The distance formula and its derivation from the Pythagorean theorem, classifying a triangle by comparing side lengths, the midpoint formula, writing the equation of a perpendicular bisector, and locating a circle's centre and diameter from three points on it.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 9 — Quadratic Relations and Conic Sections
Apply the Distance and Midpoint Formulas
Objectives
Five outcomes. Two formulas, and everything else in Chapter 9 will lean on them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 614-617 — the lesson these objectives are drawn from
Warm-up
You know the Pythagorean theorem and you can find the slope between two points.
Discussion prompt
How far is it from the point where x is 1 and y is 1 to the point where x is 6 and y is 5? Try drawing a right triangle rather than reaching for a formula.
Hint: How far across, and how far up?
Answer:
\[ \text{across } 6-1 = 5, \quad \text{up } 5-1 = 4 \]
\[ d^2 = 5^2 + 4^2 = 41 \;\Longrightarrow\; d = \sqrt{41} \approx 6.4 \]
That is the whole distance formula. Chapter 9 writes it once with letters instead of numbers, and then uses it to define every conic section — a circle is just the set of points a fixed distance from a centre.
Concept
The distance between two points is the square root of the sum of the squared coordinate differences. The midpoint is found by averaging the coordinates instead. One measures a gap; the other finds a middle.
distance formula — The distance between two points is the square root of the squared difference of the x-coordinates plus the squared difference of the y-coordinates.
\[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}; \qquad M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \]
Neither is a new idea. The first is the Pythagorean theorem in coordinates, and the second is the ordinary mean applied to each coordinate separately.
Figure (svg): Two columns comparing the distance formula with the midpoint formula
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 614-615
Section
Section 1
Concept
Join two points and complete a right triangle whose legs run horizontally and vertically. The legs measure the differences in the coordinates, so the Pythagorean theorem gives the distance directly.
\[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]
The squaring means the order of subtraction never matters. Whichever point is called the first, the differences come out opposite in sign and identical once squared.
Figure (svg): A right triangle drawn between two points, showing where the distance formula comes from
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 614-614 — The Distance Formula
Picture it
A right triangle with its legs measured in coordinates.
Figure (svg): A right triangle drawn between two points, showing where the distance formula comes from
The horizontal leg is the difference in x and the vertical leg the difference in y. Squaring, adding and rooting is the theorem you already know.
Worked example
Example 1, a multiple-choice item.
\[ \text{Find the distance between } (-2, 6) \text{ and } (3, -1). \]
Label the points
Why: Either may be first; the squaring removes the difference.
\[ (x 1, y 1) = (-2, 6) \]
Subtract the coordinates
Why: Three minus negative 2, and negative 1 minus 6.
\[ 5\text{ and } -7 \]
Square and add
Why: Twenty-five plus 49.
\[ 74 \]
Take the square root
Why: Seventy-four has no square factor.
\[ \sqrt{74} \]
Figure (svg): A right triangle drawn between two points, showing where the distance formula comes from
\[ d = \sqrt{74} \approx 8.60 \]
Verify: swap the two points
Why: Taking the other point first gives negative 5 and 7, which square to 25 and 49 — the same total. The order genuinely does not matter, and a quick sanity check confirms the size: the points are 5 across and 7 up, so the direct distance must be a little more than 7 and less than 12.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 614-614
Fill the middle
Example 1.
Fill in the blanks
d = \sqrt49 = \sqrt___}}
Why: Negative 7 squared is 49, and 25 plus 49 is 74. The squaring is what makes the negative difference harmless.
