The Central Idea
A geometric equation is a rule that selects exactly the points belonging to a geometric object.
For example,
\[ y=2x+1 \]describes a line in the Cartesian plane.
A point such as
\[ (3,7) \]lies on the line because
\[ 7=2(3)+1. \]A point that does not satisfy the equation does not lie on that line.
Equations Define Sets of Points
An equation such as
\[ y=mx+b \]does not describe just one ordered pair. It describes every ordered pair \((x,y)\) satisfying the relationship.
The full solution set of the equation is therefore a geometric line.
This is one of the central connections between algebra and geometry:
\[ \boxed{\text{equation} \longleftrightarrow \text{geometric locus}}. \]Slope-Intercept Form
One of the most familiar equations of a nonvertical line is
\[ \boxed{y=mx+b}. \]Here:
- \(m\) is the slope;
- \(b\) is the \(y\)-intercept.
The slope determines the line's direction, while the intercept determines where it crosses the \(y\)-axis.
The \(y\)-Intercept
The \(y\)-intercept occurs where
\[ x=0. \]In
\[ y=mx+b, \]substituting \(x=0\) gives
\[ y=b. \]Therefore the \(y\)-intercept is the point
\[ \boxed{(0,b)}. \]Example: Reading Slope and Intercept
Consider
\[ y=-3x+5. \]The slope is
\[ \boxed{-3}, \]and the \(y\)-intercept is
\[ \boxed{(0,5)}. \]The negative slope means the line falls from left to right.
Constructing a Line from Slope and Intercept
Suppose a line has slope
\[ \frac23 \]and \(y\)-intercept
\[ (0,-4). \]Substitute into
\[ y=mx+b. \]The equation is
\[ \boxed{y=\frac23x-4}. \]Point-Slope Form
If a line has slope \(m\) and passes through
\[ (x_1,y_1), \]then every point \((x,y)\) on the line satisfies
\[ \frac{y-y_1}{x-x_1}=m, \]provided \(x\ne x_1\).
Rearranging gives the point-slope form:
\[ \boxed{y-y_1=m(x-x_1)}. \]Why Point-Slope Form Works
The slope between a fixed point
\[ (x_1,y_1) \]and a variable point
\[ (x,y) \]on the same nonvertical line must remain constant:
\[ m=\frac{y-y_1}{x-x_1}. \]Multiplying by \(x-x_1\) gives
\[ y-y_1=m(x-x_1). \]Point-slope form therefore comes directly from the definition of slope.
Example: Equation from a Point and Slope
Find the equation of the line with slope
\[ 4 \]through
\[ (2,-3). \]Use point-slope form:
\[ y-(-3)=4(x-2). \]Thus,
\[ \boxed{y+3=4(x-2)}. \]Expanding gives
\[ y=4x-11. \]Equation from Two Points
If two distinct points are given, first compute the slope.
Suppose
\[ A=(1,3),\qquad B=(5,11). \]Then
\[ m=\frac{11-3}{5-1}=2. \]Use either point in point-slope form:
\[ y-3=2(x-1). \]Therefore,
\[ \boxed{y=2x+1}. \]Two-Point Form
When two points
\[ (x_1,y_1) \]and
\[ (x_2,y_2) \]are known and \(x_1\ne x_2\), point-slope form can be combined with the slope formula:
\[ \boxed{ y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1) }. \]This is often called the two-point form.
Standard Form of a Line
A line can also be written in the form
\[ \boxed{Ax+By=C}, \]where \(A\), \(B\), and \(C\) are constants and \(A\) and \(B\) are not both zero.
Standard form is useful because it treats horizontal, vertical, and slanted lines within one algebraic framework.
Converting Standard Form to Slope-Intercept Form
Suppose
\[ 3x+2y=12. \]Solve for \(y\):
\[ 2y=-3x+12, \] \[ y=-\frac32x+6. \]Therefore the slope is
\[ -\frac32, \]and the \(y\)-intercept is
\[ (0,6). \]Slope from Standard Form
For
\[ Ax+By=C, \]with \(B\ne0\), solve for \(y\):
\[ By=-Ax+C, \] \[ y=-\frac{A}{B}x+\frac{C}{B}. \]Therefore,
\[ \boxed{m=-\frac{A}{B}}. \]Horizontal Lines
Every point on a horizontal line has the same \(y\)-coordinate.
