Chapter 7 · Coordinate Geometry

7.2 The Section Formula

Where is the point that sits three-fifths of the way from $A$ to $B$? Where do the two trisection points of a road lie, and where does a segment cross an axis? One formula answers them all: weight each coordinate by the ratio.

A road runs in a straight line from town $A$ to town $B$, and the district office wants to build a petrol pump on it, exactly $\tfrac34$ of the way from $A$ to $B$. On the map the towns have known coordinates. How do we find the coordinates of the pump? Averaging the coordinates would give the middle of the road, which is $\tfrac12$ of the way. We need an average in which $B$ counts three times as much as $A$: a weighted average.

In this section we make that precise. We say what it means for a point to divide a segment in a ratio, prove the section formula with similar triangles, and deduce the midpoint formula. Then we use it for the standard jobs: the two trisection points of a segment, the ratio in which one of the axes (or a given point) divides a segment, and the missing endpoint of a segment when its midpoint is known. By the end you will have placed the petrol pump.

Dividing a segment in a ratio

Imagine a rope $AB$ and a knot $P$ on it. If the part $AP$ is $3$ m long and the part $PB$ is $2$ m long, we say the knot divides the rope in the ratio $3:2$. Only the comparison matters: $6$ m and $4$ m would be the same ratio. The whole rope is then cut into $3+2=5$ equal pieces, and the knot is placed after the third.

📘 Definition 7.3 — Internal division in a ratio

A point $P$ divides the segment $AB$ internally in the ratio $m_1:m_2$ (where $m_1\gt0$ and $m_2\gt0$) if $P$ lies on the segment $AB$ between $A$ and $B$ and $$AP:PB=m_1:m_2,\qquad\text{that is}\qquad AP=\frac{m_1}{m_1+m_2}\,AB.$$ The order matters: $AP:PB=2:3$ puts $P$ closer to $A$, while $AP:PB=3:2$ puts it closer to $B$. If $m_1=m_2$, then $P$ is the midpoint of $AB$.

A good habit before using any formula: guess where $P$ should be. With $AP:PB=1:3$ the point is one quarter of the way from $A$, so each of its coordinates should lie one quarter of the way from the coordinate of $A$ to the coordinate of $B$.

The section formula

📐 Theorem 7.3 — The Section Formula (internal division)

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points, and let $P(x, y)$ divide $AB$ internally in the ratio $m_1:m_2$. Then $$x=\frac{m_1x_2+m_2x_1}{m_1+m_2},\qquad y=\frac{m_1y_2+m_2y_1}{m_1+m_2}.$$

Proof. Suppose $x_1\lt x_2$ and $y_1\lt y_2$ (the other positions are handled in exactly the same way). Through $A$ draw a horizontal line and through $P$ a vertical line; they meet at $R(x, y_1)$. Through $P$ draw a horizontal line and through $B$ a vertical line; they meet at $S(x_2, y)$. Then $\triangle ARP$ has a right angle at $R$, and $\triangle PSB$ has a right angle at $S$.

The lines $AR$ and $PS$ are both horizontal, hence parallel, and $AB$ cuts them, so $\angle PAR=\angle BPS$ (corresponding angles). Both triangles also have a right angle, so $\triangle ARP\sim\triangle PSB$ by the AA criterion (Section 6.3). Corresponding sides are proportional:

$$\frac{AR}{PS}=\frac{RP}{SB}=\frac{AP}{PB}=\frac{m_1}{m_2}.$$

Here $AR=x-x_1$ and $PS=x_2-x$, so $\dfrac{x-x_1}{x_2-x}=\dfrac{m_1}{m_2}$. Cross-multiplying, $m_2(x-x_1)=m_1(x_2-x)$, that is $(m_1+m_2)\,x=m_1x_2+m_2x_1$, which gives the formula for $x$. In the same way $RP=y-y_1$ and $SB=y_2-y$, and $\dfrac{y-y_1}{y_2-y}=\dfrac{m_1}{m_2}$ gives the formula for $y$. $\blacksquare$

📐 Theorem 7.4 — The Midpoint Formula

The midpoint of the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ is $$M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right).$$

This is Theorem 7.3 with $m_1=m_2=1$: the midpoint is the plain average of the coordinates. Notice how the formula behaves: it is a weighted average in which $B$'s coordinate is weighted by $m_1$ (the number next to $A$ in the ratio, because a bigger $m_1$ pushes $P$ further from $A$, towards $B$) and $A$'s coordinate by $m_2$.

⚠️ Common mistakes

1. Attaching the weights to the wrong ends. The weight $m_1$ multiplies $x_2$ (and $y_2$), not $x_1$. Quick test: for $AP:PB=1:3$ the point must be close to $A$; the wrong version would put it close to $B$.

2. Reading the ratio in the wrong order. "$P$ divides $AB$ in the ratio $2:3$" means $AP:PB=2:3$. If a problem says $PB:AP$ or $BP:PA$, swap before you substitute.

