What is the length of the line?
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Answer:
[tex]\boxed{\sf B. \ \sqrt{61} }[/tex]
Step-by-step explanation:
The line can be made into a hypotenuse of a right triangle.
Find the length of the base and the height of the right triangle.
The base (leg) is 6 units.
The height (leg) is 5 units.
Apply Pythagorean theorem.
[tex]\sf c=\sqrt{a^2 +b^2 }[/tex]
[tex]\sf c=\sqrt{6^2 +5^2 }[/tex]
[tex]\sf c=\sqrt{36+25 }[/tex]
[tex]c=\sqrt{61}[/tex]
Answer:
Option B is the correct option
Step-by-step explanation:
Assuming center of co-ordinate axes at lower left corner at the line. So end points are:
( x1 , y1 ) = ( 0 , 0 ) and ( x2 , y2 ) = ( 6 , 5 )
Distance between two points is given by formula:
D [tex] = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } [/tex]
[tex] = \sqrt{ {6 - 0)}^{2} + {(5 - 0)}^{2} } [/tex]
[tex] = \sqrt{ {6}^{2} + {5}^{2} } [/tex]
[tex] = \sqrt{36 + 25} [/tex]
[tex] = \sqrt{61} [/tex]
Hope this helps..
Best regards!!