+0

+1
92
1
+107

A point $$(x,y)$$ is a distance of 6 units from the -axis. It is a distance of 5 units from the point$$(8,3)$$. It is a distance$$\sqrt{n}$$ from the origin. Given that $$x<8$$, what is $$n$$?

Sep 24, 2020

#1
+578
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I'll assume that given how we are given an incomplete value for x we can assume that the point is 6 distance from the x axis, leaving us with a y-value of + or - 6.

Given how the most y-negative value possible is -2 (assuming straight negative y travel), we can safely eliminate -6 as a possibiility for y, and find y to be +6.

Solving for a point with y=6 a distance of 5 units away from (8,3), by using a memorized pythagorean triple of 3-4-5, you can know that the x value is 8 plus or minus 4 units.

Given that x is less than 8, the lower answer of 4 is the only one that is possible, hence x = 4.

We now know that (x,y) is located at (3,4).

Solving for distance from the origin, we could use the pythagorean theorem, but this is another use of the same pythagorean triple. We know that it is 5 units from the origin.

Putting this back with n, we now have sqrt(n) = 5

Squaring each side to solve for n alone, we find n = 25.

If use of the pythagorean theorem is required I will be happy to supply it.

Sep 24, 2020