Calculations involving lines [ edit | edit source ] Distance between two points [ edit | edit source ] The distance between two points can be found with the Pythagorean theorem The distance between two points is given by the formula
D
i
s
t
=
Δ
x
2
+
Δ
y
2
{\displaystyle Dist={\sqrt {\Delta x^{2}+\Delta y^{2}}}}
Δ
x
{\displaystyle \Delta x}
x-values between the two points and
Δ
y
{\displaystyle \Delta y}
y-values between the two points.
This formula can also be seen as applying the Pythagorean theorem to the points, where the differences in x-values and y-values form two sides of a right-angled triangle.
The midpoint is the point halfway between two points. The midpoint is the point which is exactly halfway between two points. The coordinates of the midpoint are given by
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
{\displaystyle ({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}})}
(
x
1
,
y
1
)
{\displaystyle (x_{1},y_{1})}
(
x
2
,
y
2
)
{\displaystyle (x_{2},y_{2})}
e.g. The midpoint between
(
2
,
−
1
)
{\displaystyle (2,-1)}
(
4
,
3
)
{\displaystyle (4,3)}
(
3
,
1
)
{\displaystyle (3,1)}
You may notice that this expression states that the midpoint's x-coordinate is the average of the x-coordinates of the points, and its y-coordinate is the average of the points' y-coordinates. Essentially, this means that the midpoint is the average of the two points.
The gradient
m
{\displaystyle m}
m
=
r
i
s
e
r
u
n
{\displaystyle m={\frac {rise}{run}}}
r
i
s
e
{\displaystyle rise}
r
u
n
{\displaystyle run}
This can also be expressed as
m
=
y
2
−
y
1
x
2
−
x
1
{\displaystyle m={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}
e.g. The gradient of a line that passes through
(
2
,
4
)
{\displaystyle (2,4)}
(
3
,
6
)
{\displaystyle (3,6)}
6
−
4
3
−
2
=
2
1
=
2
{\displaystyle {\frac {6-4}{3-2}}={\frac {2}{1}}=2}
An intersection between two lines When two lines intersect , the point of intersection is where the two lines cross. The point of intersection is thus on both lines, meaning that it can be found using simultaneous equations.
e.g. The lines
y
=
2
x
+
3
{\displaystyle y=2x+3}
y
=
−
x
{\displaystyle y=-x}
y
=
2
x
+
3
y
=
−
x
−
x
=
2
x
+
3
−
3
x
=
3
x
=
−
1
y
=
−
(
−
1
)
=
1
∴
The
point of intersection is at
(
−
1
,
1
)
{\displaystyle {\begin{aligned}y&=2x+3\\y&=-x\\-x&=2x+3\\-3x&=3\\x&=-1\\y&=-(-1)=1\\\therefore {\text{The}}&{\text{ point of intersection is at }}(-1,1)\end{aligned}}}
The lines AB and CD are parallel. Parallel lines always have the same gradient, and do not intersect.
e.g. The lines
y
=
3
x
+
1
{\displaystyle y=3x+1}
y
=
3
x
−
2
{\displaystyle y=3x-2}
Sometimes we'll need to find a line which is parallel to a given line and passes through a given point.
e.g. Find the equation of a line parallel to
y
=
2
x
+
3
{\displaystyle y=2x+3}
(
5
,
6
)
{\displaystyle (5,6)}
P
a
r
a
l
l
e
l
⟹
y
=
2
x
+
c
(
5
,
6
)
⟹
6
=
2
(
5
)
+
c
6
=
10
+
c
c
=
−
4
∴
y
=
2
x
−
4
{\displaystyle {\begin{aligned}Parallel\implies &y=2x+c\\(5,6)\implies &6=2(5)+c\\&6=10+c\\&c=-4\\\therefore \quad &y=2x-4\end{aligned}}}
The blue line and red line are perpendicular Perpendicular lines are at right angles to one another. The product of the gradients of two perpendicular lines is always -1.
e.g. The lines
y
=
x
{\displaystyle y=x}
y
=
−
x
{\displaystyle y=-x}
Sometimes we'll need to find a perpendicular line that goes through a specific point.
