---
title: "Coordinate Geometry"
book: "High School Mathematics"
subject: math
language: en
chapter: 5
exercises: 10
source: https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry
---

# Chapter 5 — Coordinate Geometry

Descartes’ great idea was to describe points by pairs of numbers, turning geometry problems into computations. With just two formulas — the [midpoint](#prop-g10-coordgeom-midpoint) and the distance — one can prove that a triangle is isosceles, that a quadrilateral is a parallelogram, or that three points are aligned, without drawing a single auxiliary line.

## 5.1 Coordinates in the plane

**Definition 5.1 (Coordinate system).**

A *coordinate system* of the plane consists of an origin $O$ and two graduated axes through $O$: the horizontal $x$-axis and the vertical $y$-axis. Every point $M$ then has a unique pair of *coordinates* $(x, y)$: its *abscissa* $x$ and its *ordinate* $y$. The system is *orthonormal* when the axes are perpendicular and carry the same unit of length. All systems in this chapter are orthonormal.

![Every point of the plane is located by two numbers: first the abscissa (horizontal), then the ordinate (vertical).](https://one-course.com/images/onecourse/chapters/math-2/g10-coordgeom/fig-9451968d4277.svg)

*Every point of the plane is located by two numbers: first the [abscissa](#def-g10-coordgeom-system) (horizontal), then the [ordinate](#def-g10-coordgeom-system) (vertical).*

## 5.2 Midpoint of a segment

**Proposition 5.2 (Midpoint formula).**

Let $A(x_A, y_A)$ and $B(x_B, y_B)$. The *midpoint* $I$ of the segment $[AB]$ has [coordinates](#def-g10-coordgeom-system)

$$
I\left(\frac{x_A + x_B}{2},\ \frac{y_A + y_B}{2}\right).
$$

**Proof.** Consider the horizontal [coordinates](#def-g10-coordgeom-system). Going from $A$ to $B$, the [abscissa](#def-g10-coordgeom-system) changes by $x_B - x_A$; the point halfway has [abscissa](#def-g10-coordgeom-system)

$$
x_A + \frac{x_B - x_A}{2} = \frac{2x_A + x_B - x_A}{2}
= \frac{x_A + x_B}{2},
$$

the average of the two [abscissas](#def-g10-coordgeom-system). The same computation applies to the [ordinates](#def-g10-coordgeom-system). ∎

**Example 5.3.**

With $A(-1, 4)$ and $B(5, -2)$, the [midpoint](#prop-g10-coordgeom-midpoint) of $[AB]$ is

$$
I\left(\frac{-1 + 5}{2},\ \frac{4 + (-2)}{2}\right) = I(2, 1).
$$

The formula also runs backwards: if $A(-1, 4)$ and the [midpoint](#prop-g10-coordgeom-midpoint) is $I(2, 1)$, then $B$ satisfies $\frac{-1 + x_B}{2} = 2$ and $\frac{4 + y_B}{2} = 1$, so $x_B = 5$ and $y_B = -2$: the point $B$ is the *symmetric* of $A$ about $I$.

## 5.3 Distance between two points

**Theorem 5.4 (Distance formula).**

In an [orthonormal system](#def-g10-coordgeom-system), the distance between $A(x_A, y_A)$ and $B(x_B, y_B)$ is

$$
AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}.
$$

**Proof.** Let $C$ be the point $(x_B, y_A)$: it has the same height as $A$ and the same [abscissa](#def-g10-coordgeom-system) as $B$, so the triangle $ACB$ has a right angle at $C$. Its legs are horizontal and vertical segments, of lengths $AC = \abs{x_B - x_A}$ and $CB = \abs{y_B - y_A}$. By the Pythagorean theorem,

$$
AB^2 = AC^2 + CB^2 = (x_B - x_A)^2 + (y_B - y_A)^2,
$$

and taking the square root (both sides are nonnegative) gives the formula. ∎

![The distance formula is the Pythagorean theorem applied to the right triangle ACB built on a horizontal and a vertical leg.](https://one-course.com/images/onecourse/chapters/math-2/g10-coordgeom/fig-235b7a1fe84f.svg)

*The distance formula is the Pythagorean theorem applied to the right triangle $ACB$ built on a horizontal and a vertical leg.*

**Example 5.5.**

With $A(1, 1)$ and $B(5, 3)$:

$$
AB = \sqrt{(5-1)^2 + (3-1)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt 5 .
$$

Distances are usually kept in exact (square root) form; only round at the very end if a decimal answer is needed.

