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

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