The distance of the point P on the X-axis fro | vedclass

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(A) Let the coordinates of point $P$ on the $X-$axis be $(x, 0)$.Given that the distance $PA = 13$ units.Using the distance formula,$PA^2 = (x_2 - x_1

Vedclass · Accepted Answer

$(16, 0)$

Vedclass · Answer

(A) Let the coordinates of point $P$ on the $X-$axis be $(x, 0)$.Given that the distance $PA = 13$ units.Using the distance formula,$PA^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.$13^2 = (x - 11)^2 + (0 - 12)^2$$169 = (x - 11)^2 + (-12)^2$$169 = (x - 11)^2 + 144$$(x - 11)^2 = 169 - 144$$(x - 11)^2 = 25$Taking the square root on both sides,$x - 11 = \pm 5$.Case $1$: $x - 11 = 5 \implies x = 16$.Case $2$: $x - 11 = -5 \implies x = 6$.Therefore,the coordinates of $P$ are $(16, 0)$ or $(6, 0)$.

The distance of the point $P$ on the $X-$axis from $A(11, 12)$ is $13$ units. The coordinates of $P$ are $\ldots \ldots \ldots \ldots$.

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