Pendenza, distanza e punto medio tra due punti

Calcola distanza, punto medio, pendenza e l'equazione della retta nelle forme esplicita, generale e punto-pendenza, oltre all'intersezione di due rette.

Aggiornato Esempi verificati: 6

Prova
Distance P₁P₂
Distance P₁P₂: 5
Massimo di cifre decimali: 8; Al più vicino; a parità lontano da zero
Midpoint x
2.5
Midpoint y
4
Slope
1.33333333
Angle of inclination
53.1301°
Slope-intercept form
y = (4/3)x + 2/3
Standard form
4x − 3y = −2
Point-slope form
y − 2 = (4/3)(x − 1)
Perpendicular bisector
y = −0.75x + 5.875

P₁(1, 2) and P₂(4, 6) are 5 apart with midpoint (2.5, 4); the line through them has slope 4/3 (rising at 53.13°).

Points and lines

d = 5P₁(1, 2)P₂(4, 6)M(2.5, 4)
Come si calcola S
  1. Distance

    d=(4−1)2+(6−2)2=25=5d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{25} = 5
  2. Midpoint

    M=(x1+x22,y1+y22)=(2.5,4)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (2.5, 4)
  3. Slope

    m=y2−y1x2−x1=43=4/3m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4}{3} = 4/3

    Angle of inclination: arctan m = 53.1301°.

  4. Line through the points

    Slope-intercept: y = (4/3)x + 2/3 · Standard: 4x − 3y = −2 · Point-slope: y − 2 = (4/3)(x − 1)

  5. Perpendicular bisector

    Through M(2.5, 4) at right angles to P₁P₂: y = −0.75x + 5.875.

Informazioni su Pendenza, distanza e punto medio tra due punti

Enter two points to get the straight-line distance d = √((x₂ − x₁)² + (y₂ − y₁)²), the midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2) and the slope m = (y₂ − y₁)/(x₂ − x₁). The line through them is written in slope-intercept form y = mx + b, in standard form Ax + By = C with whole-number coefficients and in point-slope form, alongside its perpendicular bisector. Fractions stay exact, so a slope of 4/3 is not rounded to 1.3333.

The default points (1, 2) and (4, 6) are 5 apart, a 3-4-5 triangle, with midpoint (2.5, 4) and line y = (4/3)x + 2/3. Switching on a second line adds whether the two are parallel, perpendicular or intersecting, the angle between them and the crossing point.

A vertical line has no slope, so only its standard form, such as x = 3, is given. Coordinates carry no unit: the distance is in whatever unit they use.

Esempi svolti

P₁(1, 2) and P₂(4, 6)

Point 1: x
1
Point 1: y
2
Point 2: x
4
Point 2: y
6
Distance P₁P₂
5
Midpoint x
2.5
Midpoint y
4
Slope
1.33333333
Angle of inclination
53.130102 °
Slope-intercept form
y = (4/3)x + 2/3
Standard form
4x − 3y = −2
Point-slope form
y − 2 = (4/3)(x − 1)
Perpendicular bisector
y = −0.75x + 5.875

Fonte di verifica: Python 3.8 fractions: dx = 3, dy = 4 (3-4-5), m = 4/3, b = 2 − 4/3 = 2/3; bisector through (5/2, 4) with slope −3/4 has b = 47/8 = 5.875 (terminating fractions print as decimals); math.degrees(atan(4/3))

Vertical line (edge case: no slope)

Point 1: x
3
Point 1: y
1
Point 2: x
3
Point 2: y
7
Distance P₁P₂
6
Standard form
x = 3
Angle of inclination
90 °
Perpendicular bisector
y = 4
Midpoint y
4

Fonte di verifica: Equal x-coordinates: the line x = 3 is vertical; the bisector is the horizontal line through (3, 4)

Decimal coordinates

Point 1: x
-1.5
Point 1: y
2.25
Point 2: x
3.5
Point 2: y
-0.75
Distance P₁P₂
5.83095189
Slope
-0.6
Slope-intercept form
y = −0.6x + 1.35
Standard form
12x + 20y = 27

Fonte di verifica: Python 3.8 fractions: dx = 5, dy = −3, √34; −3x − 5y = −27/4 scaled by −4

Two intersecting lines

Point 1: x
1
Point 1: y
2
Point 2: x
4
Point 2: y
6
Compare with a second line
sì
Line 2, point 3: x
0
Line 2, point 3: y
6
Line 2, point 4: x
6
Line 2, point 4: y
0
The lines are
Intersecting
Angle between the lines
81.869898 °
Intersection x
2.28571429
Intersection y
3.71428571
Line 2
y = −x + 6

Fonte di verifica: tan θ = |(−1 − 4/3)/(1 − 4/3)| = 7, Python 3.8 math.degrees(atan(7)); intersection (16/7, 26/7) by Python fractions

Domande

How do you find the slope between two points?

