Question

find the distance between the points (6,-13) and (-9,7)

Answer 1

${{x}_{1}\:=\:6\:,\:{y}_{1}\:,\:-13}$

${{x}_{2}\:=\:-9\:,\:{y}_{2}\:,\:7}$

Let (6,-13) be the co - ordinate of point P and (-9,7) be the co - ordinate of point Q

by distance formula .

$PQ\:=\:{\large{\sqrt{{({x}_{2}\:-\:{x}_{1})}^{2}\:+\:{({y}_{2}\:-\:{y}_{1})}^{2}}}}$

$\large\mapsto{\sqrt{{(-9\:-6)}^{2}\:+\:{(7\:-[-13])}^{2}}}$

$\large\mapsto{\sqrt{{(-15)}^{2}\:+\:{(20)}^{2}}}$

$\large\mapsto{\sqrt{225\:+\:400}}$

$\large\mapsto{\sqrt{625}}$

$\large\mapsto{25}$

$\large{\color{chartreuse}{\underline{distance\:between\:point\:PQ\:is\:25.}}}$

Answer 2

Answer:

distance between the points (6,-13) and (-9,7)

Step-by-step explanation:

distance between the points (6,-13)and(-9,7)

consider the points as (x1,y1)and (x2,y2) then

the equation be:

$Distance,d=\sqrt{(x2-x1)^{2} +(y2-y1)^{2} } \\points (6,-13)and(-9,7)\\then,\\\\x1=6\\y1=-13\\x2=-9\\y2=7\\Distance, d=\sqrt{(-9-6 )^{2} +(7--13 )^{2} } \\ d= \sqrt{(-15)^{2}+(20)^{2} } \\ =\sqrt{225+400} \\ =\sqrt{625} \\ d =25\\distance,d=25$