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Sagot :
We need to form a right triangle so we could use the Pythagorean Theorem.
The Pythagorean Theorem states that:
[tex]a^2+b^2=c^2[/tex]
The Pythagorean Theorem applies to right triangles only. a and b are the side lengths of the legs while c is the length of the hypotenuse.
In a Cartesian plane the side lengths a and b are represented like this:
[tex](x_a-y_a)=a \\ (x_b-yb)=b[/tex]
So the Pythagorean Theorem would be:
[tex](x_a-y_a)^2+(x_b-y_b)^2=c^2[/tex]
We have [tex](x_a,y_a)[/tex] as (-1,3)
and [tex](x_b,y_b)[/tex] as (5, -1)
We substitute the values to the Pythagorean theorem:
[tex]c^2=(-1-5)^2+(3-(-1))^2 \\ =(-6)^2+4^2 \\ =36+16 \\ 50[/tex]
[tex]c^2=50 \\ c= \sqrt{50} =5 \sqrt{2} [/tex]
Therefore the distance between points is [tex]5 \sqrt{2} [/tex]
The Pythagorean Theorem states that:
[tex]a^2+b^2=c^2[/tex]
The Pythagorean Theorem applies to right triangles only. a and b are the side lengths of the legs while c is the length of the hypotenuse.
In a Cartesian plane the side lengths a and b are represented like this:
[tex](x_a-y_a)=a \\ (x_b-yb)=b[/tex]
So the Pythagorean Theorem would be:
[tex](x_a-y_a)^2+(x_b-y_b)^2=c^2[/tex]
We have [tex](x_a,y_a)[/tex] as (-1,3)
and [tex](x_b,y_b)[/tex] as (5, -1)
We substitute the values to the Pythagorean theorem:
[tex]c^2=(-1-5)^2+(3-(-1))^2 \\ =(-6)^2+4^2 \\ =36+16 \\ 50[/tex]
[tex]c^2=50 \\ c= \sqrt{50} =5 \sqrt{2} [/tex]
Therefore the distance between points is [tex]5 \sqrt{2} [/tex]
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