Worked example
Guided Practice 1.
\[ \text{Find the distance between } (3, -3) \text{ and } (-1, 5). \]
Subtract the coordinates
Why: Negative 1 minus 3, and 5 minus negative 3.
\[ -4\text{ and } 8 \]
Square and add
Why: Sixteen plus 64.
\[ 80 \]
Simplify the radical
Why: Eighty is 16 times 5.
\[ 4 \sqrt{5} \]
Approximate
Why: Four times about 2.236.
\[ \text{about } 8.94 \]
Figure (svg): The solution to Worked example one more distance shown as a ladder of expressions, one row per algebraic move
\[ d = \sqrt{80} = 4\sqrt{5} \approx 8.94 \]
Verify: check the simplification
Why: Four root 5 squared is 16 times 5, which is 80 — the original radicand. Simplifying radicals is Lesson 6.2's skill and it matters here because two lengths that look different, like root 80 and 4 root 5, are the same number and comparing them unsimplified would suggest otherwise.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 615-615
Trap
\[ (-2,6) \text{ and } (3,-1) \]
Add the coordinates and then square
Why: The two formulas of this lesson get mixed together.
\[ d = \sqrt{(3+(-2))^2+(-1+6)^2} = \sqrt{26} \quad \text{(wrong)} \]
Adding gives the midpoint's ingredients, not the distance's. A distance measures a gap, and gaps come from subtraction.
\[ d = \sqrt{(3-(-2))^2+(-1-6)^2} = \sqrt{74} \]
Subtract for distance; average for midpoint
Why: The Pythagorean legs are differences, not sums.
\[ \text{legs } 5 \text{ and } -7 \;\Longrightarrow\; d = \sqrt{74} \approx 8.6 \]
Choice B in Example 1 is exactly this error, printed as a distractor. Sketching the two points settles it immediately.
Prediction
Commit before reasoning.
Predict first
You compute a distance and then redo it with the points in the other order. What happens?
Correct: You get the same answer, because the differences are squared.
\[ (a-b)^2 = (b-a)^2 \text{ for all } a, b \]
Why: Swapping the points reverses the sign of each difference, and squaring removes the sign. So the formula is symmetric in the two points, exactly as a distance ought to be — it makes no sense for the gap from A to B to differ from the gap from B to A. This is why no rule is needed about which point to call the first.
Matching
Subtract, square, add, root.
Match the pairs
Why: Three of the four simplified, and only the first left an unsimplifiable radical since 74 is 2 times 37 with no square factor. Simplifying makes lengths comparable at a glance, which the next idea depends on.
Sorting
One subtracts, the other averages.
Sort into buckets
Sort each step by which formula it belongs to.
The midpoint formula has no square root at all, which is a quick way to remember which is which: an answer with a radical in it cannot be a midpoint.
Section
Section 2
Concept
To classify a triangle given its vertices, compute all three side lengths with the distance formula. All three equal means equilateral, exactly two equal means isosceles, and none equal means scalene.
\[ AB = 3\sqrt{2}, \; BC = \sqrt{29}, \; AC = \sqrt{29} \]
Comparing the radicands is enough. If two expressions under the roots agree, the lengths agree, and no decimals are needed at all.
Figure (svg): A triangle with vertices plotted and its three side lengths computed
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 614-614 — Classify a triangle using the distance formula
Picture it
Example 2: the triangle with vertices at three plotted points.
Figure (svg): A triangle with vertices plotted and its three side lengths computed
Both BC and AC come out as the square root of 29, from 25 plus 4 and 4 plus 25. The triangle is isosceles.
Worked example
Example 2.
\[ \text{Classify the triangle with vertices } A(4,6), \; B(7,3), \; C(2,1). \]
Find AB
Why: Three across and negative 3 up, so 9 plus 9.
\[ \sqrt{18} = 3 \sqrt{2} \]
Find BC
Why: Negative 5 across and negative 2 up, so 25 plus 4.
\[ \sqrt{29} \]
Find AC
Why: Negative 2 across and negative 5 up, so 4 plus 25.
\[ \sqrt{29} \]
Compare
Why: Two of the three lengths agree exactly.
Figure (svg): A triangle with vertices plotted and its three side lengths computed
\[ BC = AC = \sqrt{29} \;\Longrightarrow\; \text{isosceles} \]
Verify: compare without decimals
Why: The radicands 29 and 29 are identical, so the lengths are identical — no rounding needed. Root 18 is about 4.24 and root 29 about 5.39, so the third side really is shorter, which the sketch confirms. Comparing exact values avoids a rounding coincidence being mistaken for equality.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 614-614
Sorting
Compare the radicands.
