Respuesta :
A (4,-5)
B (7,-9)
[tex]AB= \sqrt{(x_A-x_B)^2+(y_A-y_B)^2} \\ x_A=4; \ \ x_B=7; \ \ y_A=-5; \ \ y_B=-9; \\ \\ AB= \sqrt{(4-7)^2+(-5-(-9))^2}= \sqrt{(-3)^2+4^2} = \sqrt{25}=5 [/tex]
B (7,-9)
[tex]AB= \sqrt{(x_A-x_B)^2+(y_A-y_B)^2} \\ x_A=4; \ \ x_B=7; \ \ y_A=-5; \ \ y_B=-9; \\ \\ AB= \sqrt{(4-7)^2+(-5-(-9))^2}= \sqrt{(-3)^2+4^2} = \sqrt{25}=5 [/tex]
If a= (4,-5) and b= (7,-9) are given then the length of AB would be 5 units.
What is the distance between two points ( p,q) and (x,y)?
The shortest distance (length of the straight line segment's length connecting both given points) between points ( p,q) and (x,y) is:
[tex]D = \sqrt{(x-p)^2 + (y-q)^2} \: \rm units.[/tex]
The length of AB
AB = [tex]\sqrt{(x_a- x_b)^2 +(y_a -y _b)^2}[/tex]
[tex]AB = \sqrt{(4-7)^2 +(-9 - (7))^2} \\\\AB = \sqrt{(-3)^2 + 4^2} \\\\AB = 5[/tex]
Learn more about the distance between two points here:
brainly.com/question/16410393
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