Analytic Geometry

Section 1.2 Analytic Geometry

In what follows, we use the notation $(x_1,y_1)$ to represent a point in the $(x,y)$ coordinate system, also called the $x$-$y$-plane. Previously, we used $(a,b)$ to represent an open interval. Notation often gets reused and abused in mathematics, but thankfully, it is usually clear from the context what we mean.

In the $(x,y)$ coordinate system we normally write the $x$-axis horizontally, with positive numbers to the right of the origin, and the $y$-axis vertically, with positive numbers above the origin. That is, unless stated otherwise, we take 鈥渞ightward鈥� to be the positive $x$-direction and 鈥渦pward鈥� to be the positive $y$-direction. In a purely mathematical situation, we normally choose the same scale for the $x$- and $y$-axes. For example, the line joining the origin to the point $(a,a)$ makes an angle of 45${}^\circ$ with the $x$-axis (and also with the $y$-axis).

In applications, often letters other than $x$ and $y$ are used, and often different scales are chosen in the horizontal and vertical directions.

Example 1.21. Data Plot.

Suppose you drop a coin from a window, and you want to study how its height above the ground changes from second to second. It is natural to let the letter $t$ denote the time (the number of seconds since the object was released) and to let the letter $h$ denote the height. For each $t$ (say, at one-second intervals) you have a corresponding height $h\text{.}$ This information can be tabulated, and then plotted on the $(t,h)$ coordinate plane, as shown in the figure below.

Solution

\begin{equation*} \begin{array}{l|ccccc} \text{seconds} \amp 0 \amp 1 \amp 2 \amp 3 \amp 4\\ \hline \text{metres} \amp 80 \amp 75.1 \amp 60.4 \amp 35.9 \amp 1.6\\ \end{array} \end{equation*}

We use the word 鈥渜uadrant鈥� for each of the four regions into which the plane is divided by the axes: the first quadrant is where points have both coordinates positive, or the 鈥渘ortheast鈥� portion of the plot, and the second, third, and fourth quadrants are counted off counterclockwise, so the second quadrant is the northwest, the third is the southwest, and the fourth is the southeast.

Suppose we have two points $A$ and $B$ in the $x$-$y$-plane. We often want to know the change in $x$-coordinate (also called the 鈥渉orizontal distance鈥� ) in going from $A$ to $B\text{.}$ This is often written $\Delta x\text{,}$ where the meaning of $\Delta$ (a capital delta in the Greek alphabet) is 鈥渃hange in鈥�. Thus, $\Delta x$ can be read as 鈥渃hange in $x$鈥� although it usually is read as 鈥渄elta $x$鈥�. The point is that $\Delta x$ denotes a single number, and should not be interpreted as 鈥渄elta times $x$鈥�. Similarly, the 鈥渃hange in $y$鈥� is written $\Delta y$ and represents the difference between the $y$-coordinates of the two points. It is the vertical distance you have to move in going from $A$ to $B\text{.}$ Using the symbol $\Delta$ in mathematical expression is referred to as delta notion.

Example 1.22. Change in $x$ and $y$.

If $A=(2,1)$ and $B=(3,3)$ the change in $x$ is

\begin{equation*} \Delta x=3-2=1 \end{equation*}

while the change in $y$ is

\begin{equation*} \Delta y= 3-1=2\text{.} \end{equation*}

The general formulas for the change in $x$ and the change in $y$ between a point $(x_1,y_1)$ and a point $(x_2,y_2)$ are:

\begin{equation*} \Delta x=x_2-x_1,\qquad\qquad\Delta y=y_2-y_1\text{.} \end{equation*}

Note that either or both of these might be negative.

Subsection 1.2.1 Lines

If we have two distinct points $A(x_1,y_1)$ and $B(x_2,y_2)\text{,}$ then we can draw one and only one straight line through both points. By the slope of this line we mean the ratio of $\Delta y$ to $\Delta x\text{.}$ The slope is often denoted by the letter $m\text{.}$

Slope Formula.

The slope of the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is:

\begin{equation*} m=\frac{\Delta y}{\Delta x}=\frac{(y_2-y_1)}{(x_2-x_1)}=\frac{\mbox{rise} }{\mbox{run} }\text{.} \end{equation*}

Example 1.23. Slope of a Line Joining Two Points.