**Remark 5.6.**

The order of the points does not matter: $(x_A - x_B)^2 =
(x_B - x_A)^2$. But do not forget the squares! The distance is *not* $\abs{x_B - x_A} + \abs{y_B - y_A}$.

## 5.4 Using the formulas in geometry

**Method 5.7 (Nature of a triangle).**

Given three points $A$, $B$, $C$ by their [coordinates](#def-g10-coordgeom-system):

1. compute the three squared lengths $AB^2$ , $AC^2$ , $BC^2$ with the distance formula (keeping squares avoids square roots);
2. two equal squared lengths $\Rightarrow$ the triangle is *isosceles* ;
3. if the largest squared length is the sum of the two others, the triangle is *right-angled* at the vertex opposite the longest side, by the converse of the Pythagorean theorem.

**Example 5.8.**

Let $A(0, 1)$, $B(4, 3)$ and $C(2, -3)$. Then

$$
\begin{align*}
AB^2 &= (4-0)^2 + (3-1)^2 = 16 + 4 = 20, \\
AC^2 &= (2-0)^2 + (-3-1)^2 = 4 + 16 = 20, \\
BC^2 &= (2-4)^2 + (-3-3)^2 = 4 + 36 = 40 .
\end{align*}
$$

First, $AB^2 = AC^2$, so $AB = AC$: the triangle is isosceles at $A$. Moreover $AB^2 + AC^2 = 40 = BC^2$, so by the converse of the Pythagorean theorem the triangle is also right-angled at $A$.

**Method 5.9 (Recognizing a parallelogram).**

A quadrilateral $ABCD$ is a parallelogram exactly when its diagonals $[AC]$ and $[BD]$ have the same [midpoint](#prop-g10-coordgeom-midpoint). So: compute both [midpoints](#prop-g10-coordgeom-midpoint) with the [midpoint](#prop-g10-coordgeom-midpoint) formula and compare them.

**Example 5.10.**

Let $A(-1, 0)$, $B(2, 2)$, $C(5, 1)$ and $D(2, -1)$. The [midpoint](#prop-g10-coordgeom-midpoint) of $[AC]$ is $\left(\frac{-1+5}{2}, \frac{0+1}{2}\right) = (2, \frac12)$; the [midpoint](#prop-g10-coordgeom-midpoint) of $[BD]$ is $\left(\frac{2+2}{2}, \frac{2-1}{2}\right) =
(2, \frac12)$. They coincide, so $ABCD$ is a parallelogram.

![ABCD is a parallelogram because its two diagonals share the same midpoint I(2, 1/2).](https://one-course.com/images/onecourse/chapters/math-2/g10-coordgeom/fig-4496c905f6f2.svg)

*$ABCD$ is a parallelogram because its two diagonals share the same [midpoint](#prop-g10-coordgeom-midpoint) $I(2, \frac12)$.*

## 5.5 Exercises

**Exercise 5.1 ★.**

Let $A(2, 5)$ and $B(-4, 1)$. Compute the [coordinates](#def-g10-coordgeom-system) of the [midpoint](#prop-g10-coordgeom-midpoint) of $[AB]$ and the distance $AB$.

**Solution of Exercise 5.1.**

[Midpoint](#prop-g10-coordgeom-midpoint): $\left(\frac{2 + (-4)}{2}, \frac{5 + 1}{2}\right) = (-1, 3)$. Distance:

$$
AB = \sqrt{(-4 - 2)^2 + (1 - 5)^2} = \sqrt{36 + 16} = \sqrt{52}
= 2\sqrt{13}.
$$

**Exercise 5.2 ★.**

Let $A(3, -2)$ and $I(1, 2)$. Find the [coordinates](#def-g10-coordgeom-system) of the point $B$ such that $I$ is the [midpoint](#prop-g10-coordgeom-midpoint) of $[AB]$.

