Line Geometry
Line Geometry calculator
▲
(1)
Line equation:
y =
x +
Two points on line:
x
_{1}
:
y
_{1}
:
x
_{2}
:
y
_{2}
:
Angle (α):
Intercepts:
(
, 0
)
:
(
0 ,
)
:
Midpoint:
x
_{d}
:
y
_{d}
:
Line length:
Point in plane:
x
_{p}
:
y
_{p}
:
The point is on the line
Distance from point (x
_{p}
, y
_{p}
) to line (1):
(2)
Equation of line passing through (x
_{p}
, y
_{p}
)
and is perpendicular to line (1):
Intersection coordinates between lines
(1) and (2):
x:
y:
Degree
Radian
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:
m
_{1}
m
_{2}
= − 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
solution by matrix:
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
y
_{intercept}
(y
_{i}
)
y
_{i}
= b
y
_{i}
= −mx
_{i}
y
_{i}
= y
_{1}
− mx
_{1}
x
_{intercept}
(x
_{i}
)
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 (x
_{1}
, y
_{1}
) and (x
_{2}
, y
_{2}
) in the ratio p:q
Point (x, y) which divides the line connecting two points (x
_{1}
, y
_{1}
) and (x
_{2}
, y
_{2}
) 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 (x
_{1}
, y
_{1}
) 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 (x
_{1}
, y
_{1}
)
y
=
mx
+ (
y
_{1}
− mx
_{1}
)
y − y
_{1}
=
m
(
x − x
_{1}
)
Equation of a line passing through two points (x
_{1}
, y
_{1}
), (x
_{2}
, y
_{2}
)
Equation of a line perpendicular to a given slope m and passing through a point (x
_{p}
, y
_{p}
)
Equation of a line perpendicular to a line which is defined by two points (x
_{1}
, y
_{1}
) and (x
_{2}
, y
_{2}
) and passing through the point (x
_{p}
, y
_{p}
)
Equation of a line passing through the intercepts x
_{i}
, y
_{i}
x
_{i}
y = −y
_{i}
x + x
_{i}
y
_{i}
Equation of a line passing through the point (x
_{p}
,y
_{p}
) and parallel to a line which is defined by two points (x
_{1}
, y
_{1}
) and (x
_{2}
, y
_{2}
)
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 x
_{i}
and y
_{i}
Distance from a line to the origin
Distance from a line given by two points (x
_{1}
,y
_{1}
),(x
_{2}
, y
_{2}
) to the origin
Distance from a line to the point
(x
_{p}
, y
_{p}
)
Distance from a line given by two points (x
_{1}
, y
_{1}
) , (x
_{2}
, y
_{2}
) to the point (x
_{p}
, y
_{p}
)
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
(
p
_{x}
, p
_{y}
)
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
= 4
x −
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
+
y
_{1}
− mx
_{1}
The equation of line PQ passing through point
(
p
_{x}
, p
_{y}
)
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 S
_{x}
and S
_{y}
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
p
_{x}
= 2
p
_{y}
=
−
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
2
x − 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
+ (
y
_{1}
− Mx
_{1}
)
x
+ 2
y −
1 = 0
If the intersection point of both lines is needed then we have to solve the equations:
x
+ 2
y −
1 = 0
2
x − y −
1 = 0
From the first equation x is
x
= 1
−
2
y
Substitute x into the second equation
2(1
−
2
y
)
− 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
2
x − 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
(P
_{x}
, P
_{y}
) = (5 , 2)
(green line) and is perpendicular to the given line is:
y
=
Mx
+ (
P
_{y}
− MP
_{x}
)
(1)
y = −
0.5
x
+ 2 + 0.5‧
5
(2)
x
+ 2
y −
9 = 0
In order to find the intersection point
(
N
_{x}
, N
_{y}
)
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
P
_{x}
= 5
P
_{y}
= 2
Solving the numerical values, we have
x
+ 2(2
x
)
−
9 = 0
x
= 9/5 = 1.8
and y is equal to
y
= 18/5 = 3.6