Amity University-Noida B.Tech Admissions 2026
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Distance between two points is considered one the most difficult concept.
25 Questions around this concept.
The locus of the centres of the circles, which touch the circle,
$x^{2}+y^{2}=1$ externally , also touch the y-axis and lie in the
first quadrant, is :
The intersection of three lines $x-y=0,x+2y=3 \; \text {and} \; 2x+y=6$ is a :
A triangle with vertices $(4, 0), (–1, –1), (3, 5)$ is
JEE Main 2026 Session 2 Memory Based Questions: April 2: Shift-1 | Shift 2
JEE Main 2026: Rank Predictor | College Predictor | Admit Card Link
JEE Main Prep: Last 10 Year's Ques | Most Repeated Questions| High Scoring Chapters
Don't Miss: Mock Test | Important Formulas | Foreign Universities in India
If in a cartesian coordinate system Point A = (3,5) and point B = (5,7) find the distance between A and B ?
If the distance between the points $(a, 0,1)$ and $(0,1,2)$ is $\sqrt{27}$, then the value of a is
If $P(\sqrt{2} \sec \theta, \sqrt{2} \tan \theta)$ is a point on the hyperbola whose distance from the origin is $\sqrt{6}$ (where P is in the first quadrant) then $\theta$ equals
The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Last Date to Apply: 29th April | Ranked #43 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements
The value of 'a' if the distance between P(a, 5) and Q(2, a) is 7 units, is
Let the range of the function
$
f(x)=6+16 \cos x. \cos \left(\frac{\pi}{3}-x\right) . \cos \left(\frac{\pi}{3}+x\right)
$
$\sin 3 x. \cos 6 x, x \in R$ be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3 x+4 y+12=0$ is:
Distance between two points
Given, Point A (x1, y1) and B (x2, y2) are two points on the cartesian plane

Using Pythagoras Theorem
$
\begin{aligned}
& c^2=a^2+b^2 \\
& A B^2=a^2+b^2 \\
& A B^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2 \\
& |A B|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}
\end{aligned}
$
Note:
The distance of a point $\mathrm{A}(\mathrm{x}, \mathrm{y})$ from the origin $\mathrm{O}(0,0)$ is given by
$
|O A|=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}
$
"Stay in the loop. Receive exam news, study resources, and expert advice!"