Sort into buckets
Sort each set of three squared lengths.
Comparing squared lengths is exactly as good as comparing lengths, since a square root preserves order — and it avoids every rounding question.
Worked example
Guided Practice 2.
\[ \text{Classify the triangle with vertices } R(-1,3), \; S(5,2), \; T(3,6). \]
Find RS
Why: Six across and negative 1 up, so 36 plus 1.
\[ \sqrt{37} \]
Find ST
Why: Negative 2 across and 4 up, so 4 plus 16.
\[ \sqrt{20} = 2 \sqrt{5} \]
Find RT
Why: Four across and 3 up, so 16 plus 9.
\[ \sqrt{25} = 5 \]
Compare
Why: Thirty-seven, 20 and 25 are all different.
Figure (svg): The solution to Worked example a second triangle shown as a ladder of expressions, one row per algebraic move
\[ \sqrt{37}, \; 2\sqrt{5}, \; 5 \;\Longrightarrow\; \text{scalene} \]
Verify: check the third side's tidy value
Why: Sixteen plus 9 is 25, a perfect square, so that side is exactly 5. Its neighbours are about 6.08 and 4.47, both clearly different. When a radicand turns out to be a perfect square the length is a whole number, which is worth noticing rather than leaving as a root.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 615-615
Error analysis
A student classifies a triangle whose side lengths are the square roots of 29, 29 and 18.
Annotate
On: \( \sqrt{29} \approx 5.4, \; \sqrt{29} \approx 5.4, \; \sqrt{18} \approx 4.2 \;\Longrightarrow\; \text{scalene} \)
Keep the exact values until the comparison is done. Rounding can make equal lengths look different and, worse, unequal lengths look the same.
Fill the middle
Example 2.
Fill in the blanks
BC = \sqrt4 = \sqrt___}}
Why: Negative 2 squared is 4, giving a radicand of 29. The third side gives 4 plus 25, the same 29 with the two contributions swapped — which is how the match arises.
Two truths and a lie
All three are about classifying with the distance formula.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The square roots of 29 and 29.1 both round to 5.4, yet the lengths differ. Rounding can hide a real difference, and it can also make an exact equality look approximate. Working with the radicands settles the question without any of that.
Prediction
Commit before reasoning.
Predict first
The triangle in Example 2 is isosceles. Is it also a right triangle?
Correct: Test it: 18 plus 29 is 47, not 29, so no.
\[ 18 + 29 = 47 \neq 29 \;\Longrightarrow\; \text{not right-angled} \]
Why: The Pythagorean test uses the squared lengths directly, which the distance formula already produces — the two shorter sides give 18 plus 29, or 47, and the longest squared is 29, so they disagree. Some triangles are both right and isosceles, such as one with squared sides 8, 8 and 16, but this is not one of them. Having the squared lengths in hand makes the extra test almost free.
Section
Section 3
Concept
The midpoint of a segment has, for each coordinate, the mean of the corresponding coordinates of the endpoints. It is the one point equidistant from both ends and lying on the segment.
\[ M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \]
No square root appears anywhere, so a midpoint never involves a radical. An answer with one in it is a distance that has been mislabelled.
Figure (svg): A segment with its midpoint marked, each coordinate the mean of the endpoints
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 615-615 — The Midpoint Formula
Picture it
Example 3: the midpoint of a segment between two plotted points.
Figure (svg): A segment with its midpoint marked, each coordinate the mean of the endpoints
The x-coordinates average to negative 3 and the y-coordinates to three and a half. Fractional coordinates are perfectly ordinary here.