The line joining the two points $(1,-2)$ and $(3,5)$ has slope $\ds m=\frac{5-(-2)}{3-1}=\frac{7}{2}\text{.}$

The most familiar form of the equation of a straight line is:

\begin{equation*} y=mx+b\text{.} \end{equation*}

Here $m$ is the slope of the line: if you increase $x$ by 1, the equation tells you that you have to increase $y$ by $m\text{;}$ and if you increase $x$ by $\Delta x\text{,}$ then $y$ increases by $\Delta y=m\Delta x\text{.}$ The number $b$ is called the y-intercept, because it is where the line crosses the $y$-axis (when $x=0$). If you know two points on a line, the formula $m=(y_2-y_1)/(x_2-x_1)$ gives you the slope. Once you know a point and the slope, then the $y$-intercept can be found by substituting the coordinates of either point in the equation: $y_1=mx_1+b\text{,}$ i.e., $b=y_1-mx_1\text{.}$ Alternatively, one can use the 鈥減oint-slope鈥� form of the equation of a straight line: start with $(y-y_1)/(x-x_1)=m$ and then multiply to get

\begin{equation*} (y-y_1)=m(x-x_1)\text{,} \end{equation*}

the point-slope form. Of course, this may be further manipulated to get $y=mx-mx_1+y_1\text{,}$ which is essentially the 鈥淺(y=mx+b\)鈥� form.

It is possible to find the equation of a line between two points directly from the relation $m=(y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1)\text{,}$ which says 鈥渢he slope measured between the point $(x_1,y_1)$ and the point $(x_2,y_2)$ is the same as the slope measured between the point $(x_1,y_1)$ and any other point $(x,y)$ on the line.鈥� For example, if we want to find the equation of the line joining our earlier points $A(2,1)$ and $B(3,3)\text{,}$ we can use this formula:

\begin{equation*} m=\frac{y-1}{x-2}=\frac{3-1}{3-2}=2,\qquad\hbox{so that}\qquad y-1=2(x-2),\qquad\hbox{i.e.,}\qquad y=2x-3\text{.} \end{equation*}

Of course, this is really just the point-slope formula, except that we are not computing $m$ in a separate step. We summarize the three common forms of writing a straight line below:

Slope-Intercept Form of a Straight Line.

An equation of a line with slope $m$ and $y$-intercept $b$ is:

\begin{equation*} y=mx+b\text{.} \end{equation*}

Point-Slope Form of a Straight Line.

An equation of a line passing through $(x_1,y_1)$ and having slope $m$ is:

\begin{equation*} y-y_1=m(x-x_1)\text{.} \end{equation*}

General Form of a Straight Line.

Any line can be written in the form

\begin{equation*} Ax+By+C=0\text{,} \end{equation*}

where $A,B,C$ are constants and $A,B$ are not both $0\text{.}$

The slope $m$ of a line in the form $y=mx+b$ tells us the direction in which the line is pointing. If $m$ is positive, the line goes into the 1st quadrant as you go from left to right. If $m$ is large and positive, it has a steep incline, while if $m$ is small and positive, then the line has a small angle of inclination. If $m$ is negative, the line goes into the 4th quadrant as you go from left to right. If $m$ is a large negative number (large in absolute value), then the line points steeply downward. If $m$ is negative but small in absolute value, then it points only a little downward. If $m=0\text{,}$ then the line is horizontal and its equation is simply $y=b\text{.}$

All of these possibilities are illustrated below.

Interactive Demonstration. Investigate the slope of the line through the point $(1,1)$ by moving the red point. Here, green indicates a positive slope, red indicates a negative slope, blue indicates a horizontal line and purple indicates a vertical line.

There is one type of line that cannot be written in the form $y=mx+b\text{,}$ namely, vertical lines. A vertical line has an equation of the form $x=a\text{.}$ Sometimes one says that a vertical line has an 鈥渋nfinite鈥� slope.

It is often useful to find the $x$-intercept of a line $y=mx+b\text{.}$ This is the $x$-value when $y=0\text{.}$ Setting $mx+b$ equal to 0 and solving for $x$ gives: $x=-b/m\text{.}$

Example 1.24. Finding $x$-intercepts.

To find $x$-intercept(s) of the line $y=2x-3$ we set $y=0$ and solve for $x\text{:}$

\begin{equation*} 0 = 2x-3\qquad\to\qquad x = \frac{3}{2}\text{.} \end{equation*}

Thus, the line has an $x$-intercept of $3/2\text{.}$

It is often necessary to know if two lines are parallel or perpendicular. Let $m_1$ and $m_2$ be the slopes of the nonvertical lines $L_1$ and $L_2\text{.}$ Then:

$L_1$ and $L_2$ are parallel if and only if $m_1=m_2\text{.}$
$L_1$ and $L_2$ are perpendicular if and only if $\ds{m_2=\frac{-1}{m_1}}\text{.}$ (Equivalently, $m_1 = \frac{-1}{m_2}$).

In the case of perpendicular lines, we say their slopes are negative reciprocals. Below is a visual representation of a pair of parallel lines and a pair of perpendicular lines.

Example 1.25. Equation of a Line.