**Solution of Exercise 5.2.**

$I$ is the [midpoint](#prop-g10-coordgeom-midpoint) of $[AB]$ means $\frac{3 + x_B}{2} = 1$ and $\frac{-2 + y_B}{2} = 2$, so $x_B = -1$ and $y_B = 6$: $B(-1, 6)$.

**Exercise 5.3 ★.**

Which of the points $P(4, 1)$, $Q(-3, 2)$ and $R(0, -5)$ is closest to the origin $O(0,0)$? Answer with exact distances.

**Solution of Exercise 5.3.**

$OP = \sqrt{16 + 1} = \sqrt{17}$; $OQ = \sqrt{9 + 4} = \sqrt{13}$; $OR = \sqrt{0 + 25} = 5 = \sqrt{25}$. Since $13 < 17 < 25$, the square root being [increasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations), $Q$ is the closest to the origin.

**Exercise 5.4 ★.**

Plot the points $A(1, 2)$, $B(5, 4)$, $C(7, 0)$ on [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) paper, then show by computation that the triangle $ABC$ is isosceles. Is it right-angled?

**Solution of Exercise 5.4.**

Squared lengths:

$$
\begin{align*}
AB^2 &= (5-1)^2 + (4-2)^2 = 16 + 4 = 20, \\
BC^2 &= (7-5)^2 + (0-4)^2 = 4 + 16 = 20, \\
AC^2 &= (7-1)^2 + (0-2)^2 = 36 + 4 = 40 .
\end{align*}
$$

$AB = BC$: the triangle is isosceles at $B$. Moreover $AB^2 + BC^2 = 40 = AC^2$: by the converse of the Pythagorean theorem it is right-angled at $B$ as well.

**Exercise 5.5 ★★.**

Let $A(-2, 1)$, $B(1, 3)$, $C(4, 1)$ and $D(1, -1)$.

1. Show that $ABCD$ is a parallelogram.
2. Compute $AB$ and $BC$ . Is $ABCD$ a rhombus (all sides equal)?
3. Compute $AC$ and $BD$ . Is $ABCD$ a rectangle (equal diagonals)?

**Solution of Exercise 5.5.**

*1.* [Midpoint](#prop-g10-coordgeom-midpoint) of $[AC]$: $\left(\frac{-2+4}{2}, \frac{1+1}{2}\right) = (1, 1)$; [midpoint](#prop-g10-coordgeom-midpoint) of $[BD]$: $\left(\frac{1+1}{2}, \frac{3+(-1)}{2}\right) = (1, 1)$. Same [midpoint](#prop-g10-coordgeom-midpoint): $ABCD$ is a parallelogram.

*2.* $AB^2 = 3^2 + 2^2 = 13$ and $BC^2 = 3^2 + (-4)^2 = 25$: $AB \neq BC$, not a rhombus.

*3.* $AC^2 = 6^2 + 0^2 = 36$ and $BD^2 = 0^2 + (-4)^2 = 16$: $AC \neq BD$, not a rectangle. $ABCD$ is a plain parallelogram.

**Exercise 5.6 ★★.**

The circle with center $\Omega(2, 1)$ and radius $5$ consists of all points at distance $5$ from $\Omega$. Which of the points $A(5, 5)$, $B(-1, -3)$, $C(6, 2)$ lie on this circle? Inside it? Outside it?

**Solution of Exercise 5.6.**

Compute squared distances to $\Omega(2,1)$ and compare with $5^2 = 25$:

$$
\Omega A^2 = 3^2 + 4^2 = 25, \qquad
\Omega B^2 = (-3)^2 + (-4)^2 = 25, \qquad
\Omega C^2 = 4^2 + 1^2 = 17 .
$$

$A$ and $B$ are on the circle; $C$ is inside it ($17 < 25$).

