Determine the value of x when (-1,-1)is equidistant from (0,2)and(x,-4)?

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Answer 1 · 2018-05-18T12:11:02

$x=0 or -2$

Using distance formula / Pythagorean theorem,

Let the distance between $(-1,-1)$ and $(0,2)$ be $d_1$ ,

$d_1=sqrt((-1-0)^2+(-1-2)^2)$
$color(white)(d_1)=sqrt10$

Let the distance between $(-1,-1)$ and $(x,-4)$ be $d_2$ ,

$d_2=sqrt((-1-x)^2+(-1+4)^2)$
$color(white)(d_2)=sqrt((-1-x)^2+9)$

Since, the distance are the same, $d_1=d_2$

$sqrt10=sqrt((-1-x)^2+9)$

Square both sides,

$10=(-1-x)^2+9$

Subtract $9$ from both sides,

$(-1-x)^2=1$

Square root both sides,

$-1-x=+-1$

Solve,

$x=0 or -2$

Answer 2 · 2018-05-18T12:14:39

$color(crimson)(x = 0$ or $color(crimson)(-2$

Given $A(-1,-1), B(0,2), C(x,-4), bar(AB) = bar(AC)$ To find x

Distance formula $d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)$

$bar(AB) = sqrt((-1-0)^2 + (-1-2)^2) = sqrt10$

$bar(AC) = sqrt((-1-x)^2 + (_1+4)^2)$

Bus $bar(AB) = bar(AC)$

$sqrt((-1-x)^2 + 3^2) = sqrt 10$

Squaring both sides,

$(x+1)^2 + 9 = 10$

$(x+1)^2 = 1 = 1^2$

$x+1 = +-1$

$color(crimson)(x = 0, -2$