For each part below, find an equation of a line satisfying the requirements:

Through the two points $(0,3)$ and $(-2,4)\text{.}$
With slope $7$ and through point $(1,-2)\text{.}$
With slope $2$ and $y$-intercept $4\text{.}$
With $x$-intercept $8$ and $y$-intercept $-3\text{.}$
Through point $(5,3)$ and parallel to the line $2x+4y+2=0\text{.}$
With $y$-intercept $4$ and perpendicular to the line $\ds{y=-\frac{2}{3}x+3}\text{.}$

Solution

We use the slope formula on $(x_1,y_1)=(0,3)$ and $(x_2,y_2)=(-2,4)$ to find m:

\begin{equation*} m=\frac{(4)-(3)}{(-2)-(0)}=\frac{1}{-2}=-\frac{1}{2}\text{.} \end{equation*}

Now using the point-slope formula we get an equation to be:

\begin{equation*} y-3=-\frac{1}{2}\left(x-0\right)\to y=-\frac{1}{2}x+3\text{.} \end{equation*}
Using the point-slope formula with $m=7$ and $(x_1,y_1)=(1,-2)$ gives:

\begin{equation*} y-(-2)=7(x-1)\to y = 7x-9\text{.} \end{equation*}
Using the slope-intercept formula with $m=2$ and $b=4$ we get $y=2x+4\text{.}$
Note that the intercepts give us two points: $(x_1,y_1)=(8,0)$ and $(x_2,y_2)=(0,-3)\text{.}$ Now follow the steps in part (a):

\begin{equation*} m=\frac{-3-0}{0-8}=\frac{3}{8}\text{.} \end{equation*}

Using the point-slope formula we get an equation to be:

\begin{equation*} y-(-3)=\frac{3}{8}\left(x-0\right)\to y=\frac{3}{8}x-3\text{.} \end{equation*}
The line $2x+4y+2=0$ can be written as:

\begin{equation*} 4y = -2x - 2 \to y=-\frac{1}{2}x-\frac{1}{2}\text{.} \end{equation*}

This line has slope $-1/2\text{.}$ Since our line is parallel to it, we have $m=-1/2\text{.}$ Now we have a point $(x_1,y_1)=(5,3)$ and slope $m=-1/2\text{,}$ thus, the point-slope formula gives:

\begin{equation*} y-3=-\frac{1}{2}\left(x-5\right)\text{.} \end{equation*}
The line $\ds{y=-\frac{2}{3}x+3}$ has slope $m=-2/3\text{.}$ Since our line is perpendicular to it, the slope of our line is the negative reciprocal, hence, $m=3/2\text{.}$ Now we have $b=4$ and $m=3/2\text{,}$ thus by the slope-intercept formula, an equation of the line is

\begin{equation*} y=\frac{3}{2}x+4\text{.} \end{equation*}

Example 1.26. Parallel and Perpendicular Lines.

Are the two lines $7x+2y+3=0$ and $6x-4y+2=0$ perpendicular? Are they parallel? If they are not parallel, what is their point of intersection?

Solution

The first line is:

\begin{equation*} 7x+2y+3=0\to 2y=-7x-3\to y=-\frac{7}{2}x-\frac{3}{2}\text{.} \end{equation*}

It has slope $m_1=-7/2\text{.}$ The second line is:

\begin{equation*} 6x-4y+2=0\to -4y=-6x-2\to y=\frac{3}{2}x+\frac{1}{2}\text{.} \end{equation*}

It has slope $m_2=3/2\text{.}$ Since $m_1\cdot m_2\neq -1$ (they are not negative reciprocals), the lines are not perpendicular. Since $m_1\neq m_2$ the lines are not parallel.

We find points of intersection by setting $y$-values to be the same and solving. In particular, we have

\begin{equation*} -\frac{7}{2}x-\frac{3}{2}=\frac{3}{2}x+\frac{1}{2}\text{.} \end{equation*}

Solving for $x$ gives $x=-2/5\text{.}$ Then substituting this into either equation gives $y=-1/10\text{.}$ Therefore, the lines intersect at the point $(-2/5,-1/10)\text{.}$

Subsection 1.2.2 Distance between Two Points and Midpoints

露

Given two points $(x_1,y_1)$ and $(x_2,y_2)\text{,}$ recall that their horizontal distance from one another is $\Delta x=x_2-x_1$ and their vertical distance from one another is $\Delta y=y_2-y_1\text{.}$ Actually, the word 鈥渄istance鈥� normally denotes 鈥減ositive distance鈥�. $\Delta x$ and $\Delta y$ are signed distances, but this is clear from context. The (positive) distance from one point to the other is the length of the hypotenuse of a right triangle with legs $|\Delta x|$ and $|\Delta y|\text{,}$ as shown in 贵颈驳耻谤别听1.1. The Pythagorean Theorem states that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides:

Figure 1.1. Distance between two points (here, $\Delta x$ and $\Delta y$ are positive).

Distance Formula.

