Analytic Geometry
Connect coordinates with distances, circles, slopes, and equations of lines.
Learning goals
- Use distance, midpoint, and section formulas.
- Classify and graph circle equations.
- Write lines through points, parallel or perpendicular to other lines.
Coordinates and distance
Coordinates let us describe a picture precisely, but the formulas should still have a geometric meaning. Join two points and imagine the horizontal and vertical changes as the legs of a right triangle. Their signs describe direction; their squares contribute to the distance. This picture is the starting point for many of our later arguments.
An ordered pair \((x,y)\) locates a point. The abscissa is \(x\); the ordinate is \(y\). Points on axes do not belong to quadrants. A real number line pairs every point with a real number.
\[d(P,Q)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]Distance between two points
Two markers lie on a coordinate map whose axes use the same unit. Find their straight-line separation.
\(A=(-2,5),\quad B=(4,-3)\)Show worked solution
The horizontal and vertical differences form the legs of a right triangle.
Midpoints and division of a segment
A point partway along a displacement
The midpoint of a line segment is the point on the segment at equal distances from its endpoints. An internal division point lies on the segment and divides its length in a specified positive ratio.
Interpolating each coordinate uses the same fraction in both directions. Check which endpoint starts the ratio, because reversing the ratio usually changes the point.
Distance measured the separation of two points. Now we want a point at a specified place between them. Think in terms of moving a fraction of the displacement from one endpoint to the other. This viewpoint helps you decide which endpoint receives which weight instead of relying on a memorized order.
\[M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\]Midpoint and missing endpoint
A segment starts at A and has midpoint M. Recover its other endpoint B.
\(A=(-6,5),\quad M=(2,-4)\)Show worked solution
Trisect a segment
Place two markers that divide the segment from P to Q into three equal lengths. Find their coordinates.
\(P=(4,5),\quad Q=(10,14)\)Show worked solution
One-third of the displacement is (2,3). Add it once and twice to the first endpoint.
\[(6,8),\qquad(8,11)\]Proving geometric properties
Coordinates are useful because they turn a geometric claim into relationships we can calculate. Choose a convenient position for the figure, but explain why moving it there does not change the property being proved. A convenient picture must still represent the general case.
An isosceles triangle
Show that the triangle with these vertices is isosceles. Identify the equal sides rather than relying on its appearance.
\(P=(-4,2),\ Q=(2,-5),\ R=(5,4)\)Show worked solution
Equal squared lengths give equal lengths, so the triangle is isosceles.
A right triangle
Determine whether this triangle is right-angled. If it is, identify the right-angle vertex.
\(A=(-4,-1),\ B=(0,7),\ C=(6,-6)\)Show worked solution
Since 80+125=205, the converse of the Pythagorean theorem gives a right angle at A.
Circles in center-radius form
We have used distance to compare pairs of points. A circle collects every point at one fixed distance from a center, so the same distance formula can describe an entire figure. We will translate that geometric condition into an equation and read the center and radius back from it.
A circle consists of all points at a fixed positive distance from its center.
\[(x-h)^2+(y-k)^2=r^2\]Center and radius
Write an equation for the circle with the given center and radius.
\(C=(-1,2),\quad r=3\)Show worked solution
A diameter determines a circle
The endpoints of a circle’s diameter are P and Q. Find its center, radius, and equation.
\(P=(2,3),\quad Q=(-6,5)\)Show worked solution
Completing the square
The same circle can appear in a clear center-radius form or in an expanded equation that hides its geometry. Completing the square lets us recover that information. Think of the algebra as a way of making the center and radius visible again.
\[x^2+y^2+Dx+Ey+F=0\]Group the x and y terms, complete both squares, and add the same amounts to the other side.
A genuine circle
Rewrite this equation in center-radius form and identify the center and radius.
\(x^2+y^2+6x-2y-15=0\)Show worked solution
Center (−3,1), radius five.
Degenerate possibilities
Decide whether this equation represents a circle, a single point, or no real points. Justify the classification.
\(x^2+y^2-4x+10y+29=0\)Show worked solution
The graph is the single point (2,−5). A negative right side instead gives an empty real locus.
Slope and special lines
To describe a line, we need to capture its direction as well as a point on it. Slope measures the change in height relative to the horizontal change, connecting a picture to an equation. Vertical lines need separate treatment because their horizontal change is zero.
\[m=\frac{y_2-y_1}{x_2-x_1},\qquad x_2\ne x_1\]Positive slope rises to the right, negative slope falls, horizontal slope is zero, and vertical slope is undefined. Distinct nonvertical parallel lines have equal slopes. For perpendicular nonvertical lines the slopes multiply to negative one.
A vertical line is perpendicular to a horizontal line; the negative-reciprocal rule alone does not cover this case.
Collinearity
Determine whether these three points lie on one straight line. Explain why the equal-slope comparison is valid here.
