Question
Show that the point $(0, 9)$ is equidistant from the points $(– 4, 1)$ and $(4, 1)$
✓
Answer
Let $P\left(x_1, y_1\right)=P(0,9), Q\left(x_2, y_2\right)=Q(-4,1), R\left(x_3, y_3\right)=R(4,1)$
By distance formula,
$ d(P, Q)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{[(-4)-0]^2+(1-9)^2}$
$=\sqrt{(-4)^2+(-8)^2}$
$=\sqrt{16+64}$
$=\sqrt{80}$
$=4 \sqrt{5} $
And
$ d(P, R)=\sqrt{\left(x_3-x_1\right)^2+\left(y_3-y_1\right)^2}$
$=\sqrt{(4-0)^2+(1-9)^2}$
$=\sqrt{4^2+(-8)^2}$
$=\sqrt{16+64}$
$=\sqrt{80}$
$=4 \sqrt{5} $
And
$d(P, R)=\sqrt{\left(x_3-x_1\right)^2+\left(y_3-y_1\right)^2}$
$=\sqrt{(4-0)^2+(1-9)^2}$
$=\sqrt{4^2+(-8)^2}$
$=\sqrt{16+64}$
$=\sqrt{80}$
$=4 \sqrt{5}$
Here, $d(P, Q)=d(P, R)$
$\therefore$ The point $(0,9)$ is equidistant from $(-4,1)$ and $(4,1)$.
By distance formula,
$ d(P, Q)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{[(-4)-0]^2+(1-9)^2}$
$=\sqrt{(-4)^2+(-8)^2}$
$=\sqrt{16+64}$
$=\sqrt{80}$
$=4 \sqrt{5} $
And
$ d(P, R)=\sqrt{\left(x_3-x_1\right)^2+\left(y_3-y_1\right)^2}$
$=\sqrt{(4-0)^2+(1-9)^2}$
$=\sqrt{4^2+(-8)^2}$
$=\sqrt{16+64}$
$=\sqrt{80}$
$=4 \sqrt{5} $
And
$d(P, R)=\sqrt{\left(x_3-x_1\right)^2+\left(y_3-y_1\right)^2}$
$=\sqrt{(4-0)^2+(1-9)^2}$
$=\sqrt{4^2+(-8)^2}$
$=\sqrt{16+64}$
$=\sqrt{80}$
$=4 \sqrt{5}$
Here, $d(P, Q)=d(P, R)$
$\therefore$ The point $(0,9)$ is equidistant from $(-4,1)$ and $(4,1)$.
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