**Exercise 5.7 ★★.**

Let $A(1, 1)$ and $B(7, 5)$. Find all points $M(x, 0)$ of the $x$-axis that are equidistant from $A$ and $B$. (Write $MA^2 = MB^2$ and solve for $x$.)

**Solution of Exercise 5.7.**

$MA^2 = (x - 1)^2 + 1$ and $MB^2 = (x - 7)^2 + 25$. Setting $MA^2 = MB^2$:

$$
x^2 - 2x + 2 = x^2 - 14x + 74
\ \Longleftrightarrow\
12x = 72
\ \Longleftrightarrow\
x = 6 .
$$

The unique such point is $M(6, 0)$.

**Exercise 5.8 ★★.**

Let $A(0, 3)$, $B(4, 1)$.

1. Compute the [coordinates](#def-g10-coordgeom-system) of the [midpoint](#prop-g10-coordgeom-midpoint) $I$ of $[AB]$ .
2. Show that $M(2, 2) = I$ , then verify by computing $MA$ , $MB$ that $M$ is equidistant from $A$ and $B$ .
3. Find a second point, on the $y$ -axis, equidistant from $A$ and $B$ .

**Solution of Exercise 5.8.**

*1.* $I = \left(\frac{0+4}{2}, \frac{3+1}{2}\right) = (2, 2)$.

*2.* $M(2,2)$ is indeed $I$. And $MA^2 = (0-2)^2 + (3-2)^2 = 5$, $MB^2 = (4-2)^2 + (1-2)^2 = 5$: the [midpoint](#prop-g10-coordgeom-midpoint) is equidistant from the two endpoints, as expected.

*3.* Look for $N(0, y)$ with $NA^2 = NB^2$:

$$
(3 - y)^2 = 16 + (1 - y)^2
\ \Longleftrightarrow\
9 - 6y = 16 + 1 - 2y
\ \Longleftrightarrow\
-4y = 8,
$$

so $y = -2$: the point $N(0, -2)$.

**Exercise 5.9 ★★.**

The points $A(-1, -1)$, $B(3, 1)$ and $C(11, 5)$ are given. Compute $AB$, $BC$ and $AC$, and deduce that $A$, $B$, $C$ are aligned. (Three points are aligned exactly when the largest of the three distances is the sum of the two others.)

**Solution of Exercise 5.9.**

$$
\begin{align*}
AB &= \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt5, \\
BC &= \sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt5, \\
AC &= \sqrt{12^2 + 6^2} = \sqrt{180} = 6\sqrt5 .
\end{align*}
$$

Then $AB + BC = 2\sqrt5 + 4\sqrt5 = 6\sqrt5 = AC$: the triangle inequality is an equality, so $B$ lies on the segment $[AC]$ — the three points are aligned.

**Exercise 5.10 ★★★.**

Let $A(a, 0)$ and $B(0, b)$ with $a, b > 0$, and let $I$ be the [midpoint](#prop-g10-coordgeom-midpoint) of $[AB]$. Show by computation that $OI = IA = IB$, where $O$ is the origin. (This proves a classical theorem: in a right triangle, the [midpoint](#prop-g10-coordgeom-midpoint) of the hypotenuse is equidistant from the three vertices.)

**Solution of Exercise 5.10.**

The [midpoint](#prop-g10-coordgeom-midpoint) of $[AB]$ is $I\left(\frac a2, \frac b2\right)$. Then

$$
OI^2 = \frac{a^2}{4} + \frac{b^2}{4},
\qquad
IA^2 = \left(a - \frac a2\right)^2 + \left(0 - \frac b2\right)^2
= \frac{a^2}{4} + \frac{b^2}{4},
$$

and $IB^2 = \left(0 - \frac a2\right)^2 + \left(b - \frac b2\right)^2 =
\frac{a^2}{4} + \frac{b^2}{4}$ as well. The three squared distances are equal, so $OI = IA = IB$: the [midpoint](#prop-g10-coordgeom-midpoint) of the hypotenuse of the right triangle $OAB$ is equidistant from all three vertices (it is the center of its circumscribed circle).