\(A=(-1,0),\ B=(-9,-2),\ C=(11,3)\)Show worked solution
The lines through A have the same slope, so all three points are collinear.
Equations of lines
A slope describes direction, but it does not tell us where a line sits. Add a point, and we can identify the line. Choose an equation form that matches the information you have rather than converting everything immediately. Remember that vertical lines need separate attention because their slope is undefined.
| Form | Equation |
|---|---|
| Point-slope | \(y-y_1=m(x-x_1)\) |
| Slope-intercept | \(y=mx+b\) |
| General | \(Ax+By+C=0\) |
General form requires A and B not both zero; it includes vertical lines. Two points can be used to find a slope unless their x-coordinates agree.
Point and slope
Find an equation for the line through the given point with the given slope, then locate its x-intercept.
\((3,-6),\quad m=4/5\)Show worked solution
Set \(y=0\) to get the x-intercept \(x=21/2\).
Parallel and perpendicular
Write the equations of both requested lines: one parallel and one perpendicular to the given line.
Through \((2,5)\), relative to \(y=\frac13x-1\).Show worked solution
Bisectors and tangents
Two constructions with different conditions
A perpendicular bisector passes through a segment’s midpoint and is perpendicular to the segment; its points are equally distant from the endpoints. A tangent to a circle meets it at one point and is perpendicular to the radius there.
First identify which object is being bisected or touched, then impose the corresponding conditions.
Distances, slopes, and circle equations now give us several tools for one problem. Start by identifying the geometric condition: equal distances, perpendicularity, or contact at one point. Then choose the algebraic description that expresses that condition most directly.
A perpendicular bisector
Find the line consisting of points equally distant from A and B.
\(A=(2,7),\quad B=(8,1)\)Show worked solution
The midpoint is (5,4) and the segment slope is −1. The perpendicular slope is 1.
\[y-4=x-5\implies y=x-1\]A tangent to a circle
Find the tangent line at the specified point, first checking that the point lies on the circle.
\((x-1)^2+(y+2)^2=25\quad\text{at }(5,1)\)Show worked solution
The point lies on the circle since 16+9=25. The radius from (1,−2) has slope 3/4, so the tangent has slope −4/3.
\[y-1=-\frac43(x-5)\implies4x+3y-23=0\]The shortest distance goes perpendicular to the line
For a line \(Ax+By+C=0\) with \(A^2+B^2>0\), a normal vector is \(\langle A,B\rangle\): it is perpendicular to every displacement along the line, since such a displacement satisfies \(A\Delta x+B\Delta y=0\). From \(P=(x_0,y_0)\), set \(t=(Ax_0+By_0+C)/(A^2+B^2)\) and \(H=(x_0-tA,y_0-tB)\). Substitution shows H is on the line, and PH is normal to it. Every other point Q on the line satisfies \(PQ^2=PH^2+HQ^2\ge PH^2\), proving H is nearest.
Consequently \(d(P,\ell)=|Ax_0+By_0+C|/\sqrt{A^2+B^2}\). Unlike a slope formula, this handles vertical and horizontal lines without separate reciprocal cases.
Will a road touch the circular boundary?
A circular boundary has center \((2,3)\) and radius \(2\). A straight road is modeled by \(3x+4y-28=0\). Does the road miss, touch or cross the boundary? Find its nearest point to the center.
Show worked solution
The signed numerator at the center is \(6+12-28=-10\); the distance is \(|-10|/5=2\). Thus the road is tangent. Here \(t=-10/25=-2/5\), so \(H=(2,3)+(6/5,8/5)=(16/5,23/5)\). Substitution gives \(3(16/5)+4(23/5)=28\). The radius to H is perpendicular to the road.
Compare the perpendicular distance with a positive radius: smaller gives two intersections, equality one, larger none. A horizontal or vertical coordinate difference alone is not the perpendicular distance to a slanted line.
Practice
A median length
In triangle ABC, find the length of the median from A to the midpoint of BC.
\(A=(2,4),\ B=(5,1),\ C=(7,3)\)Show worked solution
Circle, point, or empty?
Classify the real locus by completing both squares.
\(x^2+y^2+8x-6y+30=0\)Show worked solution
Empty real locus: a sum of squares cannot be negative.
A tangent vertical line
Show worked solution
A parameter controls the graph
Determine which values of k produce a circle, a single point, or an empty real locus.
\(x^2+y^2+8x-12y=k+3\)Show worked solution
Circle if k>−55, point if k=−55, empty if k<−55.
Additional practice: mixed exercises
Distance and midpoint practice
Show the coordinate differences or averages used.
- Find the distance from \((-9,3)\) to \((1,7)\).
- Find the distance from \((-9,x)\) to \((-5,-4)\).
- Find the midpoint of \((0,8)\) and \((-5,3)\).
