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5 votes
What you s the distance between (13,-6) and (13,12) on the coordinate plane?

2 Answers

3 votes

To find the distance, you would use the distance formula, which is:


\sqrt{(x2 - x1)^(2) + (y2 - y1)^(2)}

But, in this case, both of the x values are 13, meaning we can cancel out the x values (or just subtract y₁ from y₂, to get the distance ):


\sqrt{(13 - 13)^(2) + (y2 - y1)^(2)}


\sqrt{(0)^(2) + (y2 - y1)^(2)}


\sqrt{0 + (y2 - y1)^(2)}

Because of this, we only need to find the distance between y₁ and y₂. Plug in the -6 and 12:


\sqrt{(12 + 6)^(2)} Plug in both y terms.


\sqrt{18^(2)} Simplify as much as possible.

= 18 The √ and the ² cancel out.

Therefore, the distance between (13, -6) and (13, 12) is 18.

Hope this helps! :)

EDIT: all the subscripts and most of the exponents showed up incorrectly. should be fixed now.

User Ikram Khan Niazi
by
7.7k points
4 votes
since both x values are the same and since the y values are 18 digits apart, the distance between both points is 18 units.
User Jono Job
by
7.0k points
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