MCQ
The distance between the points $(a, b)$ and $(-a, b)$ is :
- A$\sqrt{2\text{a}\text{ units}}$
- B$2\sqrt{\text{a}^2+\text{b}^2}\text{ units}$
- ✓$2a$ units
- D$2b$ units
✓
Answer
Correct option: C.
$2a$ units
Let point $A$ be $(a,b)$ and point $B$ is $(-a-b)$
$\therefore$ distance between $A$ and $B=\sqrt{\text(a-a)^2+\text(b-b)^2}$
$=4\sqrt{4\text{a}^2+0}$
$=2a$ units
$\therefore$ distance between $A$ and $B=\sqrt{\text(a-a)^2+\text(b-b)^2}$
$=4\sqrt{4\text{a}^2+0}$
$=2a$ units
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
Choose the correct option. Justify your choice : $(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec}\theta) $
→The quadratic polynomial, the sum of whose zeroes is $-5$ and their product is $6,$ is :
→In the given figure, if M and N are points on the sides OP and OS respectively of $\triangle OPS$, such that MN || PS, then the length of OP is :
→If the distance between the points $(4, p)$ and $(1,0)$ is 5 units, then the value of $p$ is
→If $\text{HCF}\ (a, b) = 12$ and $a \times b = 1800, $ then $\text{LCM}\ (a, b$) is :
→If $8 \tan x = 15,$ then $\sin x - \cos x$ is equal to
→In $\triangle\text{ABC, }\text{AB}=6\sqrt{3 }\text{ cm},\text{AC}=12\text{ cm}$ and $\text{BC}=6\text{ cm}.$ Then, $\angle\text{B}$ is :
→If $\frac{2\text{x}+\text{y}+2}{5}=\frac{3\text{x}-\text{y}+1}{3}=\frac{3\text{x}+2\text{y}+1}{6}$ then:
→After an examination, a teacher wants to know the marks obtained by maximum number of the students in her class. She requires to calculate __________ of marks.
An observer $1.5 m$ tall is $28.5 m$ away from a tower and the angle of elevation of the top of the tower from the eye of the observer is $45^{\circ}$. The height of the tower is
→