﻿﻿ Line Geometry
 Line Geometry calculator ▲
 (1) Line equation: y = x +
 Two points on line: x1: y1: x2: y2:
 Angle (α): Intercepts:  ( , 0 ):     ( 0 , ): Midpoint: xd: yd: Line length:
 Point in plane: xp:    yp: The point is on the line
Distance from point (xp, yp) to line (1):
(2) Equation of line passing through (xp, yp)
and is perpendicular to line (1):
Intersection coordinates between lines
(1) and (2):
 x: y:
 Line basic definition Line distances Intersection point ex: 3 Basic line information Distance from point ex: 1 Line equations Perpendicular line ex: 2
 Line basic definition ▲
 The most general equation of a line is of the form: (1)
where  A, B  and  C  are any real number and  A  and  B  are not both zero. If  B ≠ 0  then we can divide
 equation (1)  by  B  to obtain the form: (2)
 This is the equation of a line whose slope is: and
 Equation (2) can be normalized to the general form: (3)
 the slope of the line (m) is defined in terms of the inclination (4)
Note: if the angle α is greater then 90 degrees then the slope is negative.
 α (0 - 90) degrees   : positive slope α (90 - 180) degree   : negative slope
Necessary condition for two lines to be perpendicular to each other is that their slopes fulfill the
 condition: m1 m2 = − 1 (5)
In order to find the intersection point of two lines we have to solve the system of linear equations
 representing the lines. A x + B y = −C D x + E y = −F
 If the determinant: then intersection point exists.
 Basic line of the form     y = ax + b     or     Ax + By + C = 0 ▲
Slope (m) of the line
 m = a
yintercept (yi)
 yi = b yi = −mxi yi = y1 − mx1
xintercept (xi)
tan θ
 tan θ = m
Line angle (θ)
from x axis (range 0 ≤ θ < π)
 θ = arctan(m)
Slope (M) of a line perpendicular to a given slope (m)
 Line midpoint
Point (x, y) which divides the line connecting two points (x1 , y1) and (x2 , y2) in the ratio p:q
Point (x, y) which divides the line connecting two points (x1 , y1) and (x2 , y2) externally at a ratio p:q
 Note: the (x,y) point is in the direction from point 1 to point 2,to get the other side extension change the point 1 with point 2 and vice versa
A point (x, y) which is located at a distance d from a point (x1 , y1) on the line
Angle θ between two lines:
Angle between two lines given by
 Ax + By + C = 0 Dx + Ey + F = 0
 Lines equations ▲
Equation of a line passing through a point (x1 , y1)
 y = mx + ( y1 − mx1) y − y1 = m(x − x1)
Equation of a line passing through two points (x1 , y1), (x2 , y2)
Equation of a line perpendicular to a given slope m and passing through a point (xp , yp)
Equation of a line perpendicular to a line which is defined by two points (x1 , y1) and (x2 , y2) and passing through the point (xp , yp)
Equation of a line passing through the intercepts xi , yi
 xiy = −yix + xiyi
Equation of a line passing through the point (xp ,yp) and parallel to a line which is defined by two points (x1 , y1) and (x2 , y2)
Equation of a line parallel to the line Ax + By + C = 0 and at a distance d from it.
Equation of the midline between the lines    Ax + By + C = 0
Dx + Ey + F = 0
Equation of a line perpendicular to the line    Ax + By + C = 0
 Bx − Ay + C = 0
Equation of a horizontal line
 y = b
Equation of a vertical line
 x = a
 Lines distances     y = ax + b    Ax + By + C = 0 ▲
Distance between two points (D)
Distance between intercepts xi and yi
Distance from a line to the origin
Distance from a line given by two points (x1 ,y1),(x2 , y2) to the origin
Distance from a line to the point
(xp , yp)
Distance from a line given by two points (x1 , y1) , (x2 , y2) to the point (xp , yp)
Distance between two parallel lines
 y = ax + b y = cx + d
or
 Ax + By + C = 0 Dx + Ey + F = 0
 Example 1 - distance from a line to a point ▲
Find the distance  d  of a point   P(px , py)   to the line given by the equation   Ax + By + C = 0.
Calculate the numerical value of the distance between point  (2 , −3)  and line  y = 4x − 5
We draw at point  P  a parallel line to the given line (dashed line).
Then we draw a line  PF  between the points  P  and  F. This line is perpendicular to the given line.
 The slope m of the given line is: m = −A / B The slope M of line PF is: M = −1 / m = B / A
In order to make calculations easier we draw another line  QS  starting from the origin and parallel to line  PF.
The general equation of a line passing through a point is given by:      y = mx + y1 − mx1
The equation of line  PQ  passing through point  (px , py)  is:
Equation of line  QS  passing through the origin is:
Point  Q  is the intersection of lines  PQ  and  QS.  Solving this equations by substituting the value of  y  we get:
 And finally, we get: and
 Point  S  is the intersection point of the two lines Ax + By + C = 0 and
Solving for  Sx  and  Sy  we get
 and
The distance  d  is the distance between the two points  Q  and  S  given by the points:
 and
The given line can be written in the standard form as       4x − y − 5 = 0
So, we have:       A = 4       B =1     and     C =5        px = 2       py =3
Substituting these values into the distance equation we get the final solution:
 Example 2 - line from point to perpendicular line ▲
Find the equation of the line passing through the point   (3 , − 1)   and is perpendicular to the line
2x − y − 1 = 0
The slope of the perpendicular line is:
 y = 2x − 1 → m = 2
The slope of the line perpendicular to the given line and passing through the point is:
The equation of the line passing through the point  (3 , − 1)   is:
 y = Mx + (y1 − Mx1) x + 2y − 1 = 0
If the intersection point of both lines is needed then we have to solve the equations:
 x + 2y − 1 = 0 2x − y − 1 = 0
 From the first equation x is x = 1 − 2y Substitute x into the second equation 2(1 − 2y) − y − 1 = 0 And the intersection coordinate is (0.6 , 0.2)
 Example 3 - intersection point of perpendicular lines ▲
Find the intersection point of the line   2x − y = 0   and the line passing through the point (5 , 2)   and is perpendicular to the given line.
If the line form is given by:     Ax + By + C = 0
 The slope of the given line is: m = − A/B
Slope of the line perpendicular to the given line is:
The equation of the line passing through the point   (Px , Py) = (5 , 2)   (green line) and is perpendicular to the given line is:
 y = Mx + (Py − MPx) (1) y = −0.5x + 2 + 0.5‧5 (2) x + 2y − 9 = 0
In order to find the intersection point (Nx , Ny) we have to solve equation (1) and the given line this can be done in direct substitution or by Cramer’s law:
 From equation (1) We have Substitute  y  into equation (1) Solving for  x  we get And  y  is equal to
The general intersection of a line and the perpendicular line that passes through a point is:
 (3)
 The given values are     A = 2       B = −1       C = 0       Px = 5       Py = 2 Solving the numerical values, we have x + 2(2x) − 9 = 0 x = 9/5 = 1.8 and y is equal to y = 18/5 = 3.6