Q1. Find the distance between (1, 2) and (4, 6).
d = √[(4−1)² + (6−2)²]
d = √[9 + 16] = √25
d = 5 units
Turn geometry into arithmetic. Compute the distance and midpoint between any two points, then master the section and area formulas.
Enter two points to get the distance between them and their midpoint.
The distance of a point (x, y) from the origin is simply √(x² + y²).
Distance-formula problems with steps.
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Q1. Find the distance between (1, 2) and (4, 6).
d = √[(4−1)² + (6−2)²]
d = √[9 + 16] = √25
d = 5 units
| Quantity | Formula |
|---|---|
| Distance | d = √[(x₂−x₁)² + (y₂−y₁)²] |
| Midpoint | ((x₁+x₂)/2, (y₁+y₂)/2) |
| Area of triangle | ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| |
The distance between (x₁,y₁) and (x₂,y₂) is √[(x₂−x₁)² + (y₂−y₁)²].
The point dividing the segment joining two points in ratio m:n is ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)).
They are collinear if the area of the triangle formed by them is zero.
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