Divide the change in y by the change in x: m = (y₂ − y₁)/(x₂ − x₁). From (1, 2) to (4, 6) the rise is 4 and the run is 3, so m = 4/3 ≈ 1.3333. A positive slope rises to the right, a negative one falls, 0 is horizontal, and when x₂ = x₁ the line is vertical and the slope is undefined.

What is the distance formula?

d = √((x₂ − x₁)² + (y₂ − y₁)²), which is Pythagoras' theorem applied to the horizontal and vertical gaps. Between (1, 2) and (4, 6) the gaps are 3 and 4, so d = √25 = 5. For points given as latitude and longitude use a great-circle formula instead, because the flat formula ignores the Earth's curvature.

How do you find the equation of a line through two points?

Find the slope m, then the intercept b = y₁ − m·x₁, and write y = mx + b. Through (1, 2) and (4, 6), m = 4/3 and b = 2 − 4/3 = 2/3, so y = (4/3)x + 2/3. Multiplying by 3 and rearranging gives the standard form 4x − 3y = −2, with whole-number coefficients.

How do you tell if two lines are parallel or perpendicular?

Compare their slopes. Parallel lines have equal slopes, and perpendicular lines have slopes whose product is −1, such as 1/2 and −2. Any other pair crosses at an angle θ with tan θ = |(m₂ − m₁)/(1 + m₁m₂)|: slopes of 4/3 and −1 give tan θ = 7, so the lines meet at 81.87°.

What is a perpendicular bisector?

The line through the midpoint of a segment at right angles to it; every point on it is equally far from both endpoints. For (1, 2) and (4, 6) it passes through (2.5, 4) with slope −3/4, the negative reciprocal of 4/3, giving y = −0.75x + 5.875. The bisectors of a triangle's three sides meet at the centre of its circumscribed circle.

Quanto è preciso «Pendenza, distanza e punto medio tra due punti»?

La precisione dipende dai dati inseriti e dalle ipotesi del metodo. Il calcolo decimale usa 50 cifre significative, ma stime, metodi numerici e dati di origine possono essere meno precisi; l’arrotondamento visualizzato non elimina questi limiti. Esempi svolti verificati con fonti indipendenti: 6. Per esempio, «P₁(1, 2) and P₂(4, 6)» viene verificato con Python 3.8 fractions: dx = 3, dy = 4 (3-4-5), m = 4/3, b = 2 − 4/3 = 2/3; bisector through (5/2, 4) with slope −3/4 has b = 47/8 = 5.875 (terminating fractions print as decimals); math.degrees(atan(4/3)).

Da dove proviene il metodo?

OpenStax College Algebra 2e, §2.1 (distance and midpoint formulas) and §2.2 (equations of lines); Weisstein, E. W. “Line”, “Perpendicular Bisector” — MathWorld.

Informazioni su questa calcolatrice

d=(x2−x1)2+(y2−y1)2,M=(x1+x22,y1+y22),m=y2−y1x2−x1,tan⁡θ=∣m2−m11+m1m2∣d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2},\quad M = \left(\tfrac{x_1 + x_2}{2}, \tfrac{y_1 + y_2}{2}\right),\quad m = \frac{y_2 - y_1}{x_2 - x_1},\quad \tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|

Fonti

  1. OpenStax College Algebra 2e, §2.1 (distance and midpoint formulas) and §2.2 (equations of lines)
  2. Weisstein, E. W. “Line”, “Perpendicular Bisector” — MathWorld

Verificato con le fonti

Questa calcolatrice include 6 esempi svolti con risposte da fonti indipendenti. Fanno parte della suite di test e puoi eseguirli anche qui.

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