The distance between points $(x_1,y_1)$ and $(x_2,y_2)$ is

\begin{equation*} \hbox{distance}=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+ (y_2-y_1)^2}\text{.} \end{equation*}

Example 1.27. Distance Between Two Points.

The distance, $d\text{,}$ between points $A(2,1)$ and $B(3,3)$ is

\begin{equation*} d=\sqrt{(3-2)^2+(3-1)^2}=\sqrt{5}\text{.} \end{equation*}

As a special case of the distance formula, suppose we want to know the distance of a point $(x,y)$ to the origin. According to the distance formula, this is

\begin{equation*} \sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}\text{.} \end{equation*}

A point $(x,y)$ is at a distance $r$ from the origin if and only if $\sqrt{x^2+y^2}=r\text{,}$ or, if we square both sides: $x^2+y^2=r^2\text{.}$ As we will see, this is the equation of the circle of radius, $r\text{,}$ centered at the origin.

Furthermore, given two points we can determine the midpoint of the line segment joining the two points.

Midpoint Formula.

The midpoint of the line segment joining two points $(x_1,y_1)$ and $(x_2,y_2)$ is the point with coordinates:

\begin{equation*} \hbox{midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\text{.} \end{equation*}

Example 1.28. Midpoint of a Line Segment.

Find the midpoint of the line segment joining the given points: $(1,0)$ and $(5,-2)\text{.}$

Solution

Using the midpoint formula on $(x_1,y_1)=(1,0)$ and $(x_2,y_2)=(5,-2)$ we get:

\begin{equation*} \left(\frac{(1)+(5)}{2},\frac{(0)+(-2)}{2}\right)=(3,-1)\text{.} \end{equation*}

Thus, the midpoint of the line segment occurs at $(3,-1)\text{.}$

Subsection 1.2.3 Conics

露

In this section we review equations of parabolas, circles, ellipses and hyperbolas. We will give the equations of various conics in standard form along with a sketch. A useful mnemonic is the following.

Mnemonic.

In each conic formula presented, the terms 鈥榎(x-h\)鈥� and 鈥榎(y-k\)鈥� will always appear. The point $(h,k)$ will alway represent either the centre or vertex of the particular conic.

Note that $h$ or $k$ (or both) may equal $0\text{.}$

Vertical Parabola: The equation of a vertical parabola is:

\begin{equation*} y-k=a(x-h)^2 \end{equation*}

Interactive Demonstration. Use the slider to investigate the effect of $a $ on the shape of the parabola and drag the point $(h,k)$ to reposition the parabola.

$(h,k)$ is the vertex of the parabola.
$a$ is the vertical stretch factor.
If $a>0\text{,}$ the parabola opens upward.
If $a\lt 0\text{,}$ the parabola opens downward.

Horizontal Parabola: The equation of a horizontal parabola is:

\begin{equation*} x-h=a(y-k)^2 \end{equation*}

$(h,k)$ is the vertex of the parabola.
$a$ is the horizontal stretch factor.
If $a>0\text{,}$ the parabola opens right.
If $a\lt 0\text{,}$ the parabola opens left.

Circle: The equation of a circle is:

\begin{equation*} (x-h)^2+(y-k)^2=r^2 \end{equation*}

Interactive Demonstration. Use the slider to investigate the effect of $r $ on the shape of the circle and drag the point $(h,k)$ to reposition the circle.

$(h,k)$ is the centre of the circle.
$r$ is the radius of the circle.

Ellipse: The equation of an ellipse is:

\begin{equation*} \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \end{equation*}

$(h,k)$ is the centre of the ellipse.
$a$ is the horizontal distance from the centre to the edge of the ellipse.
$b$ is the vertical distance from the centre to the edge of the ellipse.

Horizontal Hyperbola: The equation of a horizontal hyperbola is:

\begin{equation*} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \end{equation*}

$(h,k)$ is the centre of the hyperbola.
$a,b$ are the reference box values. The box has a centre of $(h,k)\text{.}$
$a$ is the horizontal distance from the centre to the edge of the box.
$b$ is the vertical distance from the centre to the edge of the box.

Given the equation of a horizontal hyperbola, one may sketch it by first placing a dot at the point $(h,k)\text{.}$ Then draw a box around $(h,k)$ with horizontal distance $a$ and vertical distance $b$ to the edge of the box. Then draw dotted lines (called the asymptotes of the hyperbola) through the corners of the box. Finally, sketch the hyperbola in a horizontal direction as illustrated below.

Vertical Hyperbola: The equation of a vertical hyperbola is:

\begin{equation*} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=-1 \end{equation*}

$(h,k)$ is the centre of the hyperbola.
$a,b$ are the reference box values. The box has a centre of $(h,k)\text{.}$
$a$ is the horizontal distance from the centre to the edge of the box.
$b$ is the vertical distance from the centre to the edge of the box.

