Question

If perpendicular distance of a point p from the x axis be 3 units along the negative direction of y axis, then the point p has

Answer 1

Answer:

If perpendicular distance of a point P from the x-axis be 3 units along the negative direction of y axis , then the point has positive x-coordinate( or abscissa) and negative Y co-ordinate (or ordinate).

Answer 2

Given,

The perpendicular distance of Point P from the x-axis = 3 units

Now the distance is given along the negative direction of the y-axis.

So, the abscissa of the point p is -3 as the direction is along the negative y-axis.

Now we have to find the ordinate of the p point which can be easily determined by the distance formula which is given by,

$\sqrt[2]{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 }$

here, point p coordinates are (a, -3). Assuming "a" as the ordinate. And the distance is 3 units so applying the distance formula between (3, 0) and (p, -3) we get the ordinate as:

$3 = \sqrt[2]{(3-p)^2 + (0-3)^2 }$

Taking square on both sides,

$9 = 9 + p^2-6p+9$

$0 = p^2 -6p\\p(p-6) = 0$

so p=0 or p=6,
Hence, the answer to this question is 6.

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