## 5.6 Problem: Descartes’ bridge — circle equations and how to find where you are

**Problem 5.1.**

Weekend problem — the circle becomes an equation, old theorems become computations, and three beacons locate a point: trilateration by algebra

In 1637 René Descartes built a bridge between two continents: every curve of geometry became an *[equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation)*, every geometric question a computation. This problem walks the bridge in both directions — deriving the circle’s [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) from the distance formula ([Theorem 5.4](#thm-g10-coordgeom-distance)), re-proving classical theorems in three lines of algebra, and ending where the bridge carries the most traffic today: computing a position from distance signals, the flat-earth heart of GPS.

**Part I — The circle gets an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation).**

1. A point $M(x, y)$ lies on the circle of center $\Omega(2, 1)$ and radius $5$ exactly when $\Omega M = 5$. Square this condition to obtain the circle’s [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation): $$(x - 2)^2 + (y - 1)^2 = 25 .$$
2. Test membership: which of $A(5, 5)$ , $B(6, 4)$ , $D(4, 4)$ are on the circle, inside it, outside it?
3. A circle in disguise: complete the squares ([Problem 2.1](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#pb-g10-algebra-1)) in $$x^2 + y^2 - 6x + 4y - 12 = 0$$ and give its center and radius.
4. Do the same with $x^2 + y^2 - 6x + 4y + 14 = 0$ . What does the completed form reveal? State the criterion: when does $(x - h)^2 + (y - k)^2 = c$ describe a circle, a single point, or nothing at all?
5. Intersect the circle of question 3 with the horizontal line $y = 3$ : substitute and solve. How many points, and what is the geometric name of such a line?

**Part II — Old theorems in three lines.**

6. The circle theorem of the Middle School volume’s circle theorem (a right angle in every half-circle), re-proved by algebra: let $A(-r, 0)$ and $B(r, 0)$ be the ends of a diameter and $M(x, y)$ any point of the circle $x^2 + y^2 = r^2$ . Compute $MA^2 + MB^2$ and conclude with the converse of Pythagoras that $\widehat{AMB}$ is right.
7. The median theorem: with $A(-a, 0)$, $B(a, 0)$ (so the [midpoint](#prop-g10-coordgeom-midpoint) of $[AB]$ is the origin $I$), show that for every point $M(x, y)$: $$MA^2 + MB^2 = 2\,MI^2 + \frac{AB^2}{2} .$$
8. Deduce the locus of the points $M$ with $MA^2 + MB^2 = 26$ when $AB = 6$ : what curve, which center, what radius?
9. A different locus: $A(0,0)$ , $B(3,0)$ ; find all points with $MA = 2\,MB$ . (Square, expand, complete the squares: a famous circle appears — Apollonius knew it without [coordinates](#def-g10-coordgeom-system) .)
10. In one or two sentences: what does Descartes’ bridge change about *how* theorems get proved? Compare question 6 with the rectangle proof of the Middle School volume’s circle theorem (a right angle in every half-circle).

**Part III — Three beacons find you.** Your position $M(x, y)$ is unknown. Beacon $P(0, 0)$ measures your distance as $5$; beacon $Q(6, 0)$ also measures $5$.

11. Write the two circle [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) , subtract them, and watch the squares cancel: what simple [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) survives? Combine with one circle to find the *two* candidate positions.
12. A third beacon $R(0, 8)$ measures your distance as $5$ . Compute its distance to each candidate and decide where you are.
13. Explain why subtracting two circle [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) *always* yields the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of a line (which terms cancel?), and what that line is geometrically when the circles cross at two points.
14. Real satellite positioning works in space and with imperfect clocks: each satellite gives (via signal travel time) one distance, hence one sphere. How many unknowns does a receiver have (position *and* its own clock error), and why does GPS therefore listen to at least *four* satellites?
15. Precision: suppose beacon $P$ ’s distance is really $5.1$ instead of $5$ (all else unchanged). Redo the subtraction of question 11 to find the new $x$ , and quantify how far the estimate moved. (Compare the error budget ideas of [Problem 1.1](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#pb-g10-numbers-1) .)

