Find all points on the y-axis that are 6 units from the point (4, -3).
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calendarEducator since 2012
write1,657 answers
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If we have two points (a,b) and (c,d) the distance(R) between them is given by;
`R = sqrt((a-c)^2+(b-d)^2)`
So it is given that the all the second points should be on y axis. This means the x coordinates are 0 at these places.
`R = 6`
`a = 4`
`b = -3`
`c = 0`
`d = ?`
`R = sqrt((a-c)^2+(b-d)^2)`
`6 = sqrt((4-0)^2+(-3-d)^2)`
`36 = 16+(-3-d)^2`
`20 = (3+d)^2`
`(3+d) = +-sqrt20`
`(3+d) = +-2sqrt5`
`d = +-2sqrt5-3`
`d= 2sqrt5-3` OR `d = -sqrt5-3`
So the answers that satisfies the situation are `y = 2sqrt5-3` or y = `-2sqrt5-3` . The points are `(0,2sqrt5-3 )` and `(0,-2sqrt5-3)` .
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calendarEducator since 2011
write5,348 answers
starTop subjects are Math, Science, and Business
You should remember that all points found on y axis have x coordinate 0, hence, since the problem provides the value of distance beween the points, you may evaluate the missing coordinates using distance formula such that:
`d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)`
`6 = sqrt((4-0)^2 + (3-y_1)^2)`
You need to raise to square both sides such that:
`36 = 16 + (3-y_1)^2 => (3-y_1)^2 = 36-16 => (3-y_1)^2 = 20`
`3-y_1 = +-sqrt(20) => y_1 = 3-2sqrt5 or y_1 = 3+2sqrt5`
Hence, evaluating the missing coordinates of the points that follow the given conditions yields `(0,3+2sqrt5)` and `(0,3-2sqrt5).`
We have to find all points on the y-axis that are 6 units from the point (4, -3).
For two points (x1,y1) and (x2,y2) the distance d is given by:
d = sqrt[(x2-x1)^2 + (y2-y1)^2]
d=6, x1=4, y1=-3, x2=0 being on y-axis and y2=?
6 = sqrt[(0-4)^2 + (y2-(-3))^2]
6 = sqrt[16+(y2+3)^2]
6 = sqrt[16+y2^2+6y2+9]
6 = sqrt[y2^2+6y2+25]
squaring both sides
36 = y2^2+6y2+25
y2^2+6y2-11=0
y2=[-6+sqrt{(6^2-4*1*(-11)}]/2*1 and
y2=[-6-sqrt{6^2-4*1*(-11)}]/2*1
y2 = [-6+sqrt(80)]/2 and y2 = [-6-sqrt(80)]/2
y2 = 3+2sqrt(5), 3-2sqrt(5)
Therefore the points on the y-axis that are 6 units from the point (4, -3) are:
(0 , 3+2sqrt(5)) and (0, 3-2sqrt(5))
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