Precalculus

Analytic Geometry

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Straight Line in Plane

General Equation of a Straight Line

The general equation of a straight line in a plane is given by

\[Ax + By + C = 0\]

where x, y are the coordinates of a point on the line, A, B, C are real numbers. Note that this is the standard form of the equation of a straight line, where the coefficients A and B are not both equal to zero:

\[A^2 + B^2 \ne 0.\]

This equation is also known as the canonical form of a straight line equation.

There are also other forms of the equation of a straight line, which are also commonly used. We will review them below.

Normal Vector to a Straight Line

Let the line be defined by the general equation

\[Ax + By + C = 0.\]

Then the vector n(A, B) whose coordinates are equal to the coefficients A, B is the normal vector to the straight line.

Normal vector to a straight line
Figure 1.

Slope-Intercept Form

The slope-intercept form of the equation of a straight line is given by

\[y = kx + b\]

where k is the slope of the line and b is the y-intercept (the point where the line intersects the y-axis).

This form of the equation is particularly useful for graphing a line, as the slope gives you an indication of how steep the line is and the y-intercept tells you where the line crosses the y-axis.

Slope-intercept form of the equation of a straight line
Figure 2.

Slope of a Straight Line

The slope of a straight line is determined by the formula

\[k = \tan \alpha = \frac{y_2 - y_1}{x_2 - x_1}\]

where A(x1, y1), B(x2, y2) are the coordinates of two points of the line.

The slope of a straight line
Figure 3.

Point-Slope Form

The point-slope form of the equation of a straight line is

\[y - y_0 = k\left({x - x_0}\right)\]

where P(x0, y0) is a known point on the line and k is the slope of the line.

Equation of a straight line given a point and a slope (point-slope form)
Figure 4.

To use the point-slope form, you simply substitute the values of x0, y0, and k into the equation, and then simplify it to get the equation of the line in slope-intercept form or standard form, if desired.

Two-Point Form

The two-point form of the equation of a straight line is given by

\[\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\]

where (x1, y1), (x2, y2) are two known points on the line.

This equation represents a line that passes through the points A(x1, y1) and B(x2, y2). You can use this equation to find the equation of a line given two points.

Two-point form of the equation of a straight line
Figure 5.

The two-point form equation can also be represented in matrix form:

\[\left| {\begin{array}{*{20}{c}} x&y&1\\ {x_1}&{y_1}&1\\ {x_2}&{y_2}&1 \end{array}} \right| = 0.\]

where (x, y) is an arbitrary point on the line.

To verify that a point P(x, y) lies on the line, we can substitute the x- and y-values into the determinant and check whether the result is zero. If it is zero, then P lies on the line; otherwise, it does not.

Two-Intercept Form

The two-intercept form of the equation of a straight line is written as

\[\frac{x}{a} + \frac{y}{b} = 1\]

where a and b are the x-intercept and y-intercept of the line, respectively. The x-intercept is the point at which the line intersects the x-axis, and its y-coordinate is zero. The y-intercept is the point at which the line intersects the y-axis, and its x-coordinate is zero.

Two-intercept form of the equation of a straight line
Figure 6.

Normal Form

The normal form of a straight line equation is a way to represent a line in terms of its distance from the origin (0,0). The equation of a straight line in normal form is

\[x\cos\beta + y\sin\beta - p = 0\]

where β is the angle between the line and the positive x-axis (measured counterclockwise), and p is the perpendicular distance from the origin to the line.

Normal form of a straight line equation
Figure 7.

Point-Direction Form

The point-direction form of a straight line equation is a way to represent a line in a plane (or in space) using a point on the line and the direction of the line. In the xy-plane, the equation has the form

\[\frac{x-x_0}{X} = \frac{y-y_0}{Y}\]

where the vector s(X, Y) is directed along the straight line, and the point P(x0, y0) lies on the line.

Point-Direction Form of the equation of a straight line in plane
Figure 8.