- Find the midpoint of \((-7,-2)\) and \((1+c,c)\).
- Find the distance from \(A=(2,4)\) to the midpoint of \(B=(5,1)\) and \(C=(7,3)\).
- Find all real \(y\) for which \(A=(3,5),B=(-4,y),C=(5,1)\) form an isosceles triangle.
Show worked solution
- \(d=\sqrt{10^2+4^2}=2\sqrt{29}\).
- \(d=\sqrt{4^2+(-4-x)^2}=\sqrt{x^2+8x+32}\).
- Average coordinates: \(M=(-5/2,11/2)\).
- \(M=((c-6)/2,(c-2)/2)\).
- The midpoint is \((6,2)\). Distance \(=\sqrt{4^2+(-2)^2}=2\sqrt5\).
- \(AB^2=49+(y-5)^2\), \(BC^2=81+(y-1)^2\), \(AC^2=20\). Neither \(AB\) nor \(BC\) can equal \(AC\). Equating \(AB^2=BC^2\) gives \(-8y=8\), hence \(y=-1\).
Circles and lines
Show a standard form or a slope calculation.
- Find the circle centered at \((2,3)\) tangent to \(x=6\).
- Classify \(x^2+y^2+8x-6y+30=0\).
- For which \(k\) does \(x^2+y^2+8x-12y=k+3\) give a circle, a point, or an empty set?
- Find the line with x-intercept \(-7\) and y-intercept \(1/2\).
- Find the slope and y-intercept of \(x/2+y/4=1\).
Show worked solution
- The radius is \(|6-2|=4\), so \((x-2)^2+(y-3)^2=16\).
- Completing squares gives \((x+4)^2+(y-3)^2=-5\). There are no real points.
- \((x+4)^2+(y-6)^2=k+55\). Circle: \(k>-55\); point: \(k=-55\); empty: \(k<-55\).
- Through \((-7,0)\) and \((0,1/2)\), the slope is \(1/14\). Thus \(y=x/14+1/2\).
- Multiply by \(4\): \(2x+y=4\). Thus \(y=-2x+4\), slope \(-2\), y-intercept \((0,4)\).
Midpoints and a right-angle proof
Show the calculation and justify the geometric conclusion.
- Find the midpoint of the segment with endpoints \((-2,5)\) and \((4,-3)\).
- Show that \(P(1,1)\), \(Q(2,3)\), and \(R(3,0)\) form a right triangle. Identify the right angle.
Show worked solution
- Average the corresponding coordinates: \(M=\left(\frac{-2+4}{2},\frac{5-3}{2}\right)=(1,1)\).
- The squared lengths are \(PQ^2=1^2+2^2=5\), \(PR^2=2^2+(-1)^2=5\), and \(QR^2=1^2+(-3)^2=10\). Since \(PQ^2+PR^2=QR^2\), the converse of the Pythagorean theorem gives a right angle at \(P\). Equivalently, the slopes of \(PQ\) and \(PR\) are \(2\) and \(-1/2\), whose product is \(-1\).
Summary
- Coordinate differences give distance; coordinate averages give midpoint.
- Complete squares before classifying a circle equation.
- Use perpendicular radius and tangent directions.
Reasoning, connections and deeper practice
Coordinates let us turn a geometric condition into an equation and then check whether every algebraic answer satisfies the original condition. The circle explorer can help visualize how the cases below change.
Multi-step · Classify intersections with a moving line
A circular region has boundary \(x^2+y^2=25\). A horizontal path lies on \(y=t\). For every real t, determine how many points of the path lie on the boundary, and give their coordinates when they exist.
Hint
Substitute the line equation and consider the sign of the resulting square.
Show worked solution
Substitution gives \(x^2=25-t^2\). For \(|t|<5\), the two intersections are \((\sqrt{25-t^2},t)\) and \((-\sqrt{25-t^2},t)\). At \(t=5\) or \(t=-5\), there is one tangent point, \((0,t)\). For \(|t|>5\), there is no real solution. These three cases correspond to a secant, tangent and disjoint line.
Advanced / Honors · Equidistance proves the bisector locus
Let \(A=(-a,0)\) and \(B=(a,0)\), where \(a>0\). Prove that the points equidistant from A and B are exactly the y-axis. Explain why requiring distinct endpoints matters.
Hint
Square the nonnegative distances and expand; prove the converse as well.
Show worked solution
For \(P=(x,y)\), equidistance is equivalent to \((x+a)^2+y^2=(x-a)^2+y^2\), hence \(4ax=0\). Because \(a>0\), this gives \(x=0\). Conversely, every point with \(x=0\) gives equal squared distances and thus equal distances. This is precisely the perpendicular bisector. If \(a=0\), the endpoints coincide and every point of the plane is equidistant; the locus is no longer a line.