Therefore a horizontal line has equation
\[ \boxed{y=k}, \]where \(k\) is constant.
For example,
\[ y=4 \]is the horizontal line four units above the \(x\)-axis.
Vertical Lines
Every point on a vertical line has the same \(x\)-coordinate.
Therefore a vertical line has equation
\[ \boxed{x=k}. \]For example,
\[ x=-3 \]is the vertical line three units left of the \(y\)-axis.
Vertical lines cannot be written in ordinary slope-intercept form because their slope is undefined.
The Coordinate Axes as Lines
The \(x\)-axis consists of all points whose \(y\)-coordinate is zero:
\[ \boxed{y=0}. \]The \(y\)-axis consists of all points whose \(x\)-coordinate is zero:
\[ \boxed{x=0}. \]Finding Intercepts from Standard Form
For
\[ Ax+By=C, \]set \(y=0\) to find the \(x\)-intercept and set \(x=0\) to find the \(y\)-intercept.
For example, consider
\[ 2x+3y=12. \]With \(y=0\),
\[ 2x=12, \]so the \(x\)-intercept is
\[ (6,0). \]With \(x=0\),
\[ 3y=12, \]so the \(y\)-intercept is
\[ (0,4). \]Intercept Form
A line with nonzero \(x\)- and \(y\)-intercepts
\[ (a,0) \]and
\[ (0,b) \]can be written
\[ \boxed{ \frac{x}{a}+\frac{y}{b}=1 }. \]This form makes the intercepts visible immediately.
Example: Intercept Form
A line crosses the axes at
\[ (5,0) \]and
\[ (0,2). \]Therefore,
\[ \boxed{ \frac{x}{5}+\frac{y}{2}=1 }. \]Multiplying by \(10\) gives the equivalent standard form
\[ 2x+5y=10. \]Equivalent Equations of the Same Line
A line can have many algebraically different but equivalent equations.
For example,
\[ 2x+y=6 \]and
\[ 4x+2y=12 \]describe the same line because the second equation is obtained by multiplying the first by \(2\).
Multiplying an equation by any nonzero constant preserves its solution set.
Testing Whether a Point Lies on a Line
To test whether
\[ P=(x_0,y_0) \]lies on a line, substitute its coordinates into the equation.
For
\[ 3x-y=5, \]test
\[ (2,1). \]Substitution gives
\[ 3(2)-1=5. \]Since the equation is true, the point lies on the line.
Parallel Lines from Equations
Nonvertical parallel lines have equal slopes.
For example,
\[ y=3x+1 \]and
\[ y=3x-7 \]both have slope \(3\).
They are distinct because their intercepts differ, so they are parallel.
Perpendicular Lines from Equations
Nonvertical lines with nonzero slopes are perpendicular when
\[ m_1m_2=-1. \]For example,
\[ y=2x+4 \]has slope \(2\).
A perpendicular line must have slope
\[ -\frac12. \]Equation of a Parallel Line Through a Point
Find the line through
\[ (2,5) \]parallel to
\[ y=-3x+7. \]The required slope is
\[ -3. \]Use point-slope form:
\[ y-5=-3(x-2). \]Therefore,
\[ \boxed{y=-3x+11}. \]Equation of a Perpendicular Line Through a Point
Find the line through
\[ (4,-1) \]perpendicular to
\[ y=\frac23x+5. \]The perpendicular slope is
\[ -\frac32. \]Use point-slope form:
\[ y+1=-\frac32(x-4). \]Thus,
\[ \boxed{y=-\frac32x+5}. \]Vector Form of a Line
A line can also be described using a point and a direction vector.
Let
\[ \mathbf{r}_0= \begin{pmatrix} x_0\\ y_0 \end{pmatrix} \]be a point on the line and
\[ \mathbf{v}= \begin{pmatrix} a\\ b \end{pmatrix} \]be a nonzero direction vector.
