Math Lab
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Distance Between Two Points

Coordinate Geometry · Class XI

Pick two points in the plane; see the Pythagorean distance and midpoint coordinates.

0
type a value
-1010
0
type a value
-1010
3
type a value
-1010
4
type a value
-1010
Live values
  • Distance5
  • Midpoint x1.5
  • Midpoint y2
xy
  • Distance to (x, y2) from A

Formulas in this lab

  • Distance
    d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  • Midpoint x
    xm=(x1+x2)/2x_m = (x_1+x_2)/2
  • Midpoint y
    ym=(y1+y2)/2y_m = (y_1+y_2)/2
Tip: As B slides left/right, distance traces a hyperbola , minimum is when B is directly above or below A.

Frequently asked questions

▶What is the distance formula in the plane?

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. So between $(0, 0)$ and $(3, 4)$ the distance is $5$ — the classic Pythagorean triple. It is the workhorse of all 2-D geometry.

▶How do I use the distance lab?

Slide the two points' coordinates and the lab returns $d$. Set one point at the origin to see the connection $d = \sqrt{x^2 + y^2}$. Try $(1, 1)$ and $(4, 5)$ to get $d = 5$, again a 3-4-5 in disguise.

▶Distance versus displacement: are they the same in maths?

In Class XI maths, distance is always the straight-line length, always non-negative. In Physics, displacement is a vector with direction. The formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ gives the magnitude in both cases. Don't carry sign confusion from physics into maths.

▶How does this generalize to 3-D for JEE?

In 3-D the distance between $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ — just add the $z$-term. The 3-D distance lab (xii-12) extends this idea; the 2-D version here builds the foundation.