Homework Help

Find all points on the y-axis that are 6 units from the point (4, -3).

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ashyacomes | eNoter

Posted October 22, 2012 at 4:55 PM via web

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  1. Find all points on the y-axis that are 6 units from the point (4, -3).

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 22, 2012 at 5:10 PM (Answer #1)

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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).`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted October 23, 2012 at 7:20 AM (Answer #2)

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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)` .

 

Sources:

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najm1947 | Elementary School Teacher | Valedictorian

Posted October 23, 2012 at 2:50 AM (Answer #3)

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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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