Given the equation of a vertical hyperbola, one may sketch it by following the same steps as with a horizontal hyperbola, but sketching the hyperbola going in a vertical direction.

Determining the Type of Conic.

An equation of the form

\begin{equation*} Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \end{equation*}

gives rise to a graph that can be generated by performing a conic section (parabolas, circles, ellipses, hyperbolas). Note that the $Bxy$ term involves conic rotation. The $Dx\text{,}$ $Ex\text{,}$ and $F$ terms affect the vertex and centre. For simplicity, we omit the $Bxy$ term. To determine the type of graph we focus our analysis on the values of $A$ and $C\text{.}$

If $A=C\text{,}$ the graph is a circle.
If $AC>0$ (and $A\neq C$), the graph is an ellipse.
If $AC=0\text{,}$ the graph is a parabola.
If $AC\lt 0\text{,}$ the graph is a hyperbola.

Example 1.29. Center and Radius of a Circle.

Find the centre and radius of the circle $y^2 + x^2 - 12x + 8y + 43 = 0\text{.}$

Solution

We need to complete the square twice, once for the $x$ terms and once for the $y$ terms. We'll do both at the same time. First let's collect the terms with $x$ together, the terms with $y$ together, and move the number to the other side.

\begin{equation*} (x^2-12x)+(y^2+8y)=-43 \end{equation*}

We add $36$ to both sides for the $x$ term ($-12\to \frac{-12}{2}=-6\to (-6)^2=36$), and $16$ to both sides for the $y$ term ($8\to \frac{8}{2}=4\to (4)^2=16$):

\begin{equation*} (x^2-12x+36)+(y^2+8y+16)=-43+36+16 \end{equation*}

Factoring gives:

\begin{equation*} (x-6)^2+(y+4)^2=3^2\text{.} \end{equation*}

Therefore, the centre of the circle is $(6,-4)$ and the radius is $3\text{.}$

Example 1.30. Type of Conic.

What type of conic is $4x^2-y^2-8x+8=0\text{?}$ Put it in standard form.

Solution

Here we have $A=4$ and $C=-1\text{.}$ Since $AC\lt 0\text{,}$ the conic is a hyperbola. Let us complete the square for the $x$ and $y$ terms. First let's collect the terms with $x$ together, the terms with $y$ together, and move the number to the other side.

\begin{equation*} (4x^2-8x)-y^2=-8 \end{equation*}

Now we factor out $4$ from the $x$ terms.

\begin{equation*} 4(x^2-2x)-y^2=-8 \end{equation*}

Notice that we don't need to complete the square for the $y$ terms (it is already completed!). To complete the square for the $x$ terms we add $1$ since $-2\to\frac{-2}{2}=-1\to(-1)^2=1\text{,}$ taking into consideration that the $a$ value is $4\text{:}$

\begin{equation*} 4(x^2-2x+1)-y^2=-8+4\cdot1 \end{equation*}

Factoring gives:

\begin{equation*} 4(x-1)^2-y^2=-4 \end{equation*}

A hyperbola in standard form has $\pm1$ on the right side and a positive $x^2$ on the left side, thus, we must divide by $4\text{:}$

\begin{equation*} (x-1)^2-\frac{y^2}{4}=-1 \end{equation*}

Now we can see that the equation represents a vertical hyperbola with centre $(1,0)$ (and with $a$ value $\sqrt{1}=1\text{,}$ and $b$ value $\sqrt{4}=2$).

Example 1.31. Equation of Parabola.

Find an equation of the parabola with vertex $(1,-1)$ that passes through the points $(-4, 24)$ and $(7, 35)\text{.}$

Solution

We first need to determine if it is a vertical parabola or horizontal parabola. See below for a sketch of the three points $(1,-1)\text{,}$ $(-4, 24)$ and $(7, 35)$ in the $x$-$y$-plane.

Note that the vertex is $(1,-1)\text{.}$ Given the location of the vertex, the parabola cannot open downwards. It also cannot open left or right (because the vertex is between the other two points - if it were to open to the right, every other point would need to be to the right of the vertex; if it were to open to the left, every other point would need to be to the left of the vertex). Therefore, the parabola must open upwards and it is a vertical parabola. It has an equation of

\begin{equation*} y-k=a(x-h)^2\text{.} \end{equation*}

As the vertex is $(h,k)=(1,-1)$ we have:

\begin{equation*} y-(-1)=a(x-1)^2 \end{equation*}

To determine $a\text{,}$ we substitute one of the points into the equation and solve. Let us substitute the point $(x,y)=(-4,24)$ into the equation:

\begin{equation*} 24-(-1)=a(-4-1)^2\to 25=25a\to a=1\text{.} \end{equation*}

Therefore, the equation of the parabola is:

\begin{equation*} y+1=(x-1)^2\text{.} \end{equation*}

Note that if we substituted $(7,35)$ into the equation instead, we would also get $a=1\text{.}$

Exercises for Section聽1.2.

