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# help

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A point (x,y) is a distance of 6 units from the x-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?

Jan 11, 2019

#1
+23660
0

If the distance to the x axis is 6, then y = 6  or  -6

so the point becomes   x , 6     or   x , -6

It is 5 units to 8,3   (this rules out  y = -6) so we are working with  x,6

distance formula:

5= sqrt[ ( x-8)^2 + (6-3)^2 ]

5 = sqrt [(x-8)^2 +9 ]                        square both sides

25 = (x-8)^2 + 9

16 = (x-8)^2            which shows that   x-8 = 4 or -4     so x = 12 or 4

Question states x<8   so that rules out x= 12    and we are left with  x = 4

So the point becomes  x = 4 y = 6          (4,6 )

Distance formula from the origin (0,0)  =    d = sqrt [ (4-0)^2 +(6-0)^2] =  sqrt (52)       so   n = 52

Jan 11, 2019

#1
+23660
0

If the distance to the x axis is 6, then y = 6  or  -6

so the point becomes   x , 6     or   x , -6

It is 5 units to 8,3   (this rules out  y = -6) so we are working with  x,6

distance formula:

5= sqrt[ ( x-8)^2 + (6-3)^2 ]

5 = sqrt [(x-8)^2 +9 ]                        square both sides

25 = (x-8)^2 + 9

16 = (x-8)^2            which shows that   x-8 = 4 or -4     so x = 12 or 4

Question states x<8   so that rules out x= 12    and we are left with  x = 4

So the point becomes  x = 4 y = 6          (4,6 )

Distance formula from the origin (0,0)  =    d = sqrt [ (4-0)^2 +(6-0)^2] =  sqrt (52)       so   n = 52

ElectricPavlov Jan 11, 2019
#2
+111329
0

Very nice, EP......!!!!

CPhill  Jan 11, 2019