# Find the point (0, y) that is equidistant from (4, -9) and (0, -2).

Posted on

We are given the points (4, -9) and (0, -2) and we need to find the point (0, y) which is equidistant from them.

Now we know that teh distant between two point (x1, y1) and (x2, y2) is given as sqrt[(x2-x1)^2 + (y2-y1)^2]

Therefore we have to find the point (0,y) that has the same distance from (4, -9) and (0, -2).

This is given by sqrt[(4-0)^2 + (-9-y)^2]=sqrt[(0-0)^2 + (-2-y)^2]

taking the square of both the sides

=> (4-0)^2 + (-9-y)^2 = (0-0)^2 + (-2-y)^2

=> 4^2 +(-9-y)^2 = (-2-y)^2

=> 16 + 9^2 + y^2 + 18y = 4 + y^2 + 4y

=> 16 + 81 + y^2 + 18y = 4 + y^2 + 4y

=> 93 + 14y = 0

=> y = -93 / 14

Therefore the required value for y is -93/ 14

Posted on

Let  A( 0, y) be the point that  is equidistant from B( 4, -9) and

C( 0, -2)

Then we conclufde that , the distance between A and B equals the distance between A and C

==> l AB l  = l AC l

l AB l = sqrt (xB-xA)^2 + (yB-yA)^2]

= sqrt(4-0)^2 + ( -9 - y)^2]

= sqrt( 16 + 81 + 18y + y^2)

= sqrt( 97+ 18y + y^2)

l AC l = sqrt( xc - xA)^2 = ( yC- yA)^2]

= sqrt( 0-0)^2 + ( -2-y)^2

= sqrt( 4 + 4y + y^2)

Now given AB = AC

==> sqrt( y^2 + 18y + 97) = sqrt( y^2 + 4y + 4)

We will square both sides:

==> y^2 + 18y + 97 = y^2 + 4y + 4

==> 18y + 97 = 4y + 4

Combine like terms:

==> 14y = - 93

==> y= -93/ 14

Posted on

The coordinates of all the points equidistant from two points, A and B, includes all the points that lie on a line that is its perpendicular bisector, that is the line that is perpendicular to the line AB and passes through its mid point.

A point where this line cuts the y-axis will have the coordinates of the form (0, y).

Therefore to solve this problem we need to find the point where perpendicular bisector of line segment with extreme points A(4, -9) and B(0, -2) cuts the y-axis.

Coordinates of mid point of line segment AB = [(x1 + x2)/2, (y1 + y2)/2]

Where (x1, y1) and (x2, y2) are the coordinates of the two points A and B.

Substituting the given values of x1, y1, x2, and y2 in the equation of mid point:

Coordinates of mid point of AB= [(4 + 0)/2, (-2 - 9)/2]

= (2, -11/2)

Slope of AB = m1 = (y2 - y1)/(x2 - x1)

= (-2 + 9)/(0 - 4) = -7/4

Slope of perpendicular bisector of AB = m2 = 1/m1 = 4/7

Then equation of the perpendicular bisector of AB is given by:

y = (4/7)x + c

To find value of c we substitute coordinates of mid point of AB in the above equation. Thus:

-11/2 = (4/7)*2 + c

==> c = -11/2 - 8/7 = (-77 - 16)/14 = -93/14

Thus the equation of perpendicular bisector becomes:

y = (4/7)x - 93/14

The y-coordinate of the point where it cuts the y-axis is -93/14.

Coordinates of the point are (0, -93/14)

Posted on

The point (0, y) that is equidistant from (4, -9) and (0, -2) has to be determined.

Looking at the coordinates of (0, y) and (0, -2) it is clear that both the points lie on the y axis.

The distance of (0, y) from the point (0, -2) is equal to -2 - y

The distance of (0, y) from the point (4, -9) is given by `sqrt((4 - 0)^2 + (-9-y)^2)` .

As an equidistant point is desired, equate -2 - y and `sqrt((4 - 0)^2 + (-9-y)^2)` and solve for y

`-2 - y = sqrt((4 - 0)^2 + (-9-y)^2)`

`(-2 - y)^2 = (4 - 0)^2 + (-9-y)^2`

`(-2 - y)^2 = 16 + (-9-y)^2`

4 + y^2 + 4y = 16 + 81 + y^2 + 18y

4 = 16 + 81 + 14y

14y = -93

y = -93/14

The required point (0, y) is `(0, -93/14)`

Posted on

To determine the point (0,y) equidistant from the two given points (4,-9) and(0,-2).

The distance d between the two points P(x1,y1) and P(x2,y2) is given by:

d = PQ = sqrt{(x2-x1)^2 + (y2-y1)^2}

The distance between point (0,y) and (4,-9) is sqrt{(4-0)^2+(-9-y)^2} = sqrt{4^2+(9+y)^2}...(1)

The distance between the point (o,y) and (0,-2) is sqrt{(0-4)^2+(-2-y)^2}= sqrt(2+y)^2...(2)

Since the distances at (1) and (2) are equal,

4^2+(9+y)^2 =16+ (2+y)^2.

16+81+18y+y^2=20+4y+y^2.

97 +18y = 20+4y.

97-20 = 4y-18y.

77 = -14y

y = 77/-14 = -11/2 = -5.5.

y = -5.5.

Therefore the unknown ordinate y =-5.5.