CBSE Questions for Class 9 Maths Coordinate Geometry Quiz 2 - MCQExams.com

We know, for the point $$(x,y)$$,

in $$I$$ quadrant both $$x$$ and $$y$$ are positive,
in $$II$$ quadrant  $$x$$ is negative and $$y$$ is positive,
in $$III$$ quadrant both $$x$$ and $$y$$ are negative
and in $$IV$$ quadrant $$x$$ is positive and $$y$$ is negative.

Given, the point $$(8,7)$$.

That is, $$x$$ is $$8$$, which is positive and $$y$$ is $$7$$, which is also positive.

Hence, the given point $$(8,7)$$ lies in the $$I$$st quadrant.

Therefore, the statement is true.

Hence, option $$A$$ is correct.

$${\textbf{Step - 1: Plot and identify points in the Cartesian plane.}}$$

                 $${\text{The given equation is}}$$ $$x = y.$$

                 $${\text{So, if}}$$ $$x = 1$$ $${\text{and}}$$ $$y = 1.$$

                 $${\text{If}}$$ $$x = -5$$ $${\text{and}}$$ $$y = -5.$$

                 $${\text{If}}$$ $$x = 0$$ $${\text{and}}$$ $$y = 0.$$

$${\textbf{Step - 2: By Induction.}}$$

                  $$x = a$$ $${\text{and}}$$ $$y = a.$$

                  $${\text{So, any point P on the line}}$$ $$x = y$$ $${\text{will be of the form}}$$ $$\text{P} (a, a).$$  

$${\textbf{ Hence the correct option (A) (a, a).}}$$

We know, in the $$xy$$-plane, any point on the $$y$$-axis will have $$x =0$$.
Hence, any point on the $$y$$-axis is of the form $$(0,y)$$, where $$y$$ can be any non-zero number.

But here, given the point $$(a,b)$$ and $$b=0$$.

Then, the point will be $$(a,0)$$.

Clearly, the point lies on the $$x$$-axis.

Hence, the given statement is false.

Therefore, option $$B$$ is correct

Let the required point be $$P$$.

Given, the point lies on $$y$$-axis.

Then, its abscissa is $$0$$ or $$x = 0$$.

Also, given the point is at a distance of $$6$$ units below the $$x$$-axis,

So, its ordinate will be $$-6$$.
Hence, the co-ordinates of the required point are $$(0,-6)$$. 

Let the required point be $$P$$.

Given, the point lies on $$x$$-axis.

Then, its ordinate is $$0$$ or $$y = 0$$.

Also, given the point is at a distance of $$3$$ units to the left of the origin.

So, its abscissa will be $$-3$$.
Hence, the co-ordinates of the required point are $$(-3,0)$$.