alexcasas9340 alexcasas9340
  • 21-10-2019
  • Mathematics
contestada

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

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isyllus
isyllus isyllus
  • 22-10-2019

Answer:

x = -4 , 4

Step-by-step 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: [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

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

[tex]\sqrt{(x-x)^2+(4+1)^2}=\sqrt{(x-0)^2+(4-1)^2[/tex]

[tex]\sqrt{5^2}=\sqrt{x^2+3^2[/tex]

[tex]25=x^2+9[/tex]

[tex]x^2=16[/tex]

[tex]x=\pm \sqrt{16}[/tex]

[tex]x=\pm 4[/tex]

hence, the value of x is 4 and -4

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