CBSE Questions for Class 9 Maths Coordinate Geometry Quiz 3 - MCQExams.com

Given, $$(2,7)$$.

Here, both the $$x$$-coordinate and $$y$$-coordinate are positive.

Therefore, $$(2,7)$$ lies in $$1$$st quadrant.

Hence, option $$A$$ is correct.

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.

Therefore, any point whose $$x$$-coordinate is $$0$$ and $$y$$-coordinate is non-zero will lie on the $$y$$-axis.

Here, the given point is $$(0,8)$$ is of the form $$(0,y)$$, where $$y=8$$.

and therefore lies on the $$y$$-axis.

Hence, option $$B$$ is correct.

Consider point is $$A$$.

Here, the point is at a distance of $$2$$ units right of the $$y$$-axis.

So, its abscissa will be $$2$$.

Also, the point is at a distance of $$2$$ unit above the $$x$$-axis.

So, its ordinate will be $$2$$.

Therefore, the co-ordinates of the required point are $$A(2,2)$$.

Hence, the point lies on the $$I$$st quadrant.

Consider point is $$D$$.

Here, the point is at a distance of $$6$$ units left of the $$y$$-axis.

So, its abscissa will be $$-6$$.

Also, the point is at a distance of $$4$$ unit above the $$x$$-axis.

So, its ordinate will be $$4$$.

Therefore, the co-ordinates of the required point are $$D(-6,4)$$.

Hence, the point lies on the $$II$$nd quadrant.

Consider point is $$G$$.

Here, the point is at a distance of $$7$$ units left of the $$y$$-axis.

So, its abscissa will be $$-7$$.

Also, the point is at a distance of $$5$$ unit below the $$x$$-axis.

So, its ordinate will be $$-5$$.

Therefore, the co-ordinates of the required point are $$G(-7,-5)$$.

Hence, the point lies on the $$III$$rd quadrant.

Consider point is $$H$$.

Here, the point is at a distance of $$5$$ units right of the $$y$$-axis.

So, its abscissa will be $$5$$.

Also, the point is at a distance of $$3$$ unit below the $$x$$-axis.

So, its ordinate will be $$-3$$.

Therefore, the co-ordinates of the required point are $$H(5,-3)$$.

Hence, the point lies on the $$IV$$th quadrant.

Therefore, the only point that lies in the $$II$$nd quadrant is $$D(-6,4)$$.

Hence, option $$B$$ is correct.

We know, the coordinates of the point of intersection of $$x$$ - axis and $$y$$ - axis is called the origin.

That is, the co-ordinates of the origin is $$(0,0) $$.

Therefore, option $$C$$ is correct.

Consider point is $$B$$.

Here, the point is at a distance of $$6$$ units right of the $$y$$-axis $$\implies x=6$$.

Also, the point is at a distance of $$1$$ unit above the $$x$$-axis$$\implies y=1$$.

Therefore, the co-ordinates of the required point are $$B(6,1)$$.

Here, the abscissa is $$6$$.

Hence, the magnitude of abscissa is $$6$$.

Consider point is $$D$$.

Here, the point is at a distance of $$6$$ units left of the $$y$$-axis $$\implies x=-6$$.

Also, the point is at a distance of $$4$$ unit above the $$x$$-axis$$\implies y=4$$.

Therefore, the co-ordinates of the required point are $$D(-6,4)$$.

Here, the abscissa is $$-6$$.

Hence, the magnitude of abscissa is $$6$$.

Consider point is $$C$$.

Here, the point is at a distance of $$4$$ units right of the $$y$$-axis $$\implies x=4$$.

Also, the point is at a distance of $$5$$ unit above the $$x$$-axis$$\implies y=5$$.

Therefore, the co-ordinates of the required point are $$G(4,5)$$.

Here, the abscissa is $$4$$.

Hence, the magnitude of abscissa is $$4$$.

Consider point is $$G$$.

Here, the point is at a distance of $$7$$ units left of the $$y$$-axis $$\implies x=-7$$.

Also, the point is at a distance of $$5$$ unit below the $$x$$-axis$$\implies y=-5$$.

Therefore, the co-ordinates of the required point are $$G(-7,-5)$$.

Here, the abscissa is $$-7$$.

Hence, the magnitude of abscissa is $$7$$.

Therefore, the only point which has the largest abscissa in terms of magnitude is $$G(-7,-5)$$.

Hence, option $$D$$ is correct.

The required point is $$\text{P}$$.

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

Then, its ordinate is $$0$$ i.e. $$y = 0$$.

Also, the point is at a distance of $$2$$ units right of the $$y$$-axis.

So, its abscissa will be $$+2$$.

Hence, the coordinates of the required point are $$\text{P}(2,0)$$.

Therefore, option $$\text{B}$$ is correct.

The required point is $$D$$.

Here, the point is at a distance of $$6$$ units left of the $$y$$-axis $$\implies x=-6$$.

Also, the point is at a distance of $$4$$ units above the $$x$$-axis$$\implies y=4$$.

Therefore, the co-ordinates of the required point are $$D(-6,4)$$.

Here, the ordinate is $$4$$.

Hence, option $$B$$ is correct.

Given, the point $$(-11,0)$$.

That is, $$x$$ is $$-11$$, which is negative and $$y$$ is $$0$$, i.e. the point lies on the $$x$$-axis.

Hence, the given point $$(-11,0)$$ lies on the negative direction of $$x$$-axis.

Therefore, option $$C$$ is correct.

We know, in the $$xy$$-plane, any point is of the form $$(x,y)$$, where $$x,y$$ can be any real number.

Also, for the point $$(x,y)$$,