Steigung, Abstand und Mittelpunkt zweier Punkte

Abstand, Mittelpunkt und Steigung sowie Geradengleichungen in Steigungs-, Standard- und Punkt-Steigungsform berechnen, dazu den Schnittpunkt zweier Geraden.

Aktualisiert Geprüfte Beispiele: 6

Ausprobieren
Distance P₁P₂
Distance P₁P₂: 5
Maximale Nachkommastellen: 8; Zum nächsten Wert, bei Gleichstand von null weg
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)
So wird gerechnet 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.

Über Steigung, Abstand und Mittelpunkt zweier Punkte

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.

Durchgerechnete Beispiele

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

Prüfquelle: 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

Prüfquelle: 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

Prüfquelle: 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
ja
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

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

Fragen

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.

Wie genau arbeitet „Steigung, Abstand und Mittelpunkt zweier Punkte“?

Die Genauigkeit hängt von Ihren Eingaben und den Annahmen der Methode ab. Die Dezimalrechnung nutzt 50 signifikante Stellen, doch Schätzungen, numerische Verfahren und Quelldaten können ungenauer sein. Die angezeigte Rundung beseitigt diese Grenzen nicht. Anhand unabhängiger Quellen geprüfte Rechenbeispiele: 6. Beispielsweise wird „P₁(1, 2) and P₂(4, 6)“ anhand von 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)) geprüft.

Woher stammt die Methode?

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

Über diesen Rechner

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|

Quellen

  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

Anhand von Quellen geprüft

Dieser Rechner enthält 6 Rechenbeispiele mit Ergebnissen aus unabhängigen Quellen. Sie laufen in der Testsuite und können auch hier ausgeführt werden.

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