Vertical Line Equation

A vertical line is a line that runs straight up and down, parallel to the y-axis, and has an undefined slope. The equation of a vertical line can be written as

\[x = a\]

where a is the x-coordinate of any point on the line.

This equation states that all points on the line have the same x-coordinate, which is equal to a. Therefore, we can say that the line is vertical and parallel to the y-axis, and it intersects the x-axis at the point (a,0).

Horizontal Line Equation

A horizontal line is a line that runs straight from left to right, parallel to the x-axis, and has a slope of zero. The equation of a horizontal line is given by

\[y = b\]

where b is the y-coordinate of any point on the line.

Vector Equation of a Straight Line

The vector equation of a straight line in a plane is another way to represent a line using vectors. It is based on the fact that the displacement vector between any two points on the line is parallel to the direction vector of the line.

Vector equation of a straight line
Figure 9.

Suppose we have a point A with position vector a on the line in the plane, and a direction vector b that is parallel to the line. Using vector notation, we can write the equation of the line as

\[\mathbf{r} = \mathbf{a} + t\mathbf{b}\]

The vector r is the position vector directed from the origin to any point X on this line. The number t is a parameter that varies from -∞ to ∞.

The vector equation is useful because it allows us to easily perform vector operations on the line, such as finding the intersection point of two lines or the angle between two lines.

Equation of a Straight Line in Parametric Form

To write the equation of a straight line in the plane in parametric form, we need to express the x- and y-coordinates of any point on the line in terms of a parameter, usually denoted by t.

Parametric equations of a straight line in plane
Figure 10.

Suppose we have a point A(a1, a2) on the line, and a direction vector b(b1, b2) that is parallel to the line. Then we can write the parametric equations of the line as

\[ \begin{cases} x = {a_1} + t{b_1} \\ y = {a_2} + t{b_2} \end{cases} \]

where (x, y) are the coordinates of an arbitrary point of the line.

The above parametric equations indicate that as t varies, the point (x, y) moves along the line in the direction of b. The parameter t can take any real value, so the line extends infinitely in both directions.

Solved Problems

Example 1.

Find an equation of the line through the points \(A\left({-1,0}\right)\) and \(B\left({0,2}\right).\)

Solution.

Remember that the two-point form of the equation of a straight line is

\[\frac{y - y_A}{y_B - y_A} = \frac{x - x_A}{x_B - x_A}.\]

Substituting the coordinates of points A and B, we obtain

\[\frac{y - 0}{2 - 0} = \frac{x - \left({-1}\right)}{0 - \left({-1}\right)}, \;\Rightarrow \frac{y}{2} = \frac{x + 1}{1}.\]

We can write this equation in the slope-intercept form:

\[y = 2\left({x + 1}\right) = 2x + 2.\]

Example 2.

Find the slope of the line given in parametric form \[ \begin{cases} x = 1 + 2t \\ y = 2 + t \end{cases}. \]

Solution.

We express \(t\) from the second equation and substitute it into the first equation:

\[ \begin{cases} x = 1 + 2t \\ y = 2 + t \end{cases}, \;\Rightarrow \begin{cases} x = 1 + 2t \\ t = y - 2 \end{cases}, \;\Rightarrow x = 1 + 2\left({y - 2}\right), \;\Rightarrow x = 1 + 2y - 4, \;\Rightarrow 2y = x + 3, \]

or

\[y = \frac{1}{2}x + \frac{3}{2}.\]

So, the slope of the straight line is equal to \(\frac{1}{2}.\)

Example 3.

Write an equation of the line \({3x - 4y + 4 = 0}\) in parametric and in two-intercept form.

Solution.