Exercise 1.2.1.

Find the equation of the line in the form $y=mx+b\text{:}$

through $(1,1)$ and $(-5, -3)$
Answer
$(2/3)x+(1/3)$
Solution

Given two points through which the line passes, $\left(1,1\right)$ and $\left(-5,-3\right)\text{,}$ we can set up a system of two equations:

\begin{equation*} \begin{split} 1 \amp = m + b \\ -3 \amp = -5 m + b. \end{split} \end{equation*}

To solve for $m\text{,}$ we can subtract the two equations to find that

\begin{equation*} 1+3 = (1+5)m \implies m=\frac{2}{3}\text{.} \end{equation*}

Then, we can use either equation to solve for $b\text{:}$

\begin{equation*} 1 = \frac{2}{3} + b \implies b = \frac{1}{3}\text{.} \end{equation*}

Therefore, the equation of the line is given by $y = \frac{2}{3} x + {1}{3}\text{.}$
through $(-1,2)$ with slope $-2$
Answer
$y=-2x$
Solution

Now we are given a point, $\left(-1, 2\right)$ and a slope, $m = -2\text{.}$ The equation of any line with this slope is given by

\begin{equation*} y = -2 x + b\text{.} \end{equation*}

We can now solve for $b$ by inputting the given point. This gives $b = 0\text{.}$ Putting together, we find that this line is $y = -2x\text{.}$
through $(-1,1)$ and $(5, -3)$
Answer
$y=(-2/3)x+(1/3)$
Solution

Again we are given two points, $\left(-1, 1\right)$ and $\left(5, -3\right)\text{.}$ An alternative way of solving this would be to use the slope-point formula. By definition, the slope of a line is given by

\begin{equation*} m = \dfrac{y_{1}-y_{2}}{x_{1}-x_{2}}\text{.} \end{equation*}

Substituting in the given points, we get that $m = - \frac{2}{3}\text{,}$ and the equation of the line becomes

\begin{equation*} y = - \frac{2}{3} x + b\text{.} \end{equation*}

Now $b$ can be determined using either of the two given points. We find that $b = \frac{1}{3}\text{.}$ Our final answer is thus $y = - \frac{2}{3} x + \frac{1}{3}\text{.}$
through $(2,5)$ and parallel to the line $3x+9y+6=0$
Answer
$y=-x/3+17/3$
Solution
Recall that two lines are parallel if they have the same slope. We first manipulate the line $3x + 9y + 6 = 0$ into the usual form, $y = -\frac{1}{3}x - \frac{2}{3}\text{.}$ So our line must have slope $-\frac{1}{3}\text{.}$ To find the $y$-intercept, we subsitute the point which the line is required to pass through, $\left(2,5\right)\text{.}$ We find that the desired line is given by $y = -\frac{1}{3}x + \frac{17}{3}\text{.}$
with $x$-intercept 5 and perpendicular to the line $y=2x+4$
Answer
$y=-1/2x+5/2$
Solution

If we want our line to be perpendicular to the line with slope $2\text{,}$ it must have a slope of $-\frac{1}{2}\text{.}$ Further, if the $x$-intercept is at the point $\left(5, 0\right)\text{,}$ then $b$ must satisfy

\begin{equation*} 0 = -\frac{5}{2} + b\text{.} \end{equation*}

Hence, our line is given by $y = -\frac{1}{2} x + \frac{5}{2}\text{.}$

Exercise 1.2.2.

Change the following equations to the form $y=mx+b\text{,}$ graph the line, and find the $y$-intercept and $x$-intercept.

$y-2x=2$
Answer
$y=2x+2\text{,}$ $2\text{,}$ $-1$
$x+y=6$
Answer
$y=-x+6\text{,}$ $6\text{,}$ $6$
$x=2y-1$
Answer
$y=x/2+1/2\text{,}$ $1/2\text{,}$ $-1$
$3=2y$
Answer
$y=3/2\text{,}$ $3/2\text{,}$ none
$2x+3y+6=0$
Answer
$y=(-2/3)x-2\text{,}$ $-2\text{,}$ $-3$

Exercise 1.2.3.

Determine whether the lines $3x+6y=7$ and $2x+4y=5$ are parallel.

Solution

Yes, the lines are parallel as they have the same slope of $-1/2$

Exercise 1.2.4.

Suppose a triangle in the $x$-$y$-plane has vertices $(-1,0)\text{,}$ $(1,0)$ and $(0,2)\text{.}$ Find the equations of the three lines that lie along the sides of the triangle in $y=mx+b$ form.

Answer

$y=0\text{,}$ $y=-2x+2\text{,}$ $y=2x+2$

Exercise 1.2.5.

