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Distance between two points is considered one the most difficult concept.
24 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
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If in a cartesian coordinate system Point A = (3,5) and point B = (5,7) find the distance between A and B ?
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
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}
$
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