To preserve integer coefficients it is convenient to set \({x = 4t}.\) Then we have

\[3\cdot 4t - 4y + 4 = 0, \;\Rightarrow 4y = 12t + 4, \;\Rightarrow y = 3t + 1.\]

So, the equation of the line in parametric form is given by

\[ \begin{cases} x = 4t \\ y = 3t + 1 \end{cases}. \]

The two-intercept form of a line equation looks like

\[\frac{x}{a} + \frac{y}{b} = 1.\]

We can derive it from the general equation:

\[3x - 4y + 4 = 0, \;\Rightarrow 3x - 4y = -4.\]

Dividing both sides by \(-4\) gives us the equation of the line in two-intercept form:

\[\frac{3x}{-4} + \frac{\cancel{-4}y}{\cancel{-4}} = 1, \;\Rightarrow \frac{x}{-\frac{4}{3}} + \frac{y}{1} = 1.\]

The straight line interecepts the \(x-\)axis at \(x = -\frac{4}{3}\) and the \(y-\)axis at \(y = 1\).

The straight line 3x-4y+4=0
Figure 11.

Example 4.

Find an equation of the straight line through \(A\left({3,5}\right)\) and parallel to the line \({x - 2y + 5 = 0}.\)

Solution.

If the required line is parallel to the line \({x - 2y + 5 = 0},\) then our line must have an equation of the form

\[{x - 2y + C = 0,}\]

where \(C\) is an unknown number.

The point \(A\left({3,5}\right)\) lies on the line. Therefore

\[{3 - 2\cdot5 + C = 0, \;\Rightarrow C = 7.}\]

So, the desired equation is

\[x - 2y + 7 = 0.\]

Example 5.

Find an equation of the line through the point \(A\left({-4,2}\right)\) that makes the angle \({\alpha = 135^\circ}\) with the positive \(x-\)axis.

Solution.

The slope of the line is

\[{k = \tan \alpha = \tan 135^\circ = -1.}\]

Using the point-slope form of the line equation

\[y - y_A = k\left({x - x_A}\right)\]

and substituting the coordinates of point \(A\left({x_A,y_A}\right)\) and the slope \(k\) in the equation, we get

\[y - 2 = -1\left({x - \left({-4}\right)}\right),\;\Rightarrow y - 2 = -x - 4, \;\Rightarrow y = -x - 2.\]

Example 6.

Find an equation of the straight line that passes through the point \(A\left({2,3}\right)\) and forms with the coordinate axes a triangle of area \(12\) square units.

Solution.

A line through A(2,3) that forms a triangle of area 6 units.
Figure 12.

We write the equation of the line in two-intercept form:

\[\frac{x}{a} + \frac{y}{b} = 1.\]

The area of a right triangle is one half the product of two legs. Therefore, we can write

\[\frac{ab}{2} = 12, \;\Rightarrow ab = 24.\]

Substituting the coordinates of point \(A\left({2,3}\right)\) in the equation yields:

\[\frac{2}{a} + \frac{3}{b} = 1.\]

Thus we have a system of two equations from which we can find \(a\) and \(b:\)

\[ \begin{cases} ab = 24 \\ \frac{2}{a} + \frac{3}{b} = 1 \end{cases}. \]

Let's express \(a\) in terms of \(b\) from the first equation and plug in \(a\) into the second equation:

\[ \begin{cases} a = \frac{24}{b} \\ \frac{2}{a} + \frac{3}{b} = 1 \end{cases},\;\Rightarrow \frac{2}{\frac{24}{b}} + \frac{3}{b} = 1, \;\Rightarrow \frac{b}{12} + \frac{3}{b} = 1.\]

Multiply both sides of the equation by \(12b:\)

\[b^2 + 36 = 12b, \;\Rightarrow b^2 - 12b + 36 = 0, \;\Rightarrow \left({b - 6}\right)^2 = 0, \;\Rightarrow b = 6.\]

Then

\[a = \frac{24}{b} = \frac{24}{6} = 4.\]

So the equation of our line has the form

\[\frac{x}{4} + \frac{y}{6} = 1.\]

In general form, it looks like

\[3x + 2y - 12 = 0.\]