Find the coordinates of the point on the X-ax | vedclass

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(C) Let the point on the $X-$axis be $P(x, 0)$.The distance between $P(x, 0)$ and $A(4, 4)$ is given as $5$ units.Using the distance formula: $\sqrt{(

Vedclass · Accepted Answer

Both $(1,0)$ and $(7,0)$

Vedclass · Answer

(C) Let the point on the $X-$axis be $P(x, 0)$.The distance between $P(x, 0)$ and $A(4, 4)$ is given as $5$ units.Using the distance formula: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = d$$\sqrt{(x - 4)^2 + (0 - 4)^2} = 5$Squaring both sides: $(x - 4)^2 + (-4)^2 = 5^2$$(x - 4)^2 + 16 = 25$$(x - 4)^2 = 9$Taking the square root: $x - 4 = \pm 3$Case $1$: $x - 4 = 3 \implies x = 7$Case $2$: $x - 4 = -3 \implies x = 1$Therefore,the coordinates of the points are $(1, 0)$ and $(7, 0)$.

Find the coordinates of the point on the $X-$axis which is at a distance of $5$ units from $A(4,4)$.

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