Worked example
Example 3.
\[ \text{Find the midpoint of the segment joining } (-5, 1) \text{ and } (-1, 6). \]
Average the x-coordinates
Why: Negative 5 plus negative 1, halved.
\[ -3 \]
Average the y-coordinates
Why: One plus 6, halved.
\[ \frac{7}{2} \]
Write the point
Why: Both coordinates together.
\[ (-3, 3.5) \]
Notice the form
Why: A midpoint is a point, not a length.
Figure (svg): A segment with its midpoint marked, each coordinate the mean of the endpoints
\[ M = \left(-3, \tfrac{7}{2}\right) \]
Verify: check both distances
Why: From negative 5 comma 1 to the midpoint is 2 across and 2.5 up; from the midpoint to negative 1 comma 6 is also 2 across and 2.5 up. Equal displacements mean equal distances, and the midpoint lies on the segment because both displacements point the same way.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 615-615
Fill the middle
Example 3.
Fill in the blanks
M = \left(\frac7___, \frac______\right) = \left(-3, \frac___}___\right)
Why: One plus 6 is 7, and halving gives three and a half. Fractional coordinates are entirely normal for a midpoint, since a mean of two integers is a half-integer whenever they differ in parity.
Worked example
Guided Practice 3 to 5.
\[ \text{Find the midpoints of } (0,0)-(-4,12), \; (-2,1)-(4,-7), \; (3,8)-(-5,-10). \]
First pair
Why: Zero plus negative 4 halved, and 0 plus 12 halved.
\[ (-2, 6) \]
Second pair
Why: Negative 2 plus 4 halved, and 1 plus negative 7 halved.
\[ (1, -3) \]
Third pair
Why: Three plus negative 5 halved, and 8 plus negative 10 halved.
\[ (-1, -1) \]
Note the pattern
Why: Every coordinate is a plain average, sign included.
Figure (svg): The solution to Worked example three more midpoints shown as a ladder of expressions, one row per algebraic move
\[ (-2,6), \; (1,-3), \; (-1,-1) \]
Verify: check one by doubling back
Why: For the third, doubling the midpoint's coordinates gives negative 2 and negative 2, and subtracting the first endpoint 3 comma 8 gives negative 5 comma negative 10 — the second endpoint. That reverse check works generally: twice the midpoint minus one endpoint is the other endpoint.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 616-616
Trap
\[ (-5,1) \text{ and } (-1,6) \]
Halve the differences
Why: The distance formula's subtraction is carried over.
\[ M = \left(\tfrac{-1-(-5)}{2}, \tfrac{6-1}{2}\right) = (2, 2.5) \quad \text{(wrong)} \]
The point 2 comma 2.5 is nowhere near the segment, which runs from x equal to negative 5 to x equal to negative 1.
\[ M = \left(\tfrac{-5+(-1)}{2}, \tfrac{1+6}{2}\right) = \left(-3, \tfrac{7}{2}\right) \]
Add and halve; that is what a mean is
Why: The midpoint's coordinates must lie between the endpoints' coordinates.
\[ -5 < -3 < -1 \quad \checkmark \]
A one-second sanity check catches this every time: each midpoint coordinate has to sit between the two it came from.
Matching
Average each coordinate.
Match the pairs
Why: In every row each midpoint coordinate lies between the two it came from, which is the fastest check available. Only the first produced a fraction, because only there did two coordinates differ in parity.
Comparison
Fill the blanks. Same inputs, opposite operations.
Comparison matrix
| Question | Distance | Midpoint |
|---|---|---|
| Combine coordinates by | subtracting | adding |
| Then | square, add, take the root | divide each sum by 2 |
| The answer is | one number, a length | a point, two numbers |
| Radicals possible? | yes, usually | never |
The last row is the quickest diagnostic of all: if a midpoint answer contains a square root, the wrong formula was used.
Prediction
Commit before reasoning.
Predict first
A segment has midpoint (1, 5) and one endpoint (-3, 4). Where is the other?