Then the line can be written
\[ \boxed{ \mathbf{r} = \mathbf{r}_0+t\mathbf{v} }. \]The parameter \(t\) ranges over the real numbers.
Parametric Equations of a Line
The vector equation
\[ \mathbf{r} = \begin{pmatrix} x_0\\ y_0 \end{pmatrix} + t \begin{pmatrix} a\\ b \end{pmatrix} \]is equivalent to the pair of parametric equations
\[ \boxed{x=x_0+at} \]and
\[ \boxed{y=y_0+bt}. \]These equations generate every point on the line as \(t\) varies.
Example: Parametric Line
A line passes through
\[ (1,-2) \]with direction vector
\[ \begin{pmatrix} 3\\ 4 \end{pmatrix}. \]Its parametric equations are
\[ \boxed{x=1+3t} \]and
\[ \boxed{y=-2+4t}. \]The direction vector gives slope
\[ \frac43. \]Normal Form of a Line
In standard form,
\[ Ax+By=C, \]the vector
\[ \boxed{ \mathbf{n}= \begin{pmatrix} A\\ B \end{pmatrix} } \]is perpendicular to the line.
This vector is called a normal vector.
The coefficients of \(x\) and \(y\) therefore encode a perpendicular direction.
Why the Standard-Form Coefficients Give a Normal
Suppose
\[ Ax+By=C. \]A direction vector for the line can be chosen as
\[ \mathbf{v}= \begin{pmatrix} B\\ -A \end{pmatrix}. \]Then
\[ \begin{pmatrix} A\\ B \end{pmatrix} \cdot \begin{pmatrix} B\\ -A \end{pmatrix} = AB-AB=0. \]Their dot product is zero, so the vectors are perpendicular.
From Lines in Two Dimensions to Planes in Three
In the plane, a single linear equation in \(x\) and \(y\) usually describes a line.
In three-dimensional space, a single linear equation in \(x\), \(y\), and \(z\) usually describes a plane:
\[ \boxed{Ax+By+Cz=D}. \]The coefficients determine the plane's orientation.
Normal Vector of a Plane
For the plane
\[ Ax+By+Cz=D, \]the vector
\[ \boxed{ \mathbf{n}= \begin{pmatrix} A\\ B\\ C \end{pmatrix} } \]is perpendicular to the plane.
This normal vector plays the same structural role that
\[ \begin{pmatrix} A\\ B \end{pmatrix} \]plays for a line in two dimensions.
Point-Normal Form of a Plane
Suppose a plane passes through
\[ P_0=(x_0,y_0,z_0) \]and has normal vector
\[ \mathbf{n}= \begin{pmatrix} A\\ B\\ C \end{pmatrix}. \]For any point
\[ P=(x,y,z) \]on the plane, the displacement
\[ \overrightarrow{P_0P} = \begin{pmatrix} x-x_0\\ y-y_0\\ z-z_0 \end{pmatrix} \]lies within the plane and is therefore perpendicular to the normal.
Hence,
\[ \boxed{ A(x-x_0)+B(y-y_0)+C(z-z_0)=0 }. \]Example: Plane from a Point and Normal
Find the plane through
\[ (1,2,-1) \]with normal vector
\[ \begin{pmatrix} 2\\ -3\\ 4 \end{pmatrix}. \]Use point-normal form:
\[ 2(x-1)-3(y-2)+4(z+1)=0. \]Expanding gives
\[ 2x-3y+4z+8=0. \]Equivalently,
\[ \boxed{2x-3y+4z=-8}. \]Testing Whether a Point Lies on a Plane
Consider the plane
\[ 2x-y+3z=7. \]Test the point
\[ (1,1,2). \]Substitute:
\[ 2(1)-1+3(2)=7. \]Since the equation is true, the point lies on the plane.
Parallel Planes
Parallel planes have parallel normal vectors.
For example,
\[ 2x-y+3z=4 \]and
\[ 4x-2y+6z=9 \]have normal vectors
\[ (2,-1,3) \]and
\[ (4,-2,6)=2(2,-1,3). \]Their normals are parallel, so the planes are parallel unless the equations represent the same plane.
