Question

10. IfA (17, 5) and B (3,0) are points then distance AB =​__

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{Solution:-}}}}}}$.

$\large{\underline{\underline{\mathfrak{\green{\sf{Given\:Here:-}}}}}}$.

✴Point A(17,5) , and B(3,0).

$\large{\underline{\underline{\mathfrak{\pink{\sf{Find\::Here:-}}}}}}$.

✴$\red{\:Distance\:between\:AB}$.

$\large{\underline{\underline{\mathfrak{\green{\sf{Explanation:-}}}}}}$.

➡We Know that,

✴$\:Distance\:Between\:A(x,y)\:And\:B(p,q)\:=\sqrt{(x-p)^2+(y-q)^2}$.

➡So,

$\implies\:Distance\:=\sqrt{(17-3)^2+(5-0)^2}$.

$\implies\:Distance\:=\sqrt{(14^2+5^2}$.

$\implies\:Distance\:=\sqrt{196+25}$.

$\implies\:Distance\:=\sqrt{221}$.

$\implies\:Distance\:=\:11$.

________________

$\red{\:Follow}$.

Answer 2

Step-by-step explanation:

Given:- A(17,5) and B(3,0)

so, X1=17

X2=3

y1=5

y2=0

by using distance formula...

we have...,

AB = √(x2 -x1)² + (y2 - y1)²

AB = √(3 - 17)² + (0 - 5)²

AB = √(-10)² + (-5)²

AB = √100 + 25

AB = √125

AB = 5 sq.units

therefore, the distance between AB is 5 sq.units