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1. Distance and midpoint in the coordinate plane

Warm-up

Three quick questions from earlier lessons.

Discover

From $(1, 1)$ to $(4, 5)$ you move 3 right and 4 up. How long is the straight segment?

Put A at (0, 0) and B at (4, 6). Where is M?

(2, 3)
(4, 6)

Key idea

Pythagoras on the grid, and averaging for the middle.

Distance:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

. Midpoint:

\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)

Worked example

$A(-2, 3)$ , $B(4, -5)$ .

d = \sqrt{36 + 64} = 10

Δx = 6, Δy = −8.

Midpoint

(1, -1)

Average the coordinates.

Your turn, guided

Your turn: the distance from $(0, 0)$ to $(5, 12)$ .

13

√(25 + 144).

Practice ★

Find the distance between $(-1, -3)$ and $(2, 1)$ .
Find the distance between $(-3, 3)$ and $(0, 7)$ .

Tap the midpoint of the segment from $(-6, -5)$ to $(-4, -1)$ .
Tap the midpoint of the segment from $(-7, -8)$ to $(1, -4)$ .

Watch out!

Practice ★★

Find the exact distance between $(0, -2)$ and $(1, -4)$ .
Find the exact distance between $(2, -2)$ and $(3, -6)$ .

Exit ticket

Three questions to check you have got it.

Find the distance between $(-1, -3)$ and $(2, 1)$ .
Find the distance between $(-3, 3)$ and $(0, 7)$ .
Find the distance between $(2, 5)$ and $(11, 17)$ .

Summary card

Distance and midpoint

$d = \sqrt{\Delta x^2 + \Delta y^2}$ ; midpoint = average of the coordinates.

Watch out: average, not half the difference.

2. Circles: equation, center and radius

Warm-up

Three quick questions from earlier lessons.

Discover

A point $(x, y)$ is on the circle with center $(2, 1)$ and radius 3 when its distance to $(2, 1)$ is 3. Which equation says that?

$(x - 2)^2 + (y - 1)^2 = 9$
$(x + 2)^2 + (y + 1)^2 = 9$
$(x - 2)^2 + (y - 1)^2 = 3$

Key idea

A circle is the set of points at a fixed distance $r$ from a center $(h, k)$ .

(x - h)^2 + (y - k)^2 = r^2

. From the expanded form

x^2 + y^2 + Dx + Ey + F = 0

, complete the square in

x

and in

y

to read the center and radius.

Worked example

Find the center and radius of $x^2 + y^2 - 4x + 6y - 3 = 0$ .

(x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0

Group and complete the squares: x² − 4x = (x − 2)² − 4, y² + 6y = (y + 3)² − 9.

(x - 2)^2 + (y + 3)^2 = 16

Move the numbers to the right.

Center

(2, -3)

, radius

4

Read h, k and r = √16.

Your turn, guided

Your turn: $(x + 5)^2 + (y - 1)^2 = 49$ .

Center

(-5, 1)

x + 5 = x − (−5).

Radius

7

√49.

Practice ★

What is the radius of the circle $(x - 9)^2 + (y + 1)^2 = 16$ ?
What is the radius of the circle $(x - 1)^2 + (y + 7)^2 = 64$ ?
What is the radius of the circle $(x - 4)^2 + (y + 9)^2 = 16$ ?

Watch out!

Practice ★★

Find the center and radius of the circle $x^2 + y^2 - 10x + 4y + 25 = 0$ .
Find the center and radius of the circle $x^2 + y^2 - 2x + 4y + -11 = 0$ .

Which is the equation of the circle with center $(2, -8)$ and radius 7?
1. $(x - 2)^2 + (y + 8)^2 = 49$
2. $(x + 2)^2 + (y - 8)^2 = 49$
3. $(x - 2)^2 + (y + 8)^2 = 7$
Which is the equation of the circle with center $(7, -2)$ and radius 9?
1. $(x - 7)^2 + (y + 2)^2 = 81$
2. $(x + 7)^2 + (y - 2)^2 = 81$
3. $(x - 7)^2 + (y + 2)^2 = 9$

Exit ticket

Three questions to check you have got it.

Which is the equation of the circle with center $(2, -8)$ and radius 7?
1. $(x - 2)^2 + (y + 8)^2 = 49$
2. $(x + 2)^2 + (y - 8)^2 = 49$
3. $(x - 2)^2 + (y + 8)^2 = 7$
Which is the equation of the circle with center $(7, -2)$ and radius 9?
1. $(x - 7)^2 + (y + 2)^2 = 81$
2. $(x + 7)^2 + (y - 2)^2 = 81$
3. $(x - 7)^2 + (y + 2)^2 = 9$
Which is the equation of the circle with center $(6, -1)$ and radius 6?
1. $(x - 6)^2 + (y + 1)^2 = 36$
2. $(x + 6)^2 + (y - 1)^2 = 36$
3. $(x - 6)^2 + (y + 1)^2 = 6$

Summary card

Circles

$(x - h)^2 + (y - k)^2 = r^2$ : center $(h, k)$ , radius $r$ .
Expanded form: complete the square in $x$ and in $y$ .

