Divisors — Mathslive.ie

Section 1 of 4

Splitting a segment in a ratio

A divisor is the point that cuts a segment $[PQ]$ into a given ratio.

Say a point $R$ divides $[PQ]$ internally in the ratio $a:b$ — that means $|PR|:|RQ| = a:b$.

The formula is in your tables. For $P(x_1,y_1)$ and $Q(x_2,y_2)$ split in ratio $a:b$:

$R = \left(\dfrac{b\,x_1 + a\,x_2}{a+b}, \; \dfrac{b\,y_1 + a\,y_2}{a+b}\right)$

Watch the cross-over: the $b$ multiplies the first point, the $a$ multiplies the second. Get that backwards and you land on the wrong side.

(i) Worked example

What point divides $P(1,-5)$ and $Q(6,-10)$ in the ratio $3:2$?

$a = 3, \quad b = 2$

$R_x = \dfrac{2(1) + 3(6)}{3+2} = \dfrac{2 + 18}{5} = \dfrac{20}{5} = 4$

$R_y = \dfrac{2(-5) + 3(-10)}{3+2} = \dfrac{-10 - 30}{5} = \dfrac{-40}{5} = -8$

$R = (4, -8)$

(ii) The parts method — a quick check

Ratio $3:2$ is $5$ parts. The point sits $3$ parts along, measured from $P$.

$x$: gap from $1$ to $6$ is $5$. $5$ parts $= 5$, so $1$ part $= 1$.

$3$ parts $= 3$, so $R_x = 1 + 3 = 4$ ✓

$y$: gap from $-5$ to $-10$ is $-5$. $1$ part $= -1$, $3$ parts $= -3$.

$R_y = -5 + (-3) = -8$ ✓

Must learn

1.$R = \left(\dfrac{b\,x_1 + a\,x_2}{a+b}, \dfrac{b\,y_1 + a\,y_2}{a+b}\right)$ for ratio $a:b$.

2.Cross-multiply: $b$ on the first point, $a$ on the second.

3.Check with parts — move $a$ parts of the gap on from $P$.

YOU TRY · 1

Find the point dividing $[PQ]$ in ratio $1:1$, where $P(2,4)$ and $Q(8,10)$.

$1:1$ is just the midpoint — but do it with the formula.

$a=1,\ b=1$

$x = \dfrac{1(2)+1(8)}{2} = 5$

$y = \dfrac{1(4)+1(10)}{2} = 7$

$(5, 7)$

YOU TRY · 2

Find the point dividing $[AB]$ in ratio $2:3$, where $A(1,2)$ and $B(11,17)$.

$a=2,\ b=3$. Mind the cross-over.

$x = \dfrac{3(1)+2(11)}{5} = \dfrac{25}{5} = 5$

$y = \dfrac{3(2)+2(17)}{5} = \dfrac{40}{5} = 8$

$(5, 8)$

YOU TRY · 3

Find the point dividing $[PQ]$ in ratio $3:1$, where $P(-3,4)$ and $Q(9,-8)$.

$a=3,\ b=1$. The point lands close to $Q$.

$x = \dfrac{1(-3)+3(9)}{4} = \dfrac{24}{4} = 6$

$y = \dfrac{1(4)+3(-8)}{4} = \dfrac{-20}{4} = -5$

$(6, -5)$

Section 2 of 4

Finding a missing endpoint

Sometimes you are given the dividing point and one end, and asked for the other end.

Same formula — you just leave the unknown end as $x$ and $y$ and solve.

(i) Worked example

$A(2,-5)$ divides $P(8,-29)$ and $Q$ in the ratio $5:1$. Find $Q$.

$P(8,-29)$ is the first point, $Q(x,y)$ is the second. $a=5,\ b=1$.

$x$-equation: $\dfrac{1(8) + 5x}{6} = 2$

$8 + 5x = 12 \;\Rightarrow\; 5x = 4 \;\Rightarrow\; x = \dfrac{4}{5}$

$y$-equation: $\dfrac{1(-29) + 5y}{6} = -5$

$-29 + 5y = -30 \;\Rightarrow\; 5y = -1 \;\Rightarrow\; y = -\dfrac{1}{5}$

$Q = \left(\dfrac{4}{5}, \; -\dfrac{1}{5}\right)$

(ii) The parts check

Walk from $P$ to $A$ — that journey is the first $5$ parts. One more part takes you to $Q$.

$x$: gap $8 \to 2$ is $-6$. $5$ parts $= -6$, so $1$ part $= -\dfrac{6}{5}$.

$Q_x = 2 - \dfrac{6}{5} = \dfrac{4}{5}$ ✓

$y$: gap $-29 \to -5$ is $24$. $5$ parts $= 24$, so $1$ part $= \dfrac{24}{5}$.

$Q_y = -5 + \dfrac{24}{5} = -\dfrac{1}{5}$ ✓

YOU TRY · 4

The point $(4,3)$ divides $[AB]$ in ratio $1:2$, where $A(1,0)$. Find $B$.

