Question

examine that the points (20,3),(19,8)and (2,-9) are equidistant from point(7,3)​

Answer 1

HEY MATE HERE IS YOUR ANSWER

YES,ALL THE POINTS ARE IN EQUAL DISTANCE FROM THE POINT (7,3).

HERE IS YOUR SOLUTIONSEE THE ATTACHMENT

HOPE THIS MAY HELP YOU ...

Answer 2

Yes , all the points (20,3), (19,8) and (2,-9) are equidistant from point (7,3)​ .

Explanation:

Distance formula : The distance between two points (a,b) and (c,d) is given by :-

$D=\sqrt{(c-a)^2+(d-b)^2}$

So distance between (20,3) and (7,3)​ will be :-

$D_1=\sqrt{(7-20)^2+(3-3)^2}$

$D_1=\sqrt{(-13)^2+(0)^2}=\sqrt{13^2}=13$ units.

Distance between (19,8) and (7,3)​ will be :-

$D_2=\sqrt{(7-19)^2+(3-8)^2}$

$D_2=\sqrt{(-12)^2+(-5)^2}$

$D_2=\sqrt{144+25}=\sqrt{169}=13$ units.

Distance between (2,-9) and (7,3)​ will be :-

$D_3=\sqrt{(7-2)^2+(3-(-9))^2}$

$D_3=\sqrt{(5)^2+(3+(9))^2}$

$D_3=\sqrt{25+(12)^2}$ units.

$D_3=\sqrt{25+144}=\sqrt{169}=13$

Since, $D_1=D_2=D_3$

Therefore , all the points (20,3), (19,8) and (2,-9) are equidistant from point (7,3)​ .

#Learn more :

Examine whwther the point(1,-1),(-5,7)and(2,5)are equidistant from the point(-2,3)

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