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ಇಳಿಜಾರು, ದೂರ ಮತ್ತು ಮಧ್ಯಬಿಂದು ಕ್ಯಾಲ್ಕುಲೇಟರ್

ಎರಡು ಬಿಂದುಗಳ ದೂರ, ಮಧ್ಯಬಿಂದು, ಇಳಿಜಾರು ಮತ್ತು ರೇಖೆಯ ಸಮೀಕರಣವನ್ನು ಇಳಿಜಾರು-ಛೇದ, ಪ್ರಮಾಣಿತ ಹಾಗೂ ಬಿಂದು-ಇಳಿಜಾರು ರೂಪಗಳಲ್ಲಿ ನೋಡಿ. ಎರಡು ರೇಖೆಗಳ ಛೇದನಬಿಂದುವನ್ನೂ ಲೆಕ್ಕಿಸಿ.

ನವೀಕರಿಸಲಾಗಿದೆ ಪರಿಶೀಲಿಸಿದ ಉದಾಹರಣೆಗಳು: 6

ಪ್ರಯತ್ನಿಸಿ
Distance P₁P₂
Distance P₁P₂: 5
ಗರಿಷ್ಠ ದಶಮಾಂಶ ಸ್ಥಾನಗಳು: 8; ಹತ್ತಿರದ ಮೌಲ್ಯಕ್ಕೆ; ಸಮದೂರದಲ್ಲಿದ್ದರೆ ಸೊನ್ನೆಯಿಂದ ದೂರಕ್ಕೆ
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)
ಲೆಕ್ಕಾಚಾರದ ವಿಧಾನ 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.

ಇಳಿಜಾರು, ದೂರ ಮತ್ತು ಮಧ್ಯಬಿಂದು ಕ್ಯಾಲ್ಕುಲೇಟರ್ ಬಗ್ಗೆ

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.

ಪರಿಹರಿಸಿದ ಉದಾಹರಣೆಗಳು

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

ಪರಿಶೀಲನೆಯ ಮೂಲ: 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

ಪರಿಶೀಲನೆಯ ಮೂಲ: 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

ಪರಿಶೀಲನೆಯ ಮೂಲ: 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
ಹೌದು
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

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

ಪ್ರಶ್ನೆಗಳು

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.

“ಇಳಿಜಾರು, ದೂರ ಮತ್ತು ಮಧ್ಯಬಿಂದು ಕ್ಯಾಲ್ಕುಲೇಟರ್” ಎಷ್ಟು ನಿಖರವಾಗಿದೆ?

ನಿಖರತೆ ನಿಮ್ಮ ಇನ್‌ಪುಟ್‌ಗಳು ಮತ್ತು ವಿಧಾನದ ಊಹೆಗಳನ್ನು ಅವಲಂಬಿಸಿದೆ. ದಶಮಾಂಶ ಗಣನೆ 50 ಸಾರ್ಥಕ ಅಂಕೆಗಳನ್ನು ಬಳಸುತ್ತದೆ. ಆದರೆ ಅಂದಾಜುಗಳು, ಸಂಖ್ಯಾತ್ಮಕ ವಿಧಾನಗಳು ಮತ್ತು ಮೂಲ ದತ್ತಾಂಶ ಕಡಿಮೆ ನಿಖರವಾಗಿರಬಹುದು; ಪ್ರದರ್ಶಿತ ಮೌಲ್ಯಗಳನ್ನು ರೌಂಡ್ ಮಾಡುವುದರಿಂದ ಈ ಮಿತಿಗಳು ನಿವಾರಣೆಯಾಗುವುದಿಲ್ಲ. ಸ್ವತಂತ್ರ ಮೂಲಗಳ ಪರಿಹಾರಗಳೊಂದಿಗೆ ಪರಿಶೀಲಿಸಿದ ಉದಾಹರಣೆಗಳು: 6. ಉದಾಹರಣೆಗೆ, “P₁(1, 2) and P₂(4, 6)” ಅನ್ನು 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)) ಜೊತೆಗೆ ಪರಿಶೀಲಿಸಲಾಗುತ್ತದೆ.

ಈ ವಿಧಾನದ ಮೂಲ ಯಾವುದು?

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

ಈ ಕ್ಯಾಲ್ಕುಲೇಟರ್ ಬಗ್ಗೆ

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|

ಮೂಲಗಳು

  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

ಆಧಾರಗಳೊಂದಿಗೆ ಪರಿಶೀಲಿಸಲಾಗಿದೆ

ಈ ಕ್ಯಾಲ್ಕುಲೇಟರ್‌ನಲ್ಲಿ ಸ್ವತಂತ್ರ ಆಧಾರಗಳಿಂದ ಉತ್ತರಗಳನ್ನು ಪಡೆದ 6 ಬಿಡಿಸಿದ ಉದಾಹರಣೆಗಳಿವೆ. ಇವು ಪರೀಕ್ಷಾ ಸಮೂಹದಲ್ಲಿ ನಡೆಯುತ್ತವೆ; ನೀವು ಇಲ್ಲಿಯೂ ನಡೆಸಬಹುದು.

ಸಂಬಂಧಿತ ಕ್ಯಾಲ್ಕುಲೇಟರ್‌ಗಳು