Dividing a segment in a given ratio

Analytic geometry, grade 12 (5 units) Lesson 2 of 21Learning

Dividing a segment in a given ratio

Dividing internally in the ratio

\displaystyle k:l

The point $\displaystyle C$ is on the segment $\displaystyle AB$ , and $\displaystyle {AC:CB}={k:l}$ . We say that $\displaystyle C$ divides the segment $\displaystyle AB$ in the ratio $\displaystyle k:l$ (internally – $\displaystyle C$ is between $\displaystyle A$ and $\displaystyle B$ ).

Cut $\displaystyle AB$ into $\displaystyle k+l$ equal parts: $\displaystyle AC$ is $\displaystyle k$ of them, and $\displaystyle CB$ is $\displaystyle l$ of them.
So $\displaystyle C$ is $\displaystyle \frac{k}{k+l}$ of the way from $\displaystyle A$ to $\displaystyle B$ .
By Thales’ theorem, the projections on the axes are divided in the same ratio: the difference of the $\displaystyle x$ -coordinates and the difference of the $\displaystyle y$ -coordinates both go $\displaystyle \frac{k}{k+l}$ of the way.

For $\displaystyle A(x_1,y_1)$ and $\displaystyle B(x_2,y_2)$ :

x_C=x_1+\frac{k}{k+l}(x_2-x_1)=\frac{lx_1+kx_2}{k+l}

and $\displaystyle y_C$ the same way. The formula, as on the formula sheet:

\displaystyle C\left(\frac{lx_1+kx_2}{k+l},\frac{ly_1+ky_2}{k+l}\right)

Note: $\displaystyle l$ , the part far from $\displaystyle A$ , multiplies $\displaystyle A$ ’s coordinate. The midpoint is the case $\displaystyle {k}={l}={1}$ .

An example: the ratio

\displaystyle 1:2

$\displaystyle A(1,2)$ and $\displaystyle B(7,-7)$ .
The point $\displaystyle C$ divides $\displaystyle AB$ in the ratio $\displaystyle {AC:CB}={1:2}$ .

Here $\displaystyle {k}={1}$ and $\displaystyle {l}={2}$ :

x_C=\frac{2\cdot1+1\cdot7}{1+2}=\frac{9}{3}=3,\qquad y_C=\frac{2\cdot2+1\cdot(-7)}{1+2}=\frac{-3}{3}=-1

$\displaystyle C(3,-1)$ – a third of the way from $\displaystyle A$ to $\displaystyle B$ : $\displaystyle x$ went up $\displaystyle 2$ out of $\displaystyle 6$ , and $\displaystyle y$ went down $\displaystyle 3$ out of $\displaystyle 9$ .

\displaystyle y=-\frac{3}{2}x+\frac{7}{2}

\displaystyle A=(1,2)

\displaystyle C=(3,-1)

\displaystyle B=(7,-7)

Where the medians meet

In grade 10 we saw that the medians’ meeting point $\displaystyle G$ divides each median in the ratio $\displaystyle 2:1$ from the vertex. The dividing formula turns this into $\displaystyle G$ ’s coordinates.

In the triangle $\displaystyle A(x_1,y_1)$ , $\displaystyle B(x_2,y_2)$ , $\displaystyle C(x_3,y_3)$ , the midpoint of $\displaystyle BC$ is $\displaystyle M\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$ , and $\displaystyle G$ divides $\displaystyle AM$ in the ratio $\displaystyle 2:1$ :

x_G=\frac{1\cdot x_1+2\cdot\frac{x_2+x_3}{2}}{2+1}=\frac{x_1+x_2+x_3}{3}

The medians’ meeting point is the mean of the three vertices:

\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)

Exercise 1

$\displaystyle A(-2,5)$ and $\displaystyle B(10,-1)$ .
The point $\displaystyle C$ divides $\displaystyle AB$ in the ratio $\displaystyle {AC:CB}={1:3}$ .

Work out

\displaystyle x_C

/ – 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

Work out

\displaystyle y_C

/ – 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

Exercise 2

The triangle’s vertices: $\displaystyle A(1,1)$ , $\displaystyle B(7,3)$ , $\displaystyle C(4,8)$ .
Find the medians’ meeting point $\displaystyle G$ .

\displaystyle x_G

/ – 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

\displaystyle y_G

/ – 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

Exercise 3

Write the dividing formula with

\displaystyle x

in place of

\displaystyle x_B

, and make it equal to

\displaystyle C

’s known coordinate.

$\displaystyle A(2,3)$ .
The point $\displaystyle C(4,1)$ divides the segment $\displaystyle AB$ in the ratio $\displaystyle {AC:CB}={1:2}$ .
Find the coordinates of the point $\displaystyle B$ .

An equation for

\displaystyle x_B

An equation for

\displaystyle y_B

Exercise 4

$\displaystyle A(-3,-1)$ and $\displaystyle B(6,5)$ .
The $\displaystyle y$ -axis cuts the segment $\displaystyle AB$ at the point $\displaystyle D$ .

In which ratio does $\displaystyle D$ divide $\displaystyle AB$ ? Use the $\displaystyle x$ -coordinates.

Find $\displaystyle y_D$ , and mark $\displaystyle D$ on the board.

\displaystyle y=\frac{2}{3}x+1

\displaystyle A=(-3,-1)

\displaystyle B=(6,5)

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
$\displaystyle y=\pm\frac{b}{a}x$