VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Distance between two points is considered one the most difficult concept.
30 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 2025: City Slip Link | Study Plan | Official Question Paper (Session 1)
JEE Main 2025: Sample Papers | Mock Tests | PYQs | High Scoring Topics | College Predictor
New: Meet Careers360 experts in your city and get guidance on shortlisting colleges
Apply to TOP B.Tech /BE Entrance exams: VITEEE | MET | AEEE | BITSAT
If in a cartesian coordinate system Point A = (3,5) and point B = (5,7) find the distance between A and B ?
What is the distance b/w points A(3,-6 ) and B (2,1) along y-axis ?
What is the distance b/w points P(2,7) and Q (7,8) along x-axis
If the distance between the points $(a, 0,1)$ and $(0,1,2)$ is $\sqrt{27}$, then the value of a is
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements | Last Date to Apply: 28th March
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
The value of 'a' if the distance between P(a, 5) and Q(2, a) is 7 units, 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!"