Correct: (5, 6).
\[ (x_2, y_2) = (2 \cdot 1 - (-3), \; 2 \cdot 5 - 4) = (5, 6) \]
Why: Doubling the midpoint gives 2 and 10, and subtracting the known endpoint gives 5 and 6. The reasoning is that the midpoint is the average, so twice the average minus one value is the other — the same algebra as recovering a missing test score from a known mean. Checking: the midpoint of negative 3 comma 4 and 5 comma 6 is indeed 1 comma 5, which is Example 4's segment.
Section
Section 4
Concept
The perpendicular bisector of a segment passes through its midpoint and is perpendicular to it. Finding it means computing the midpoint, computing the slope, taking the negative reciprocal, and writing point-slope form.
\[ y - 5 = -4(x-1) \;\Longrightarrow\; y = -4x+9 \]
Both formulas of this lesson appear, along with Lesson 2.3's slope and Lesson 2.4's point-slope form. Nothing here is new; the assembly is.
Figure (svg): A segment with the perpendicular bisector drawn through its midpoint
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 615-615 — Find a perpendicular bisector
Picture it
Example 4, with the segment and its bisector drawn.
Figure (svg): A segment with the perpendicular bisector drawn through its midpoint
The midpoint gave the point and the negative reciprocal gave the slope. Point-slope form then produced the equation in one line.
Worked example
Example 4.
\[ \text{Write the perpendicular bisector of the segment joining } A(-3,4) \text{ and } B(5,6). \]
Find the midpoint
Why: Negative 3 plus 5 halved, and 4 plus 6 halved.
\[ (1, 5) \]
Find the slope of the segment
Why: Six minus 4, over 5 minus negative 3.
\[ \frac{2}{8} = \frac{1}{4} \]
Take the negative reciprocal
Why: The negative of the reciprocal of one quarter.
\[ -4 \]
Write point-slope form
Why: Through the midpoint with that slope.
\[ y = -4 x + 9 \]
Figure (svg): A segment with the perpendicular bisector drawn through its midpoint
\[ y = -4x+9 \]
Verify: check both conditions
Why: At x equal to 1 the line gives y equal to 5, so it passes through the midpoint. And one quarter times negative 4 is negative 1, so the two lines are perpendicular. A perpendicular bisector must satisfy both, and checking only one of them is the usual way to miss an error.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 615-615
Ranking
Writing a perpendicular bisector.
Put in order
Why: The first two are independent and could be done in either order, but both are needed before step four: point-slope form requires a point and a slope, and here the point comes from one formula and the slope from two more steps.
Worked example
Guided Practice 3 to 5.
\[ \text{Write perpendicular bisectors for } (0,0)-(-4,12), \; (-2,1)-(4,-7), \; (3,8)-(-5,-10). \]
First: midpoint and slope
Why: Midpoint negative 2 comma 6; slope 12 over negative 4.
\[ m = -3,\text{ perp } \frac{1}{3} \]
First: write the line
Why: Through negative 2 comma 6 with slope one third.
\[ y = \frac{x}{3} + \frac{20}{3} \]
Second: midpoint and slope
Why: Midpoint 1 comma negative 3; slope negative 8 over 6.
\[ m = -\frac{4}{3},\text{ perp } \frac{3}{4} \]
Third: midpoint and slope
Why: Midpoint negative 1 comma negative 1; slope 9 over 4.
\[ \text{perp } -\frac{4}{9} \]
Figure (svg): The solution to Worked example three more bisectors shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{x}{3}+\tfrac{20}{3}; \; y = \tfrac{3}{4}x-\tfrac{15}{4}; \; y = -\tfrac{4}{9}x-\tfrac{13}{9} \]
Verify: check the second at its midpoint
Why: At x equal to 1 the line gives three quarters minus fifteen quarters, which is negative twelve quarters, or negative 3 — the midpoint's y-coordinate. And negative four thirds times three quarters is negative 1, confirming perpendicularity. Both checks take seconds and between them catch every kind of slip in this idea.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 616-616
Error analysis
A student writes the perpendicular bisector of a segment of slope one quarter.