Let $x$ stand for temperature in degrees Celsius (centigrade), and let $y$ stand for temperature in degrees Fahrenheit. A temperature of $0^\circ$C corresponds to $32^\circ$F, and a temperature of $100^\circ$C corresponds to $212^\circ$F. Find the equation of the line that relates temperature Fahrenheit $y$ to temperature Celsius $x$ in the form $y=mx+b\text{.}$ Graph the line, and find the point at which this line intersects $y=x\text{.}$ What is the practical meaning of this point?

Answer

$y=(9/5)x+32\text{,}$ $(-40,-40)$

Exercise 1.2.6.

A car rental firm has the following charges for a certain type of car: $25 per day with 100 free miles included, $0.15 per mile for more than 100 miles. Suppose you want to rent a car for one day, and you know you'll use it for more than 100 miles. What is the equation relating the cost $y$ to the number of miles $x$ that you drive the car?

Answer

$y=0.15x+10$

Exercise 1.2.7.

A photocopy store advertises the following prices: 5c聽per copy for the first 20 copies, 4c聽per copy for the 21st through 100th copy, and 3c聽per copy after the 100th copy. Let $x$ be the number of copies, and let $y$ be the total cost of photocopying. Graph the cost as $x$ goes from 0 to 200 copies. Find the equation in the form $y=mx+b$ that tells you the cost of making $x$ copies when $x$ is more than 100.

Solution

Let $x$ be the number of copies and $y$ be the cost of photocopying in cents. Then we know that the graph of the cost function $y$ (for $0\leq x \leq 200$) will be given by three line segments (for simplicity, we graph $y$ as a continuous function):

The first line segment has slope 5, the second has slope 4, and the third has slope 3. It remains to work out the values on the $y$-axis. We therefore consider the three cases:

For $x \leq 20\text{:}$

\begin{equation*} y = 5\left(\text{ number of copies } \right) = 5x\text{.} \end{equation*}

In particular, $y(20) = 100\text{.}$
For $21 \leq x \leq 100\text{:}$

\begin{equation*} \begin{split} y \amp= \left(\text{ cost of first 20 copies } \right) + 4\left(\text{ number of additional copies } \right)\\ \amp = 100 + 4(x-20) = 4x + 20. \end{split} \end{equation*}

In particular, $y(100) = 420\text{.}$
For $x \geq 101\text{:}$

\begin{equation*} \begin{split} y \amp= \left(\text{ cost of first 100 copies } \right) + 3\left(\text{ number of additional copies } \right)\\ \amp = 420 + 3(x-100) = 3x + 120 \end{split} \end{equation*}

In particular, $y(200) = 720\text{.}$

Therefore, the cost $y$ of making $x$ copies for $x \geq 101$ is $y = 3x + 120$ cents. Our final graph is shown below.

Exercise 1.2.8.

Market research tells you that if you set the price of an item at $1.50, you will be able to sell 5000 items; and for every 10 cents you lower the price below $1.50 you will be able to sell another 1000 items. Let $x$ be the number of items you can sell, and let $P$ be the price of an item.

Express $P$ linearly in terms of $x\text{,}$ in other words, express $P$ in the form $P=mx+b\text{.}$
Answer
$P=-0.0001x+2$
Solution

We wish to express the price $P$ of an item as a function of the number of items which will be sold, $x\text{.}$ Let $P$ have units of dollars. We are told that we will sell $5000$ items when the price is $ $1.50\text{.}$ This means that

\begin{equation*} P(5000) = 1.50\text{.} \end{equation*}

The question next tells us that for every $ $0.10$ the price is lowered from $ $1.50\text{,}$ the number of items which will be sold goes up by $1000$ units. We can express this mathematically as,

\begin{equation*} P(5000 + 1000n) = 1.50 - 0.10n \ \ \ \ n = 0,1,2,...,15 \end{equation*}

In order to find an equation of a line, we need (any) two data points. So, for example, let's take

\begin{equation*} P(5000) = 1.50 \ \ \text{ and } \ \ P(6000) = 1.40\text{.} \end{equation*}

We find the slope to be

\begin{equation*} m = \dfrac{1.50 - 1.40}{5000 - 6000} = -0.0001 \end{equation*}

and the $y$-intercept to be

\begin{equation*} b = (0.0001)(5000) + 1.50 = 2 \end{equation*}

Therefore, we find $P(x) = -0.0001x + 2\text{.}$ $x$ and $P$ must both be positive, so the domain of $P(x)$ is $0 \leq x \leq 20 000$ units.
Express $x$ linearly in terms of $P\text{.}$
Answer
$x=-10000P+20000$
Solution

We can manipulate the relationship above to find $x(P)\text{.}$

\begin{equation*} \begin{split} \amp P = -0.0001x + 2 \\ \amp \implies 10000P = -x + 20000 \implies x(P) = -10000P + 20000.\end{split} \end{equation*}

The domain of $x(P)$ is $0 \leq P \leq 2$ dollars.

