Question

Find the values of x for which the point (x, 4) is equidistant from (0,1), and the line y = −1.

Answer 1

x = -4 , 4

Explanation:

The point (x,4) is equidistant from (0,1) and line y=-1

y=-1, it is horizontal line.

Let point on line be (x,-1)

Distance formula:
`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

Distance between (x,4) and (x,-1) = Distance between (x,4) and (0,1)

`√((x-x)^2+(4+1)^2)=\sqrt{(x-0)^2+(4-1)^2`

`√(5^2)=\sqrt{x^2+3^2`

`25=x^2+9`

`x^2=16`

`x=\pm √(16)`

`x=\pm 4`

hence, the value of x is 4 and -4