Annotate
On: \( m = \tfrac{1}{4} \;\Longrightarrow\; m_{\perp} = 4, \; y - 5 = 4(x-1) \)
The product test is the check: multiply the two slopes and insist on negative 1. It catches a missing sign instantly.
Fill the middle
Example 4.
Fill in the blanks
m = \tfrac-4___ \;\Longrightarrow\; m____ = ___
Why: The reciprocal of one quarter is 4, and changing the sign gives negative 4. The product one quarter times negative 4 is negative 1, which is the definition of perpendicular from Lesson 2.2.
Sorting
A perpendicular bisector needs a point and a direction.
Sort into buckets
Sort each ingredient by what it supplies.
A line needs exactly one of each, which is why point-slope form is the natural template for this problem.
Prediction
Commit before reasoning.
Predict first
What do all the points on the perpendicular bisector of a segment have in common?
Correct: Each is equidistant from the two endpoints.
\[ \text{on the bisector} \;\Longleftrightarrow\; \text{equidistant from } A \text{ and } B \]
Why: That is the defining property, and it is what makes the next idea work: the centre of a circle is equidistant from every point on it, so it must lie on the perpendicular bisector of any chord. Take the point 1 comma 9 on Example 4's bisector — its distance to negative 3 comma 4 is the square root of 16 plus 25, and to 5 comma 6 it is the square root of 16 plus 9 plus... in fact both give the square root of 41, as the property promises.
Section
Section 5
Concept
The perpendicular bisector of any chord of a circle passes through the centre. So given three points on a circle, write the bisectors of two chords, solve the system, and the intersection is the centre.
\[ y = -x+34 \text{ and } y = 3x+110 \;\Longrightarrow\; C(-19, 53) \]
Once the centre is known, the distance formula gives the radius, using any one of the three points. That the three give the same answer is a check on the whole calculation.
Figure (svg): Three points on a circular crater with two perpendicular bisectors meeting at the centre
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 616-616 — Finding a circle's center
Picture it
Example 5: an asteroid crater outlined by three points.
Figure (svg): Three points on a circular crater with two perpendicular bisectors meeting at the centre
The two bisectors meet at negative 19 comma 53. The distance from there to the origin is about 56.3 miles, so the crater is about 113 miles across.
Worked example
Example 5.
\[ \text{Two chord bisectors are } y=-x+34 \text{ and } y=3x+110. \text{ Find the centre and the diameter.} \]
Solve the system
Why: Substitute the first into the second.
\[ 3 x + 110 = -x + 34 \]
Find x
Why: Four x equals negative 76.
\[ x = -19 \]
Find y
Why: Substitute back into the first equation.
\[ y = 53 \]
Find the radius and double it
Why: Distance from the centre to the origin.
\[ \sqrt{3170}\text{ about } 56.3 \]
Figure (svg): Three points on a circular crater with two perpendicular bisectors meeting at the centre
\[ C(-19,53), \quad d \approx 112.6 \text{ mi} \]
Verify: check the radius against a second point
Why: The centre must be the same distance from all three of the given points, so computing that distance twice from different points is a genuine test of the whole calculation rather than of the last step. If two of them disagreed, the error would be in the bisectors rather than in the arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 616-616
Fill the middle
Example 5.
Fill in the blanks
3x+110 = -x+34 \;\Longrightarrow\; 4x = -76
Why: Thirty-four minus 110 is negative 76, so x is negative 19. Substituting into either equation then gives y equal to 53.
Worked example
Guided Practice 6.
\[ \text{The points } (0,0), (6,-2), (16,8) \text{ lie on a circle. Find its diameter.} \]
Bisector of the first chord
Why: Midpoint 3 comma negative 1; slope negative one third, so perpendicular slope 3.
\[ y = 3 x - 10 \]
Bisector of the second chord
Why: Midpoint 8 comma 4; slope one half, so perpendicular slope negative 2.
\[ y = -2 x + 20 \]
Solve the system
Why: Three x minus 10 equals negative 2x plus 20.
\[ x = 6, y = 8 \]
Find the radius
Why: From 6 comma 8 to the origin.