e.g. Find the equation of the line perpendicular to
y
=
−
x
2
+
3
{\displaystyle y=-{\frac {x}{2}}+3}
P
e
r
p
e
n
d
i
c
u
l
a
r
⟹
y
=
−
1
−
1
/
2
x
+
c
y
=
2
x
+
c
T
h
r
o
u
g
h
o
r
i
g
i
n
⟹
0
=
2
(
0
)
+
c
c
=
0
∴
y
=
2
x
{\displaystyle {\begin{aligned}Perpendicular\implies &y={\frac {-1}{-1/2}}x+c\\&y=2x+c\\Through\ origin\implies &0=2(0)+c\\&c=0\\\therefore \quad &y=2x\end{aligned}}}
Different forms of the equation of a line [ edit | edit source ] There are three main ways to write an equation of a line:
y
=
m
x
+
c
{\displaystyle y=mx+c}
y
−
y
1
x
−
x
1
=
m
{\displaystyle {\frac {y-y_{1}}{x-x_{1}}}=m}
a
x
+
b
y
+
c
=
0
{\displaystyle ax+by+c=0}
Finding the equation of a line from a point and the gradient [ edit | edit source ] The equation of a line can be found using a point and the gradient using the second equation
y
−
y
1
x
−
x
1
=
m
{\displaystyle {\frac {y-y_{1}}{x-x_{1}}}=m}
y
=
m
x
+
c
{\displaystyle y=mx+c}
e.g. A line with gradient
2
{\displaystyle 2}
(
2
,
3
)
{\displaystyle (2,3)}
y
−
3
x
−
2
=
2
y
−
3
=
2
(
x
−
2
)
y
=
2
x
−
4
+
3
y
=
2
x
−
1
{\displaystyle {\begin{aligned}{\frac {y-3}{x-2}}&=2\\y-3&=2(x-2)\\y&=2x-4+3\\y&=2x-1\end{aligned}}}
Finding the equation of a line from two points [ edit | edit source ] When given two points, we can find the gradient using
m
=
y
2
−
y
1
x
2
−
x
1
{\displaystyle m={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}
e.g. A line travels through the points
(
2
,
−
3
)
{\displaystyle (2,-3)}
(
1
,
1
)
{\displaystyle (1,1)}
m
=
1
+
3
1
−
2
=
4
−
1
=
4
y
+
3
x
−
2
=
4
y
+
3
=
4
(
x
−
2
)
y
+
3
=
4
x
−
8
y
=
4
x
−
11
{\displaystyle {\begin{aligned}m={\frac {1+3}{1-2}}&={\frac {4}{-1}}=4\\{\frac {y+3}{x-2}}&=4\\y+3&=4(x-2)\\y+3&=4x-8\\y&=4x-11\end{aligned}}}
A circle consists of all points that are a given distance from its centre. The distance between two points can be defined using the Pythagorean theorem
a
2
+
b
2
=
c
2
{\displaystyle a^{2}+b^{2}=c^{2}}
x
2
+
y
2
=
r
2
{\displaystyle x^{2}+y^{2}=r^{2}}
r
{\displaystyle r}
e.g. A circle centred at the origin with radius
5
{\displaystyle 5}
x
2
+
y
2
=
25
{\displaystyle x^{2}+y^{2}=25}
If the circle is not centred at the origin, we can translate this equation to a different point. Thus, the equation becomes
(
x
−
x
c
)
2
+
(
y
−
y
c
)
2
=
r
2
{\displaystyle (x-x_{c})^{2}+(y-y_{c})^{2}=r^{2}}
(
x
c
,
y
c
)
{\displaystyle (x_{c},y_{c})}
e.g. A circle centred at
(
1
,
2
)
{\displaystyle (1,2)}
5
{\displaystyle {\sqrt {5}}}
(
x
−
1
)
2
+
(
y
−
2
)
2
=
5
{\displaystyle (x-1)^{2}+(y-2)^{2}=5}
Interactions between Lines & Circles [ edit | edit source ] When given a problem where a line and a circle intersect, it is useful to use a substitution method of solving simultaneous equations.
e.g. The line
y
=
1
−
x
{\displaystyle y=1-x}
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1}
x
2
+
y
2
=
1
y
=
1
−
x
x
2
+
(
1
−
x
)
2
=
1
x
2
+
1
−
2
x
+
x
2
=
1
2
x
2
−
2
x
=
0
2
x
(
x
−
1
)
=
0
x
=
{
0
,
1
}
y
=
{
1
,
0
}
∴
Intersections are at (0,1) and (1,0)
{\displaystyle {\begin{aligned}x^{2}+y^{2}&=1\\y&=1-x\\x^{2}+(1-x)^{2}&=1\\x^{2}+1-2x+x^{2}&=1\\2x^{2}-2x&=0\\2x(x-1)&=0\\x&=\{0,1\}\\y&=\{1,0\}\\\therefore &\ {\text{Intersections are at (0,1) and (1,0)}}\end{aligned}}}
Interactions between Lines & Quadratics [ edit | edit source ] When a quadratic and a line intersect, we can again use substitution to find the points of intersection.
e.g. The line
y
=
3
x
−
8
{\displaystyle y=3x-8}
y
=
x
2
−
2
x
−
2
{\displaystyle y=x^{2}-2x-2}
y
=
x
2
−
2
x
−
2
y
=
3
x
−
8
3
x
−
8
=
x
2
−
2
x
−
2
x
2
−
5
x
+
6
=
0
(
x
−
2
)
(
x
−
3
)
=
0
x
=
{
2
,
3
}
y
=
{
−
2
,
1
}
∴
The points of
intersection are at
(
2
,
−
2
)
and
(
3
,
1
)
{\displaystyle {\begin{aligned}y&=x^{2}-2x-2\\y&=3x-8\\3x-8&=x^{2}-2x-2\\x^{2}-5x+6&=0\\(x-2)(x-3)&=0\\x&=\{2,3\}\\y&=\{-2,1\}\\\therefore {\text{The points of }}&{\text{intersection are at }}(2,-2){\text{ and }}(3,1)\end{aligned}}}
In some cases, we need to find a constant that ensures there is only one point of intersection. In said cases, we should use the discriminant.
e.g. Find the value
k
{\displaystyle k}
y
=
x
+
k
{\displaystyle y=x+k}
y
=
x
2
−
3
x
−
2
{\displaystyle y=x^{2}-3x-2}
T
a
n
g
e
n
t
⟹
o
n
e
r
o
o
t
⟹
Δ
=
0
y
=
x
2
−
3
x
−
2
y
=
x
+
k
x
+
k
=
x
2
−
3
x
−
2
x
2
−
4
x
−
2
−
k
=
0
Δ
=
(
−
4
)
2
−
4
(
1
)
(
−
2
−
k
)
=
0
16
+
8
+
4
k
=
0
4
k
=
−
24
k
=
−
6
{\displaystyle {\begin{aligned}Tangent\implies one\ &root\implies \Delta =0\\y&=x^{2}-3x-2\\y&=x+k\\x+k&=x^{2}-3x-2\\x^{2}-4x-2-k&=0\\\Delta =(-4)^{2}-4(1)(-2-k)&=0\\16+8+4k&=0\\4k&=-24\\k&=-6\end{aligned}}}
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