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

Question

a.
(–2, 0)
b.
(0, –2)
c.
(0, 0)
d.
(0, 3)
e.
(0, –4)

Answer 1

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)

 

Answer 2

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.