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?

Guest Jan 11, 2019

#1**+2 **

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

#1**+2 **

Best Answer

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