Exercise 1.2.9.

An instructor gives a 100-point final exam, and decides that a score 90 or above will be a grade of 4.0, a score of 40 or below will be a grade of 0.0, and between 40 and 90 the grading will be linear. Let $x$ be the exam score, and let $y$ be the corresponding grade. Find a formula of the form $y=mx+b$ which applies to scores $x$ between 40 and 90.

Answer

$(2/25)x-(16/5)$

Exercise 1.2.10.

Find the distance between the pairs of points:

$(-1,1)$ and $(1,1)\text{.}$
Answer
$2$
$(5,3)$ and $(-7,-2)\text{.}$
Answer
$\sqrt{2}$
$(1,1)$ and the origin.
Answer
$\sqrt{2}$

Exercise 1.2.11.

Find the midpoint of the line segment joining the point $(20,-10)$ to the origin.

Exercise 1.2.12.

Find the equation of the circle of radius 3 centered at:

$(0,0)$
Answer
$x^2+y^2=9$
$(5,6)$
Answer
$(x-5)^2+(y-6)^2=9$
$(-5,-6)$
Answer
$(x+5)^2+(y+6)^2=9$
$(0,3)$
Answer
$x^2+(y-3)^2=9$
$(0,-3)$
Answer
$x^2+(y+3)^2=9$
$(3,0)$
Answer
$(x-3)^2+y)^2=9$

Exercise 1.2.13.

For each pair of points $A(x_1,y_1)$ and $B(x_2,y_2)$ find an equation of the circle with center at $A$ that goes through $B\text{.}$

$A(2,0)\text{,}$ $B(4,3)$
Answer
$\left(x-2\right)^{2} + y^{2} = 13$
Solution

$A = \left(2,0\right)$ and $B = \left(4,3\right)\text{.}$ The radius of the circle will just be the distance between the two points,

\begin{equation*} r = d(A,B) = \sqrt{(4-2)^{2}+3^{2}} = \sqrt{13} \end{equation*}

The equation of the circle is thus given by $\left(x-2\right)^{2} + y^{2} = 13\text{.}$
$A(-2,3)\text{,}$ $B(4,3)$
Answer
$\left(x+2\right)^{2} + \left(y-3\right)^{2} = 36$
Solution
$A = \left(-2,3\right)$ and $B = \left(4,3\right)\text{.}$ We find that $r = \sqrt{6^{2}} = 6\text{.}$ We therefore find the equation of the circle to be $\left(x+2\right)^{2} + \left(y-3\right)^{2} = 36\text{.}$

Exercise 1.2.14.

Determine the type of conic and sketch it.

$x^2+y^2+10y=0$
Answer
circle
$9x^2-90x+y^2+81=0$
Answer
ellipse
$6x+y^2-8y=0$
Answer
horizontal parabola

Exercise 1.2.15.

Find the standard equation of the circle passing through $(-2,1)$ and tangent to the line $3x-2y =6$ at the point $(4,3)\text{.}$ Sketch. Hint

The line through the center of the circle and the point of tangency is perpendicular to the tangent line.

Answer

$(x+2/7)^2+(y-41/7)^2=1300/49$

丁香园AV

Section 1.2 Analytic Geometry

Example 1.21. Data Plot.

Example 1.22. Change in \(x\) and \(y\).

Subsection 1.2.1 Lines

Slope Formula.

Example 1.23. Slope of a Line Joining Two Points.

Slope-Intercept Form of a Straight Line.

Point-Slope Form of a Straight Line.

General Form of a Straight Line.

Example 1.24. Finding \(x\)-intercepts.

Example 1.25. Equation of a Line.

Example 1.26. Parallel and Perpendicular Lines.

Subsection 1.2.2 Distance between Two Points and Midpoints

Distance Formula.

Example 1.27. Distance Between Two Points.

Midpoint Formula.

Example 1.28. Midpoint of a Line Segment.

Subsection 1.2.3 Conics

Mnemonic.

Determining the Type of Conic.

Example 1.29. Center and Radius of a Circle.

Example 1.30. Type of Conic.

Example 1.31. Equation of Parabola.

Exercises for Section聽1.2.

Exercise 1.2.1.

Exercise 1.2.2.

Exercise 1.2.3.

Exercise 1.2.4.

Exercise 1.2.5.

Exercise 1.2.6.

Exercise 1.2.7.

Exercise 1.2.8.

Exercise 1.2.9.

Exercise 1.2.10.

Exercise 1.2.11.

Exercise 1.2.12.

Exercise 1.2.13.

Exercise 1.2.14.

Exercise 1.2.15.