$A$ is the first point, $B(x,y)$ the second. $a=1,\ b=2$.

$\dfrac{2(1)+1\cdot x}{3} = 4 \Rightarrow x = 10$

$\dfrac{2(0)+1\cdot y}{3} = 3 \Rightarrow y = 9$

$B = (10, 9)$

YOU TRY · 5

The point $(0,1)$ divides $[PQ]$ in ratio $1:3$, where $P(-3,-2)$. Find $Q$.

$a=1,\ b=3$. The dividing point is close to $P$.

$\dfrac{3(-3)+1\cdot x}{4} = 0 \Rightarrow x = 9$

$\dfrac{3(-2)+1\cdot y}{4} = 1 \Rightarrow y = 10$

$Q = (9, 10)$

Section 3 of 4

Finding the ratio

If you know all three points, you can read the ratio straight off the gaps.

No formula needed. Compare the gap from $P$ to the point with the gap from the point to $Q$.

(i) Worked example

$A(3,5)$ divides $P(8,0)$ and $Q(0,8)$ in what ratio?

Track the $x$-coordinate from $P$ through $A$ to $Q$:

$8 \;\longrightarrow\; 3 \;\longrightarrow\; 0$

$|PA|$ gap $= 8 - 3 = 5$, $|AQ|$ gap $= 3 - 0 = 3$.

$|PA| : |AQ| = 5 : 3$

The $y$-coordinates give the same answer: $0 \to 5 \to 8$, gaps $5$ and $3$. Always worth a quick double-check.

YOU TRY · 6

In what ratio does $A(4,7)$ divide $[PQ]$, where $P(1,1)$ and $Q(10,19)$?

Gaps on $x$: $1 \to 4 \to 10$. Then simplify.

$|PA| = 4-1 = 3$

$|AQ| = 10-4 = 6$

$3 : 6$

$1 : 2$

YOU TRY · 7

In what ratio does $A(0,1)$ divide $[PQ]$, where $P(-4,-7)$ and $Q(2,5)$?

$x$: $-4 \to 0 \to 2$. Mind the signs in the gaps.

$|PA| = 0-(-4) = 4$

$|AQ| = 2-0 = 2$

$4 : 2$

$2 : 1$

Section 4 of 4

Extending a segment

Now the point lands beyond the segment. The middle point becomes the divisor of the whole run.

(i) Worked example

$A(1,5)$ and $B(6,-5)$ are extended to $P$ such that $|AB| : |BP| = 5:6$. Find $P$.

So $B$ divides $[AP]$ in the ratio $5:6$. Put $A$ first, $P(x,y)$ second, $a=5,\ b=6$.

$x$-equation: $\dfrac{6(1) + 5x}{11} = 6$

$6 + 5x = 66 \;\Rightarrow\; 5x = 60 \;\Rightarrow\; x = 12$

$y$-equation: $\dfrac{6(5) + 5y}{11} = -5$

$30 + 5y = -55 \;\Rightarrow\; 5y = -85 \;\Rightarrow\; y = -17$

$P = (12, -17)$

(ii) The parts check

From $A$ to $B$ is $5$ parts. From $B$ to $P$ is $6$ more parts.

$x$: gap $1 \to 6$ is $5$. $5$ parts $= 5$, so $1$ part $= 1$, $6$ parts $= 6$.

$P_x = 6 + 6 = 12$ ✓

$y$: gap $5 \to -5$ is $-10$. $1$ part $= -2$, $6$ parts $= -12$.

$P_y = -5 + (-12) = -17$ ✓

YOU TRY · 8

$A(2,1)$ and $B(5,7)$ are extended to $P$ such that $|AB| : |BP| = 1:2$. Find $P$.

$B$ divides $[AP]$ in $1:2$. $A$ first, $P(x,y)$ second.

$\dfrac{2(2)+1\cdot x}{3} = 5 \Rightarrow x = 11$

$\dfrac{2(1)+1\cdot y}{3} = 7 \Rightarrow y = 19$

$P = (11, 19)$

YOU TRY · 9

$A(-1,0)$ and $B(3,2)$ are extended to $P$ such that $|AB| : |BP| = 2:3$. Find $P$.

$a=2,\ b=3$. Parts check: $A \to B$ is $2$ parts.

$\dfrac{3(-1)+2x}{5} = 3 \Rightarrow x = 9$

$\dfrac{3(0)+2y}{5} = 2 \Rightarrow y = 5$

$P = (9, 5)$

SUM

The lot in one box

Divisors toolkit

1.Ratio $a:b$ $\to$ $\left(\dfrac{b\,x_1 + a\,x_2}{a+b}, \dfrac{b\,y_1 + a\,y_2}{a+b}\right)$.

2.Cross-over: $b$ on the first point, $a$ on the second.

3.Missing end: put the unknown as $x,y$ and solve each equation.

4.Missing ratio: read the gaps, $|PA| : |AQ|$, then simplify.

5.Extended: the middle point divides the whole segment $[AP]$.

6.Always sanity-check with the parts method.

End of lesson

Divisors — HL · Mathslive.ie