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?
+37157 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
+37157 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