Watch out: $(x + 3)$ means $h = -3$ , and the right side is $r^2$ .

3. Parabolas: focus and directrix

Warm-up

Three quick questions from earlier lessons.

Discover

A satellite dish is a parabola: every signal coming straight in bounces to one point, the focus. For $x^2 = 8y$ , where is it?

$(0, 2)$
$(0, 8)$

Key idea

A parabola is the set of points as far from a point (the focus) as from a line (the directrix).

x^2 = 4py

: vertex

(0, 0)

, focus

(0, p)

, directrix

y = -p

, opens up if

p > 0

y^2 = 4px

: focus

(p, 0)

, directrix

x = -p

, opens right if

p > 0

Worked example

Find the focus and directrix of $y^2 = -12x$ .

4p = -12

, so

p = -3

Match with y² = 4px.

Focus

(-3, 0)

: it opens to the left

Focus (p, 0).

Directrix

x = 3

Directrix x = −p.

Your turn, guided

Your turn: $x^2 = 20y$ .

p = 5

, focus

(0, 5)

4p = 20.

Directrix

y = -5

y = −p: type the number.

Practice ★

The parabola $x^2 = 8y$ has its focus at $(0, c)$ . What is $c$ ?
The parabola $x^2 = 24y$ has its focus at $(0, c)$ . What is $c$ ?
The parabola $x^2 = 4y$ has its focus at $(0, c)$ . What is $c$ ?

Watch out!

Practice ★★

The parabola $y^2 = 8x$ has the directrix $x = d$ . What is $d$ ?
The parabola $y^2 = 20x$ has the directrix $x = d$ . What is $d$ ?
The parabola $y^2 = 4x$ has the directrix $x = d$ . What is $d$ ?

A parabola has focus $(0, 3)$ and directrix $y = -3$ . Its equation is $x^2 = ky$ . What is $k$ ?
A parabola has focus $(0, 2)$ and directrix $y = -2$ . Its equation is $x^2 = ky$ . What is $k$ ?

Exit ticket

Three questions to check you have got it.

The parabola $y^2 = 8x$ has the directrix $x = d$ . What is $d$ ?
The parabola $y^2 = 20x$ has the directrix $x = d$ . What is $d$ ?
The parabola $y^2 = 4x$ has the directrix $x = d$ . What is $d$ ?

Summary card

Parabolas

$x^2 = 4py$ : focus $(0, p)$ , directrix $y = -p$ .
$y^2 = 4px$ : focus $(p, 0)$ , directrix $x = -p$ .

Watch out: divide by 4 to get $p$ .

4. Slopes of parallel and perpendicular lines

Warm-up

Three quick questions from earlier lessons.

Discover

Rotate a line of slope 2 by $90^\circ$ . Its "1 right, 2 up" becomes "2 left, 1 up". What is the new slope?

$-\tfrac{1}{2}$
$\tfrac{1}{2}$

Key idea

Slopes tell parallel and perpendicular apart.

Slope

m = \dfrac{y_2 - y_1}{x_2 - x_1}

. Parallel lines have equal slopes. Perpendicular lines (neither vertical) have slopes with product

-1

m_2 = -\dfrac{1}{m_1}

Worked example

Line through $(1, 2)$ and $(5, 4)$ .

m = \tfrac{1}{2}

2/4.

Parallel:

\tfrac{1}{2}

Same slope.

Perpendicular:

-2

Negative reciprocal.

Your turn, guided

Your turn: perpendicular to slope $\tfrac{3}{4}$ .

-\tfrac{4}{3}

Flip and change sign.

Practice ★

A line has slope $1/8$ . Find the slope of a line perpendicular to it.
A line has slope $8/4$ . Find the slope of a line perpendicular to it.

Are the lines $y = 6x + 1$ and $y = -\tfrac{1}{ 6 }x + 5$ parallel, perpendicular or neither?
1. Perpendicular: the product of the slopes is −1
2. Parallel
3. Neither

Watch out!

Practice ★★

Find the slope of the line through $(0, 5)$ and $(-5, -3)$ .
Find the slope of the line through $(-5, -2)$ and $(5, 4)$ .

Exit ticket

Three questions to check you have got it.

A line has slope $1/8$ . Find the slope of a line perpendicular to it.
A line has slope $8/4$ . Find the slope of a line perpendicular to it.
A line has slope $4/7$ . Find the slope of a line perpendicular to it.

Summary card

Parallel and perpendicular slopes

Parallel: equal. Perpendicular: $m_1 m_2 = -1$ .

Watch out: flip and change the sign.

5. Equations of lines through a point

Warm-up

Three quick questions from earlier lessons.

Discover

A line has slope 2 and passes through $(3, 1)$ . Does it pass through $(4, 3)$ ?

Key idea

A point and a slope fix a line.

Point-slope form:

y = y_1 + m(x - x_1)

. To get

y = mx + b

, substitute the point to find

b

. A line parallel (perpendicular) to a given line uses the same (negative reciprocal) slope, with its own intercept.

