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    Amrita University BTech Admission Through JEE Main 2026 and AEEE

    Distance Between Two Points Formula - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • Distance between two points is considered one the most difficult concept.

    • 24 Questions around this concept.

    Solve by difficulty

    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

    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

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    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:

    Concepts Covered - 1

    Distance between two points

    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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    Books

    Reference Books

    Distance between two points

    Mathematics for Joint Entrance Examination JEE (Advanced) : Coordinate Geometry

    Page No. : 1.4

    Line : 58

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