Question: The perpendicular distance of a point from the \[x\]–axis is 3 units and the perpendicular distance ...

Question

The perpendicular distance of a point from the $x$ –axis is 3 units and the perpendicular distance of a point from the $y$ –axis is 2 units. Write the coordinates of the point if it lies in
(a) Quadrant I
(b) Quadrant II
(c) Quadrant III
(d) Quadrant IV

Answer 1

Solution

Here, we need to find which of the given options is correct. We will use the given perpendicular distances to find the abscissa and ordinate of the point when it lies in the different quadrants, and find the coordinates of the point if it lies in the different quadrants.

Complete step by step solution:
We will use the given perpendicular distances to find the abscissa and ordinate of the point when it lies in the different quadrants.
The perpendicular distance of a point from the $x$ –axis can be used to get the ordinate of the point.
It is given that the perpendicular distance of a point from the $x$ –axis is 3 units.
Therefore, the ordinate of the point will be 3 if the point lies above the $x$ –axis, or $- 3$ if the point lies below the $x$ –axis.
The perpendicular distance of a point from the $y$ –axis can be used to get the abscissa of the point.
It is given that the perpendicular distance of a point from the $y$ –axis is 2 units.
Therefore, the abscissa of the point will be 2 if the point lies to the right of the $y$ –axis, or $- 2$ if the point lies to the left of the $y$ –axis.
(a) Quadrant I
All points lying in Quadrant I are above the $x$ –axis.
Therefore, we get the ordinate of the given point as 3.
All points lying in Quadrant I are to the right of the $y$ –axis.
Therefore, we get the abscissa of the given point as 2.
Thus, we get the point in Quadrant I as $\left( {2,3} \right)$ .
(b) Quadrant II
All points lying in Quadrant II are above the $x$ –axis.
Therefore, we get the ordinate of the given point as 3.
All points lying in Quadrant II are to the left of the $y$ –axis.
Therefore, we get the abscissa of the given point as $- 2$ .
Thus, we get the point in Quadrant II as $\left( { - 2,3} \right)$ .
(c) Quadrant III
All points lying in Quadrant III are below the $x$ –axis.
Therefore, we get the ordinate of the given point as $- 3$ .
All points lying in Quadrant III are to the left of the $y$ –axis.
Therefore, we get the abscissa of the given point as $- 2$ .
Thus, we get the point in Quadrant III as $\left( { - 2, - 3} \right)$ .
(d) Quadrant IV
All points lying in Quadrant IV are below the $x$ –axis.
Therefore, we get the ordinate of the given point as $- 3$ .
All points lying in Quadrant IV are to the right of the $y$ –axis.
Therefore, we get the abscissa of the given point as 2.

Thus, we get the point in Quadrant IV as $\left( {2, - 3} \right)$ .

Note:
We used the terms ‘abscissa’ and ‘ordinate’ in the solution. The abscissa of a point $\left( {x,y} \right)$ is $x$ , and the ordinate of a point $\left( {x,y} \right)$ is $y$ .
We can observe the points with their perpendicular distances in the graph below.

In the first quadrant, all the values of $x$ and $y$ are positive. In the second quadrant, the values of $x$ are negative but the values of $y$ are positive. In the third quadrant, all the values of $x$ and $y$ are negative. In the second quadrant, the values of $x$ are positive but the values of $y$ are negative.