**Part IV — The surveyor’s toolkit.**

16. The shortest watering path: a camp at $A(1, 5)$ , a barn at $B(7, 3)$ , a straight river along the axis $y = 0$ . To go from $A$ to the river and then to $B$ with the least walking: reflect $B$ across the river into $B'$ , and find where the segment $[AB']$ crosses the river. Give the crossing point and the minimal total length. (The idea is the billiard-bounce exercise of the Middle School volume, now fully computable.)
17. Confirm minimality on a rival: compute the total path via $Q(3, 0)$ and check it loses to your answer.
18. Classify the quadrilateral $A(1,1)$ , $B(4,2)$ , $C(5,5)$ , $D(2,4)$ completely: parallelogram? rhombus? rectangle? square? ( [Method 5.9](#met-g10-coordgeom-parallelogram) , [Method 5.7](#met-g10-coordgeom-triangle) — compare diagonals too.)
19. Given $A(-1, 2)$ , $B(3, 4)$ , $C(6, 0)$ , find $D$ such that $ABCD$ is a parallelogram (the [diagonal-midpoint](#prop-g10-coordgeom-midpoint) trick of the half-turns weekend problem of the Middle School volume, now in formulas).
20. Finale: three tools — [midpoint](#prop-g10-coordgeom-midpoint) formula, distance formula, subtraction of circle [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) . For each, name the kind of question it settled in this problem, and state what the next two chapters add to the kit (directions and [slopes](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) : vectors and line [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) ).

**Solution of Problem 5.1.**

**1.** $\Omega M = 5 \iff \Omega M^2 = 25 \iff
(x - 2)^2 + (y - 1)^2 = 25$ by the distance formula ([Theorem 5.4](#thm-g10-coordgeom-distance)) — squaring is harmless, both sides being positive.

**2.** $A$: $(5-2)^2 + (5-1)^2 = 9 + 16 = 25$: on the circle. $B$: $16 + 9 = 25$: on it too. $D$: $4 + 9 = 13 < 25$: inside.

**3.** $x^2 - 6x = (x - 3)^2 - 9$ and $y^2 + 4y = (y + 2)^2 - 4$, so the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) reads $(x - 3)^2 + (y + 2)^2 = 25$: center $(3, -2)$, radius $5$.

**4.** $(x - 3)^2 + (y + 2)^2 = -1$: a sum of squares is never negative — no point satisfies it: the empty set. In general $(x - h)^2 + (y - k)^2 = c$ is a circle of radius $\sqrt c$ if $c > 0$, the single point $(h, k)$ if $c = 0$, and empty if $c < 0$.

**5.** With $y = 3$: $(x - 3)^2 + 25 = 25$, so $(x - 3)^2 = 0$: the single point $(3, 3)$. One contact point: the line is *tangent* to the circle.

**6.** $MA^2 + MB^2 = (x + r)^2 + y^2 + (x - r)^2 + y^2
= 2x^2 + 2y^2 + 2r^2 = 2r^2 + 2r^2 = 4r^2$ (using $x^2 + y^2 = r^2$). Since $AB^2 = (2r)^2 = 4r^2$, the triangle $AMB$ satisfies $MA^2 + MB^2 = AB^2$: right-angled at $M$ (converse of Pythagoras) — three lines, as promised.

**7.** $MA^2 + MB^2 = (x + a)^2 + y^2 + (x - a)^2 + y^2
= 2x^2 + 2y^2 + 2a^2 = 2\,MI^2 + 2a^2$, and $\frac{AB^2}{2} = \frac{(2a)^2}{2} = 2a^2$. Identity proved.

