Find the distance between the two points in simplest radical form.
(-1,9) and (4, -3)

Answer:
GIVEN BY POINTS :-
( -1, 9) , ( 4 , -3)
Here ,
[tex]\bullet \: \: x _{1} = - 1 \\ \bullet \: \: x _{2} = 4 \\ \bullet \: \: y _{1} = 9 \\ \bullet \: \: y _{2} = - 3 \\[/tex]
Using Distance formula :
[tex]= > \sf d = \sqrt{( {x _{2} - x _{1}) }^{2} + {( {y _{2} - y _{1}) }^{2} } } \\ \\ = > \sf d = \sqrt{ ({ 4 + 1)}^{2} + { ( - 3 - 9)}^{2} } \\ \\ = > \sf d = \sqrt{ {5}^{2} + {( - 12)}^{2} } \\ \\ = > \sf d = \sqrt{25 + 144} \\ \\ = > \sf d = \sqrt{169} \\ \\ = > \boxed{ \sf{ d = 13\:units }}\\ [/tex]
Answer: 13
Step-by-step explanation:
(-1, 9) \text{ and } (4, -3)
(−1,9) and (4,−3)
(x_1,y_1)\text{ and }(x_2,y_2)
(x
1
,y
1
) and (x
2
,y
2
)
\text{Distance Formula: }\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
Distance Formula:
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
\sqrt{(-1-4)^2+(9-(-3))^2}
(−1−4)
2
+(9−(−3))
2
Plug in.
\sqrt{(-1-4)^2+(9+3)^2}
(−1−4)
2
+(9+3)
2
Negative of a negative is a positive.
\sqrt{(-5)^2+(12)^2}
(−5)
2
+(12)
2
Add/Subtract
\sqrt{25+144}
25+144
Square
\sqrt{169}
169
Perfect Square
13
13
Final Answer.