# Find a point on the y-axis that is equidistant from the points (1, –7) and (5, –1).?

a.
(–2, 0)
b.
(0, –2)
c.
(0, 0)
d.
(0, 3)
e.
(0, –4)
asked Mar 3, 2013 in GEOMETRY

+1 vote

Point is on y-axis so x-coordinate value =0, so the point is (0, y)
Find the distance between the point (0,y) and (1, –7) and (5, –1).
Find the distance using the formula = sqrt((y2-y1)^2 + (x2-x1)^2)
the point is equidistant from the point (0,y). so, the equate the distance to two points.

sqrt((y-(-7))^2+(0-1)^2 = sqrt((y-(-1))^2+(0-5)^2
(y-(-7))^2+(0-1)^2 = (y-(-1))^2+(0-5)^2
y^2 +14y + 49 +1 = y^2 + 2y + 1 +25
14y + 50 = 2y + 26
12y = 26 - 50
2y = -24
y = -2
point (0, -2)

selected Mar 3, 2013 by mathgirl

Point is on y-axis so x-coordinate value =0, so the point is (0, y).
Find the distance between the point (0,y) and (1, –7) and (5, –1).
Find the distance using the formula = sqrt((y2-y1)^2 + (x2-x1)^2)
the point is equidistant from the point (0,y). so, the equate the distance to two points
sqrt((y-(-7))^2+(0-1)^2 = sqrt((y-(-1))^2+(0-5)^2
(y-(-7))^2+(0-1)^2 = (y-(-1))^2+(0-5)^2
y^2 +14y + 49 +1 = y^2 + 2y + 1 +25
14y + 50 = 2y + 26
12y = 26 - 50
2y = -24
y = -2
point (0, -2)

Right option is b.