Question

P is a point on the X-axis. It is equidistant from the points $(2,6)$ and $(-4,0)$. The co-ordinates of P are

• A. $P(-2,0)$
• B. $P(2,0)$
• C. $P(-6,0)$
• D. $P(6,0)$

Answer 1

Answer: (B)

$P(2,0)$

Given that-

P is a point on the X-axis.

P is equidtstant from two points $Q(x_1,y_1)=(2,6)$ & $R(x_2,y_2)=(-4,0)$.

To find out-

The coordinates of P.

Solution-

Since P is a point on the X-axis its coordinates are $P(x,0)$.

Also, the distances $PQ=PR$, as P is equidistant from the given two points Q & R.

Now the distance formula for two points,

$(x_a,y_a)$ & $(x_b,y_b)$ is $d_{ab}=\sqrt{{(x_b-x_a)}^2+{(y_b-y_a)}^2}$

$\therefore PQ=d_{PQ}=\sqrt{{(x-2)}^2+{(0-6)}^2}$ units $=\sqrt{x^2-4x+40}$ units And $PR=d_{pr}=\sqrt{{(x+4)}^2+{(0-0)}^2}$ units $=\sqrt{{(x+4)}^2}$ units $=\sqrt{x^2+8x+16}$ units.

We have $PQ=PR⟹x^2-4x+40=x^2+8x+16$

$⟹12x=24⟹x=2.$

$\therefore P(x,0)=(2,0)$