حاسبة الميل والمسافة ونقطة المنتصف

احسب المسافة ونقطة المنتصف والميل بين نقطتين، ومعادلة المستقيم بصيغ الميل والمقطع، والصيغة القياسية، والنقطة والميل، وتقاطع مستقيمين.

آخر تحديث أمثلة تم التحقق منها: 6

جرّب
Distance P₁P₂
Distance P₁P₂: 5
الحد الأقصى للمنازل العشرية: 8؛ إلى الأقرب، وعند التعادل بعيدًا عن الصفر
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)
طريقة الحساب 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.

حول حاسبة الميل والمسافة ونقطة المنتصف

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.

أمثلة محلولة

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

مصدر التحقق: ⁨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

مصدر التحقق: ⁨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

مصدر التحقق: ⁨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
نعم
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

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

الأسئلة

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.

ما مدى دقة «⁨حاسبة الميل والمسافة ونقطة المنتصف⁩»؟

تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 6. مثلًا، يجري التحقق من «⁨P₁(1, 2) and P₂(4, 6)⁩» بالرجوع إلى ⁨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))⁩.

ما مصدر هذه الطريقة؟

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

حول هذه الحاسبة

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|

المصادر

  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

تم التحقق بالرجوع إلى المصادر

تتضمن هذه الحاسبة أمثلة محلولة بإجابات من مصادر مستقلة، وعددها 6. تُشغّل ضمن مجموعة الاختبارات، ويمكنك تشغيلها هنا أيضًا.

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