**8.** $2\,MI^2 + \frac{36}{2} = 26$ gives $MI^2 = 4$: the circle of center the [midpoint](#prop-g10-coordgeom-midpoint) $I$ and radius $2$.

**9.** $MA^2 = 4\,MB^2$: $x^2 + y^2 = 4\left((x - 3)^2 + y^2\right)$, i.e. $3x^2 + 3y^2 - 24x + 36 = 0$, or $x^2 + y^2 - 8x + 12 = 0$: completing, $(x - 4)^2 + y^2 = 4$ — the circle of center $(4, 0)$ and radius $2$ (the *circle of Apollonius* of ratio $2$).

**10.** The bridge converts constructions into computations: no auxiliary rectangle, no case analysis — one expansion and the theorem falls out. The price: computations carry less geometric insight; the classical proof shows *why*, the algebraic one shows *that* — a working mathematician keeps both.

**11.** $x^2 + y^2 = 25$ and $(x - 6)^2 + y^2 = 25$. Subtracting: $x^2 - (x - 6)^2 = 0$, i.e. $12x - 36 = 0$: $x = 3$. Then $9 + y^2 = 25$: $y = \pm 4$. Candidates $(3, 4)$ and $(3, -4)$.

**12.** From $(3, 4)$ to $R(0, 8)$: $\sqrt{9 + 16} = 5$ — matches. From $(3, -4)$: $\sqrt{9 + 144} \approx 12.4$ — wrong. You are at $(3, 4)$.

**13.** Both [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) carry the same $x^2 + y^2$ (or with equal coefficients after expansion): subtraction cancels the squares and leaves an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of degree one — a line. When the circles meet twice, that line passes through both meeting points: it is the line of the common chord.

**14.** Four unknowns: three [coordinates](#def-g10-coordgeom-system) plus the receiver’s clock error (a cheap watch, corrected by the mathematics). Each satellite supplies one [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation), so four satellites give four [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) for four unknowns — with more satellites improving the precision.

**15.** Subtraction now gives $x = \frac{5.1^2 - 5^2 + 36}{12} = \frac{37.01}{12} \approx
3.084$: a $10$ cm error in one distance moved the estimate about $8$ cm here. Errors propagate through every formula — the interval-arithmetic caution of [Problem 1.1](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#pb-g10-numbers-1), now with [coordinates](#def-g10-coordgeom-system).

**16.** $B' (7, -3)$. The line $(AB')$ has [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $\frac{-3 - 5}{7 - 1} = -\frac43$ and crosses $y = 0$ at $x = 1 + \frac{5}{8} \times 6 = 4.75$: drink at $P(4.75,\ 0)$. Minimal length: $AP + PB = AP + PB' = AB' =
\sqrt{6^2 + 8^2} = 10$.

**17.** Via $Q(3, 0)$: $AQ + QB = \sqrt{4 + 25} + \sqrt{16 + 9} = \sqrt{29} + 5
\approx 10.39 > 10$. The reflected straight line wins.

**18.** Diagonal [midpoints](#prop-g10-coordgeom-midpoint): $[AC]$: $(3, 3)$; $[BD]$: $(3, 3)$ — equal: parallelogram. Sides: $AB = \sqrt{10} = AD$: rhombus. Diagonals: $AC = \sqrt{32} \neq BD = \sqrt{8}$: not a rectangle, hence not a square. Verdict: a rhombus, and nothing more.

**19.** $D = A + C - B = (-1 + 6 - 3,\ 2 + 0 - 4) =
(2, -2)$: then $[AC]$ and $[BD]$ share the [midpoint](#prop-g10-coordgeom-midpoint) $\left(\frac52, 1\right)$.

**20.** [Midpoint](#prop-g10-coordgeom-midpoint) formula: parallelograms and centers (questions 18, 19). Distance formula: circles, loci, triangle natures, shortest paths (questions 1–9, 16). Subtraction of circle [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation): the line that locates a position (questions 11–15). Missing from the kit: a clean algebra of *directions* — vectors — and of *[slopes](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine)* — line [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) and systems: the next two chapters supply exactly that.