Recap: points, segments and lines

Analytic geometry, grade 12 (5 units) Lesson 1 of 21Learning

Recap: points, segments and lines

What grade 10 taught

In grade 10 we described points and lines with numbers: each point has two coordinates, and each line an equation. This course adds curves: the circle, the parabola, the ellipse and the hyperbola. The tools from grade 10:

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$ – by Pythagoras’ theorem:

\displaystyle d

\displaystyle =

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

The midpoint of the segment – the mean of the coordinates: $\displaystyle \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .
The slope of the line through the two points:

\displaystyle m

\displaystyle =

\displaystyle \frac{y_2-y_1}{x_2-x_1}

\displaystyle x_1\neq x_2

The equation of the line with slope $\displaystyle m$ through $\displaystyle (x_1,y_1)$ :

\displaystyle y-y_1=m(x-x_1)

The explicit and the implicit form

The same line has two forms of equation:

Explicit form – $\displaystyle y$ alone on one side: $\displaystyle {y}={mx+b}$ . From it you read the slope $\displaystyle m$ , and $\displaystyle b$ – the $\displaystyle y$ -coordinate of the point where the line meets the $\displaystyle y$ -axis.
Implicit form – all the terms on one side: $\displaystyle {Ax+By+C}={0}$ . A line perpendicular to the $\displaystyle x$ -axis, such as $\displaystyle {x-3}={0}$ , which has no slope, is written in this form too.

This course uses the implicit form a lot: the formula for the distance of a point from a line needs it.

Parallel and perpendicular lines:

parallel lines – the same slope: $\displaystyle {m_1}={m_2}$ ;
perpendicular lines – the product of the slopes is $\displaystyle -1$ :

\displaystyle m_1\cdot m_2

\displaystyle =

\displaystyle -1

A worked example

$\displaystyle A(-1,2)$ and $\displaystyle B(5,-6)$ .

The length: $\displaystyle {AB}={\sqrt{(5+1)^{2}+(-6-2)^{2}}}={\sqrt{36+64}}={\sqrt{100}}={10}$ .
The midpoint: $\displaystyle {\left(\frac{-1+5}{2},\frac{2-6}{2}\right)}={(2,-2)}$ .
The slope: $\displaystyle {m}={\frac{-6-2}{5+1}}={\frac{-8}{6}}={-\frac{4}{3}}$ .
The equation: $\displaystyle {y-2}={-\frac{4}{3}(x+1)}$ , in explicit form $\displaystyle {y}={-\frac{4}{3}x+\frac{2}{3}}$ .
In implicit form: we multiply by $\displaystyle 3$ and bring everything to one side: $\displaystyle {4x+3y-2}={0}$ .

A check: $\displaystyle A$ satisfies the equation – $\displaystyle {4\cdot(-1)+3\cdot2-2}={0}$ .

\displaystyle 4x+3y-2=0

\displaystyle A=(-1,2)

\displaystyle B=(5,-6)

\displaystyle M=(2,-2)

Exercise 1

$\displaystyle A(2,-3)$ and $\displaystyle B(8,5)$ .

Work out the length of the segment

\displaystyle AB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

What are the coordinates of the midpoint of $\displaystyle AB$ ?

Write the equation of the line $\displaystyle AB$ in a row of your own on the board.

\displaystyle A=(2,-3)

\displaystyle B=(8,5)

Exercise 2

The line $\displaystyle {2x-y+3}={0}$ and the point $\displaystyle P(1,4)$ .
Write the equation of the line through $\displaystyle P$ perpendicular to the given line.

\displaystyle 2x-y+3=0

\displaystyle P=(1,4)

Exercise 3

A parallel line has the same slope. First write the given line in explicit form.

Write the equation of the line through $\displaystyle Q(-2,1)$ parallel to the line $\displaystyle {3x+2y}={6}$ .

\displaystyle 3x+2y=6

\displaystyle Q=(-2,1)

Exercise 4

Find where the lines $\displaystyle {y}={2x-1}$ and $\displaystyle {x+3y}={11}$ meet.

Formulas: analytic geometry, grade 12 (5 units)

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Points and lines

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$
The slope of the line through the two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
The line with slope $\displaystyle m$ through $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1=m(x-x_1)$
Perpendicular lines – $\displaystyle m_1$ , $\displaystyle m_2$ their slopes
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The point $\displaystyle C$ dividing the segment $\displaystyle AB$ in the ratio $\displaystyle {AC:CB}={k:l}$ – $\displaystyle A(x_1,y_1)$ , $\displaystyle B(x_2,y_2)$
$\displaystyle C\left(\frac{lx_1+kx_2}{k+l},\frac{ly_1+ky_2}{k+l}\right)$
The distance of the point $\displaystyle (x_0,y_0)$ from the line $\displaystyle {Ax+By+C}={0}$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \frac{|Ax_0+By_0+C|}{\sqrt{A^{2}+B^{2}}}$
The distance between the parallel lines $\displaystyle {Ax+By+C_1}={0}$ and $\displaystyle {Ax+By+C_2}={0}$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \frac{|C_1-C_2|}{\sqrt{A^{2}+B^{2}}}$

Circle

A circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$
The expanded equation – a circle when, after completing the square, the right side is positive
$\displaystyle x^{2}+y^{2}+Cx+Dy+E=0$
The tangent to the circle at the point $\displaystyle (x_0,y_0)$ on it
$\displaystyle (x_0-a)(x-a)+(y_0-b)(y-b)=R^{2}$

Parabola

The canonical parabola
$\displaystyle y^{2}=2px$
The focus
$\displaystyle F\left(\frac{p}{2},0\right)$
The directrix
$\displaystyle x=-\frac{p}{2}$
The tangent at the point $\displaystyle (x_0,y_0)$ on the parabola
$\displaystyle y\cdot y_0=p(x+x_0)$

Ellipse

The canonical ellipse – the foci on the $\displaystyle x$ -axis when $\displaystyle a>b$
$\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
The distance of a focus from the origin
$\displaystyle c$ $\displaystyle =$ $\displaystyle \sqrt{a^{2}-b^{2}}$
The sum of the distances of a point $\displaystyle P$ on the ellipse from the foci
$\displaystyle PF_1+PF_2$ $\displaystyle =$ $\displaystyle 2a$

Hyperbola

The canonical hyperbola – the foci on the $\displaystyle x$ -axis
$\displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
The distance of a focus from the origin
$\displaystyle c$ $\displaystyle =$ $\displaystyle \sqrt{a^{2}+b^{2}}$
The difference of the distances of a point $\displaystyle P$ on the hyperbola from the foci
$\displaystyle |PF_1-PF_2|$ $\displaystyle =$ $\displaystyle 2a$
The asymptotes
$\displaystyle y=\pm\frac{b}{a}x$