Perpendicular Planes
Two planes are perpendicular when their normal vectors are perpendicular.
If
\[ \mathbf{n}_1\cdot\mathbf{n}_2=0, \]then the planes meet at a right angle.
This turns a three-dimensional geometric relationship into a dot-product test.
Vector Form of a Line in Three Dimensions
A line in three-dimensional space can be described by a point
\[ \mathbf{r}_0= \begin{pmatrix} x_0\\ y_0\\ z_0 \end{pmatrix} \]and a nonzero direction vector
\[ \mathbf{v}= \begin{pmatrix} a\\ b\\ c \end{pmatrix}. \]The line is
\[ \boxed{ \mathbf{r} = \mathbf{r}_0+t\mathbf{v} }. \]Parametric Form of a Three-Dimensional Line
The vector equation
\[ \mathbf{r} = \begin{pmatrix} x_0\\ y_0\\ z_0 \end{pmatrix} + t \begin{pmatrix} a\\ b\\ c \end{pmatrix} \]gives
\[ \boxed{x=x_0+at}, \] \[ \boxed{y=y_0+bt}, \] \[ \boxed{z=z_0+ct}. \]This representation is especially natural in higher-dimensional analytic geometry.
A Plane Requires Two Independent Directions
A line is generated by moving in one direction.
A plane requires two nonparallel directions.
If a plane contains a point \(\mathbf{r}_0\) and independent direction vectors \(\mathbf{u}\) and \(\mathbf{v}\), then it can be written
\[ \boxed{ \mathbf{r} = \mathbf{r}_0+s\mathbf{u}+t\mathbf{v} }. \]The parameters \(s\) and \(t\) vary independently.
Different Forms Reveal Different Information
No single equation form is best for every purpose.
- Slope-intercept form reveals slope and \(y\)-intercept.
- Point-slope form is convenient when a point and slope are known.
- Standard form treats vertical and nonvertical lines uniformly.
- Intercept form displays axis intercepts.
- Vector and parametric forms emphasize direction and motion.
- Point-normal form emphasizes perpendicular orientation.
Choosing a useful representation is part of analytic geometry.
Common Error: Using Slope-Intercept Form for a Vertical Line
A vertical line such as
\[ x=4 \]cannot be written as
\[ y=mx+b \]with a real-number slope \(m\).
Its slope is undefined, so its natural equation is simply
\[ \boxed{x=4}. \]Common Error: Forgetting the Sign in Point-Slope Form
If the known point is
\[ (3,-2), \]then point-slope form uses
\[ y-(-2)=m(x-3), \]which simplifies to
\[ y+2=m(x-3). \]The subtraction is part of the formula and must be applied to the actual coordinate value.
Common Error: Changing the Line When Rearranging
Algebraic transformations must preserve the solution set.
From
\[ 2x+3y=12, \]dividing every term by \(3\) gives
\[ \frac23x+y=4. \]Dividing only selected terms would produce a different equation and therefore a different geometric set.
Common Error: Confusing Direction and Normal Vectors
For a line
\[ Ax+By=C, \]the vector
\[ (A,B) \]is normal to the line, not a direction along the line.
A direction vector can instead be taken as
\[ (B,-A). \]Common Error: Assuming Equivalent Equations Must Look Alike
The equations
\[ y=2x+3 \]and
\[ 2x-y=-3 \]look different but describe the same line.
Equivalence depends on having the same solution set, not on having the same visual form.
A Reliable Procedure for Finding a Line Equation
- Identify the information given: points, slope, intercepts, or geometric relationships.
- If needed, compute the slope.
- Choose a form suited to the available information.
- Substitute the known values carefully.
- Simplify only as far as useful.
- Check that the known point or points satisfy the equation.
- Verify that the resulting slope or direction matches the geometry.
A Reliable Procedure for Finding a Plane Equation
- Identify a point on the plane.
- Determine a normal vector.
- Use point-normal form.
- Expand into \(Ax+By+Cz=D\) if desired.
