Question

Find the distance of the point \[P(-1,5)\] from x-axis.

Answer 1

Hint: To solve this type of problem we have to know the concept of the distance of any point from either x-axis or y-axis. The distance of any point from the x-axis is the point on y-axis. By drawing graphs we can easily solve this problem.

Complete step-by-step answer:
Consider a point \[A\left( {{x}_{1}},{{y}_{1}} \right)\]in coordinate axis then the distance from
 x-axis is \[\left| {{y}_{1}} \right|\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
So similarly we can find the distance for the given point \[P(-1,5)\] from x-axis.
From (a) we can directly say that the distance of a point \[P(-1,5)\] from the x-axis is 5units.
Now drawing graph
In the coordinate axis first locate the given point.
The given point has x as negative value and y as positive that means the point lies in the second quadrant.
Now plotting the point in the coordinate axis.

Now we have to calculate the distance from x-axis to this point.
From the graph we can say that the distance from x-axis to the point P is 5 units.

Note: This is a direct problem which can be solved easily by plotting points in the coordinate axis. Here we found the distance. Distance can never be negative, If we get a negative value then we have to apply modulus.