Question

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

Answer 1

Answer: 13

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.

Answer 2

GIVEN BY POINTS :-

( -1, 9) , ( 4 , -3)

Here ,

`\bullet \: \: x _(1) = - 1 \\ \bullet \: \: x _(2) = 4 \\ \bullet \: \: y _(1) = 9 \\ \bullet \: \: y _(2) = - 3 \\`

Using Distance formula :

`= > \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 = √(25 + 144) \\ \\ = > \sf d = √(169) \\ \\ = > \boxed{ \sf{ d = 13\:units }}\\`