\[ \sqrt{36 + 64} = 10 \]
Figure (svg): The solution to Worked example a circle through three points shown as a ladder of expressions, one row per algebraic move
\[ C(6,8), \; r = 10, \; d = 20 \]
Verify: check the radius from all three points
Why: From 6 comma 8 to 6 comma negative 2 is 10 straight down. To 16 comma 8 is 10 straight across. To the origin is the square root of 36 plus 64, which is 10. All three agree exactly, which confirms both bisectors and the intersection at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 616-616
Trap
\[ \text{chord from } (0,0) \text{ to } (16,8) \]
Take its midpoint as the circle's centre
Why: The midpoint is central to the chord, so it is taken as central to the circle.
\[ C = (8,4) \quad \text{(wrong)} \]
From 8 comma 4 the distance to the origin is about 8.9 but to 6 comma negative 2 it is about 6.3. A centre must be equidistant from every point on the circle.
\[ C = (6,8), \text{ where two bisectors meet} \]
Intersect two perpendicular bisectors
Why: The centre lies on the bisector of EVERY chord, so two of them locate it.
\[ \text{distances } 10, 10, 10 \quad \checkmark \]
A chord's midpoint lies on the right line but at the wrong place along it. Only the intersection of two such lines pins the centre down.
Ranking
Finding a circle through three points.
Put in order
Why: Only two chords are needed, since two lines meet in one point. Computing the radius from all three given points in step five costs almost nothing and checks every earlier step at once.
Two truths and a lie
All three are about circles through three points.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false in general — although it is true for one special chord, the diameter, whose midpoint IS the centre. For any other chord the midpoint lies on the right line but nearer the chord than the centre. In Guided Practice 6 the chord from the origin to 16 comma 8 has midpoint 8 comma 4, while the centre is 6 comma 8.
Prediction
Commit before reasoning.
Predict first
Three points that are not in a straight line always lie on exactly one circle. Why exactly one?
Correct: Because two perpendicular bisectors meet at exactly one point, fixing the centre and hence the radius.
\[ \text{3 points, 3 constants} \;\Longrightarrow\; \text{one circle} \]
Why: The centre must lie on all three bisectors, and two non-parallel lines determine a single point — so there is at most one candidate. That point is genuinely equidistant from all three, so there is at least one. Two lines are parallel only when the chords are parallel, which happens exactly when the three points are collinear, and then no circle passes through them. The counting also matches: a circle's equation has three constants, so three points determine it, exactly as three points determined a parabola in Lesson 4.10.
Comparison
Fill the blanks. All five ideas use two or three of these.
Comparison matrix
| Tool | Formula | Gives |
|---|---|---|
| Distance | sqrt of the squared differences | a length, one number |
| Midpoint | the mean of each coordinate | a point, two numbers |
| Slope | rise over run | a direction |
| Perpendicular slope | the negative reciprocal | the direction at a right angle |
The perpendicular bisector uses all four, and the circle problem uses all four twice and then solves a system on top.
Pattern
One routine each, and the last one uses all of them.
Keep exact values until the last step. Rounding early can make equal lengths look different and unequal ones look the same.
OpenStax Algebra and Trigonometry 2e, §2.1 The Rectangular Coordinate Systems and Graphs §2.1
Check
Distance. Subtract, then square.
Check your understanding
What is the distance between (-2, 6) and (3, -1)?
Answer: A
Why: The differences are 5 and -7, giving 25 plus 49.
Check
Midpoint. Average, do not subtract.
Check your understanding
What is the midpoint of the segment joining (-5, 1) and (-1, 6)?
Answer: A
Why: Average each coordinate: -6 over 2 and 7 over 2.
Check
Perpendicular bisector. Two ingredients.
Check your understanding
What is the perpendicular bisector of the segment joining (-3, 4) and (5, 6)?
Answer: A
Why: Midpoint (1, 5) and perpendicular slope -4 give y - 5 = -4(x - 1).