- Substitute known points to verify the equation.
- Check that the normal vector matches the intended orientation.
Practice 1: Slope-Intercept Form
Write the equation of the line with slope
\[ 3 \]and \(y\)-intercept
\[ (0,-5). \]Practice 2: Point-Slope Form
Write an equation for the line with slope
\[ -2 \]through
\[ (4,7). \]Practice 3: Two Points
Find an equation of the line through
\[ (1,2) \]and
\[ (5,14). \]Practice 4: Vertical Line
Write the equation of the vertical line through
\[ (-6,3). \]Practice 5: Standard Form
Convert
\[ 4x+5y=20 \]to slope-intercept form.
Practice 6: Parallel Line
Find the equation of the line through
\[ (2,-1) \]parallel to
\[ y=\frac12x+6. \]Practice 7: Perpendicular Line
Find the equation of the line through
\[ (3,5) \]perpendicular to
\[ y=-4x+1. \]Practice 8: Test a Point
Does
\[ (2,3) \]lie on
\[ 5x-2y=4? \]Practice 9: Plane from Point and Normal
Find the equation of the plane through
\[ (2,-1,3) \]with normal vector
\[ (1,4,-2). \]Practice 10: Identify the Error
A student says the line through \((2,4)\) with slope \(3\) is
\[ y-2=3(x-4). \]Explain the error and give a correct equation.
Practice Answers
Practice 1
\[ \boxed{y=3x-5}. \]Practice 2
\[ \boxed{y-7=-2(x-4)}. \]Equivalent slope-intercept form:
\[ \boxed{y=-2x+15}. \]Practice 3
First find the slope:
\[ m=\frac{14-2}{5-1}=3. \]Then
\[ y-2=3(x-1), \]so
\[ \boxed{y=3x-1}. \]Practice 4
A vertical line has constant \(x\):
\[ \boxed{x=-6}. \]Practice 5
\[ 5y=-4x+20, \] \[ \boxed{y=-\frac45x+4}. \]Practice 6
The parallel slope is
\[ \frac12. \]Using point-slope form:
\[ y+1=\frac12(x-2). \]Therefore,
\[ \boxed{y=\frac12x-2}. \]Practice 7
The perpendicular slope to \(-4\) is
\[ \frac14. \]Therefore,
\[ y-5=\frac14(x-3). \]One correct form is
\[ \boxed{y=\frac14x+\frac{17}{4}}. \]Practice 8
Substitute:
\[ 5(2)-2(3)=10-6=4. \]Yes,
\[ \boxed{(2,3)\text{ lies on the line}}. \]Practice 9
Use point-normal form:
\[ (x-2)+4(y+1)-2(z-3)=0. \]Simplify:
\[ x+4y-2z+8=0. \]Therefore,
\[ \boxed{x+4y-2z=-8}. \]Practice 10
The coordinates were reversed inside point-slope form. For the point \((2,4)\),
\[ x_1=2,\qquad y_1=4. \]Therefore,
\[ \boxed{y-4=3(x-2)}. \]Equivalently,
\[ \boxed{y=3x-2}. \]What to Remember
A line can be represented in several equivalent ways:
\[ \boxed{y=mx+b} \] \[ \boxed{y-y_1=m(x-x_1)} \] \[ \boxed{Ax+By=C}. \]Horizontal and vertical lines have the special forms
\[ \boxed{y=k} \]and
\[ \boxed{x=k}. \]A plane in three dimensions can be represented by
\[ \boxed{Ax+By+Cz=D}. \]- line equations describe sets of points in two dimensions;
- plane equations describe sets of points in three dimensions;
- different algebraic forms can describe the same geometric object;
- direction vectors run along lines;
- normal vectors are perpendicular to lines or planes;
- substitution verifies whether a point belongs to the geometric set.
Where We Go Next
An equation describes all points belonging to a line or plane. When two or more such equations are considered together, their shared solutions reveal where the corresponding geometric objects meet.
The next lesson, Intersections and Solution Sets, develops the connection between solving simultaneous equations and finding geometric intersections.