Worked example

Through $(2, 5)$ , parallel to $y = 3x - 4$ .

m = 3

Same slope.

b = -1

5 = 3·2 + b.

y = 3x - 1

The line.

Your turn, guided

Your turn: slope $-2$ through $(1, 4)$ . The intercept $b$ .

b = 6

4 = −2 + b.

Practice ★

Write the line with slope $3$ through $(4, -1)$ : enter the right-hand side of $y = \ldots$
Write the line with slope $1$ through $(2, -1)$ : enter the right-hand side of $y = \ldots$

Find the $y$ -intercept of the line parallel to $y = -3x + -1$ through $(-3, -5)$ .
Find the $y$ -intercept of the line parallel to $y = 1x + -1$ through $(-1, 2)$ .

Watch out!

Practice ★★

Find the $y$ -intercept of the line perpendicular to $y = 4x$ through $(-3, -3)$ .

Boss problem. $A(-2, -2)$ and $B(2, 0)$ . Find the perpendicular bisector of $AB$ .

Exit ticket

Three questions to check you have got it.

Write the line with slope $3$ through $(4, -1)$ : enter the right-hand side of $y = \ldots$
Write the line with slope $1$ through $(2, -1)$ : enter the right-hand side of $y = \ldots$
Write the line with slope $5$ through $(4, -1)$ : enter the right-hand side of $y = \ldots$

Summary card

Equations of lines

$y = y_1 + m(x - x_1)$ ; find $b$ by substituting the point.

Watch out: parallel lines share the slope, not the intercept.

6. Partitioning a segment

Warm-up

Three quick questions from earlier lessons.

Discover

From $A(0, 0)$ to $B(6, 3)$ . Where is the point one third of the way along?

$(2, 1)$
$(3, 1.5)$

Key idea

Move a fraction of the way, or take a weighted average.

The point a fraction

t

of the way from

A

B

A + t(B - A)

. A point dividing

AB

in the ratio

m : n

(from

A

) is

\left(\dfrac{n x_A + m x_B}{m + n}, \dfrac{n y_A + m y_B}{m + n}\right)

, a weighted average.

Worked example

$A(1, 2)$ , $B(9, 6)$ , ratio $1 : 3$ .

t = \tfrac{1}{4}

1 part of 4.

P = (3, 3)

¼ of (8, 4) = (2, 1).

Your turn, guided

Your turn: x-coordinate halfway between 2 and 10.

6

Average.

Practice ★

Tap the point one third of the way from $A(-3, -2)$ to $B(3, 4)$ .
Tap the point one third of the way from $A(-5, -6)$ to $B(4, 3)$ .

Point $P$ divides segment $AB$ in the ratio $1 : 2$ (closer to $A$ ). $A$ has $x = -4$ and $B$ has $x = 2$ . Find the $x$ -coordinate of $P$ .

Watch out!

Practice ★★

Point $P$ divides $AB$ in the ratio $3 : 1$ from $A$ . $A$ has $x = -2$ , $B$ has $x = 1$ . Find $P$ 's $x$ -coordinate.
Point $P$ divides $AB$ in the ratio $1 : 4$ from $A$ . $A$ has $x = -2$ , $B$ has $x = 2$ . Find $P$ 's $x$ -coordinate.

Exit ticket

Three questions to check you have got it.

Tap the point one third of the way from $A(-3, -2)$ to $B(3, 4)$ .
Tap the point one third of the way from $A(-5, -6)$ to $B(4, 3)$ .
Tap the point one third of the way from $A(-7, -2)$ to $B(2, 1)$ .

Summary card

Partitioning a segment

$A + t(B - A)$ ; ratio $m : n$ gives $t = \dfrac{m}{m + n}$ .

Watch out: count the parts of the ratio.

Answers: Coordinate geometry

1. Distance and midpoint in the coordinate plane

$(-5, -3)$
$(-3, -6)$

$\sqrt{5}$
$\sqrt{17}$

$5$
$5$
$15$

2. Circles: equation, center and radius

$(x - 2)^2 + (y + 8)^2 = 49$
$(x - 7)^2 + (y + 2)^2 = 81$

$(x - 2)^2 + (y + 8)^2 = 49$
$(x - 7)^2 + (y + 2)^2 = 81$
$(x - 6)^2 + (y + 1)^2 = 36$

3. Parabolas: focus and directrix

$-2$
$-5$
$-1$

$12$
$8$

$-2$
$-5$
$-1$

4. Slopes of parallel and perpendicular lines

$-8$
$- \frac{1}{2}$

Perpendicular: the product of the slopes is −1

$\frac{8}{5}$
$\frac{3}{5}$

$-8$
$- \frac{1}{2}$
$- \frac{7}{4}$

5. Equations of lines through a point

$3 x - 13$
$x - 3$

$-14$
$3$

$- \frac{15}{4}$

$-2$

$3 x - 13$
$x - 3$
$5 x - 21$

6. Partitioning a segment

$(-1, 0)$
$(-2, -3)$

$-2$

$\frac{1}{4}$
$- \frac{6}{5}$

$(-1, 0)$
$(-2, -3)$
$(-4, -1)$