Real world
A phone reaches three towers and the delay of each signal gives its distance. Suppose the phone is 5 miles from a tower at the origin, 5 miles from one at 6 comma minus 2, and 5 miles from one at 16 comma 8, in a plane where each unit is a mile.
Discussion prompt
Explain why those three distances cannot all be 5, and describe how the real system locates the phone.
Hint: Use Guided Practice 6.
Answer:
The only point equidistant from those three towers is 6 comma 8, and its distance to each is 10 miles, not 5. So no phone can be 5 miles from all three — the data would be inconsistent.
\[ \text{centre } (6,8), \quad r = \sqrt{36+64} = 10 \]
A real positioning system runs the argument the other way. Each tower's delay puts the phone somewhere on a circle around that tower; two circles meet at two points, and the third tower picks which. That is why three towers are the minimum, and why the calculation is essentially the one in Example 5 with the roles of the known and unknown exchanged. Satellite navigation does the same thing in three dimensions with spheres, needing four satellites rather than three — one extra to fix the receiver's clock.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the centre of a circle the midpoint of one of its chords?
Correct: Only for a diameter; for any other chord the midpoint is not the centre.
\[ \text{midpoint of a chord} \in \text{ its bisector}, \text{ but } \neq \text{ centre} \]
Why: A chord's midpoint lies on that chord's perpendicular bisector, which does pass through the centre — so the midpoint is on the right LINE but at the wrong place along it. Only when the chord is a diameter do the two coincide. In Guided Practice 6 the chord from the origin to 16 comma 8 has midpoint 8 comma 4, while the centre is 6 comma 8: close, but the distances from 8 comma 4 to the three given points are 8.9, 6.3 and 8.9, not equal. That is exactly why two bisectors are needed rather than one midpoint.
Explain it
They know the Pythagorean theorem and have never seen the distance formula.
Discussion prompt
In four sentences or fewer, explain the distance formula without asking them to memorise it.
Hint: Draw a triangle.
Answer:
Mark your two points and draw a horizontal line from one and a vertical line from the other until they meet. You now have a right triangle whose legs you can count off: how far across, and how far up.
The distance you want is the hypotenuse, so square the two legs, add, and take the square root. The formula is just that sentence written with x's and y's instead of counting.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For keeping them straight, remember that a midpoint never has a square root in it. For radicals, look for the largest perfect square dividing the radicand. For negative reciprocals, multiply the two slopes and insist on negative 1. For assembly, write the four steps down the margin before starting: midpoint, slope, negative reciprocal, point-slope.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a coordinate-geometry page. Top left: plot two points of your own, draw the right triangle between them, label both legs, and derive the distance formula from the Pythagorean theorem in one line. Top right: write both formulas side by side with a note on which subtracts and which averages, and one sentence on why a midpoint never has a radical. Middle: plot Example 2's triangle, compute all three squared lengths, and classify it, then apply the Pythagorean test to see whether it is also right-angled. Bottom left: work Example 4 in full with all four steps labelled, and check both conditions at the end. Bottom right: plot the three points of Guided Practice 6, draw both perpendicular bisectors, mark the centre, and verify the radius from all three points.
If your radius came out different from the three points, the error is in a bisector rather than in the arithmetic — recheck each midpoint and each negative reciprocal.
Recap
Five things, and Chapter 9 will lean on the first two constantly.
| If you see | Then |
|---|---|
| Two points and the word distance | Subtract, square, add, take the root |
| Two points and the word midpoint | Average each coordinate |
| A radical in a midpoint answer | The wrong formula was used |
| The word perpendicular | Take the negative reciprocal of the slope |
| Three points on a circle | Intersect two perpendicular bisectors |
| A centre and a point on the circle | Their distance is the radius |
Lesson 9.2 defines a parabola by distance — every point equally far from a fixed point and a fixed line — which is where the distance formula starts paying off.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.1 Apply the Distance and Midpoint Formulas §9.1, pp. 614-617 — everything on these slides traces back here
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