$    

BC=√13 UNITS

Answer 2

Step-by-step explanation:

$by \: using \: distance \: formula$

$ac = \sqrt{ {(x2 - x1) + (y2 - y1)}^{2} }$

$ac = \sqrt{ {( - 4 + 2)}^{2} + {(3 - 3)}^{2} }$

$ac = \sqrt{ {(2)}^{2} + (0) } = \sqrt{4} = 2units$

$ab = \sqrt{ {( - 2 - ( - 2)}^{2} + {(6 - 3)}^{2} }$

$ab = \sqrt{ {(0)}^{2} + {(3)}^{2} } = \sqrt{9} = 3units$

$bc = \sqrt{( { - 4 + 2)}^{2} + {(3 - 6)}^{2} }$

$bc = \sqrt{( {2}^{2}) + ( {3}^{2} )}$

$bc = \sqrt{ 4 + 9 } = \sqrt{13}units$