MCQ
The point $O(0, 0)$ lies on:
- A$x-$axis
- B$y-$axis
- ✓both $x-$axis and $y-$axis
- DAny quadrant.
✓
Answer
Correct option: C.
both $x-$axis and $y-$axis
Point $(0, 0)$ is the co-ordinate of origin and origin is the point of intersection of $x$ and $y-$axis.
So, point $O(0, 0)$ lies on both axis.
So, point $O(0, 0)$ lies on both axis.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The empirical relation between mean, mode and median is:
For the equation $5x - 7y = 35$, if $y = 5$, then the value of $‘x’$ is:
→Which of the following is a linear polynomial?
Let $f(x)$ be a polynomial such that $\text{f}\Big(-\frac{1}{2}\Big)=0,$ then a factor of $f(x)$ is:
→In a bar graph, length of a bar is 6.4 cm and it represents 256 units. The number of units represented by a bar of length 5.3 cm is
Three statements are given below:
$i.$ In a $\| gm$, the angle bisectors of two adjacent angles enclose a right angle.
$ii.$ The angle bisectors of a $\| gm$ form a rectangle.
$iii.$ The triangle formed by joining the mid$-$points of the sides of an isosceles triangle is not necessarily an isosceles triangle.
Which is true?
→$i.$ In a $\| gm$, the angle bisectors of two adjacent angles enclose a right angle.
$ii.$ The angle bisectors of a $\| gm$ form a rectangle.
$iii.$ The triangle formed by joining the mid$-$points of the sides of an isosceles triangle is not necessarily an isosceles triangle.
Which is true?
A hemispherical bowl is made of steel, $0.25\ cm$ thick. If the inner radius of the bowl is $3.25\ cm,$ then the outer curved surface area of the bowl is:
→For a given data, the difference between the maximum and minimum observation is known as its.
If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, the ratio of the volumes of upper and lower part is:
In the given figure, $O$ is the centre of the circle and $\angle\text{BDC} = 42^\circ.$ The measure of $\angle\text{ACB}$ is:
→