3. Dividing by the wrong total. The denominator is $m_1+m_2$ (the number of equal parts), not $m_1$ or $m_2$ alone.

4. Using the midpoint formula for other ratios, or using the section formula with ratio $1:2$ when the question says the point is "$\tfrac13$ of the way" — that is $AP:PB=1:2$, because the other part is $\tfrac23$.

✏️ Example 1 — Dividing a segment

(a) Find the point $P$ that divides the segment joining $A(-5, 4)$ and $B(10, -1)$ internally in the ratio $2:3$. (b) Find the point $Q$ on the segment joining $A(2, 1)$ and $B(10, 9)$ with $AQ:QB=3:1$.

💡 Show solution▶

(a) Set up $(x_1, y_1)=(-5, 4)$, $(x_2, y_2)=(10, -1)$, $m_1=2$, $m_2=3$, $m_1+m_2=5$. Expect $P$ closer to $A$.

(a) Substitute $x=\dfrac{2\times10+3\times(-5)}{5}=\dfrac{20-15}{5}=1$, $\;y=\dfrac{2\times(-1)+3\times4}{5}=\dfrac{-2+12}{5}=2$. So $P(1, 2)$.

(b) $m_1=3$, $m_2=1$: $\;x=\dfrac{3\times10+1\times2}{4}=8$, $\;y=\dfrac{3\times9+1\times1}{4}=7$. So $Q(8, 7)$, three-quarters of the way from $A$.

Check In (a), $A\to P$ is $(6, -2)$ and $P\to B$ is $(9, -3)$; $6:9=2:3$ and $-2:-3=2:3$ ✓. In (b), $A\to Q$ is $(6, 6)$ and $Q\to B$ is $(2, 2)$; $6:2=3:1$ ✓.

✏️ Example 2 — Midpoint, and the other end

(a) Find the midpoint of the segment joining $(-6, 3)$ and $(4, 9)$. (b) $M(2, -1)$ is the midpoint of $AB$ and $A$ is $(-3, 5)$. Find $B$.

💡 Show solution▶

(a) $\left(\dfrac{-6+4}{2},\ \dfrac{3+9}{2}\right)=(-1, 6)$.

(b) Equations Let $B=(p, q)$. The midpoint formula gives $\dfrac{-3+p}{2}=2$ and $\dfrac{5+q}{2}=-1$.

(b) Solve $-3+p=4$ gives $p=7$; $\;5+q=-2$ gives $q=-7$. So $B(7, -7)$.

Check The midpoint of $(-3, 5)$ and $(7, -7)$ is $\left(\tfrac{4}{2}, \tfrac{-2}{2}\right)=(2, -1)$ ✓. Shortcut: $B$ is as far beyond $M$ as $A$ is behind it, so $B=2M-A$.

The segment $AB$ is cut into $m_1+m_2$ equal pieces. The first $m_1$ pieces (from $A$) are blue, the remaining $m_2$ pieces are red, and the green point $P$ sits where the colours change. Change the ratio with the sliders, choose a pair of points, or press Play to cycle through ratios. The readouts apply the section formula.

Points:$m_1$:2$m_2$:3

$x=\tfrac{m_1x_2+m_2x_1}{m_1+m_2}$—

$y=\tfrac{m_1y_2+m_2y_1}{m_1+m_2}$—

$AP:PB$—

✏️ Example 3 — The trisection points

Find the points $P$ and $Q$ that divide the segment joining $A(-3, 6)$ and $B(9, -3)$ into three equal parts.

💡 Show solution▶

Ratios Let $P$ be nearer $A$ and $Q$ nearer $B$. Then $AP:PB=1:2$ (one part against two) and $AQ:QB=2:1$.

P $m_1=1$, $m_2=2$: $\;x=\dfrac{1\times9+2\times(-3)}{3}=1$, $\;y=\dfrac{1\times(-3)+2\times6}{3}=3$. So $P(1, 3)$.

Q $m_1=2$, $m_2=1$: $\;x=\dfrac{2\times9+1\times(-3)}{3}=5$, $\;y=\dfrac{2\times(-3)+1\times6}{3}=0$. So $Q(5, 0)$.

Check $Q$ is the midpoint of $PB$: $\left(\tfrac{1+9}{2}, \tfrac{3-3}{2}\right)=(5, 0)$ ✓. Also $AP=\sqrt{16+9}=5$, $PQ=\sqrt{16+9}=5$, $QB=\sqrt{16+9}=5$, three equal parts ✓. Notice that $Q$ lies on the $x$-axis.

Ratios, axes and missing endpoints

Often the question runs the other way: the point is known (or known to lie on an axis) and we must find the ratio. Since we do not know $m_1$ and $m_2$ separately, only their comparison, we write the ratio as $k:1$, where $k=m_1/m_2$. The section formula then has the single unknown $k$.

🧭 Problem-solving strategy — find the ratio

Let the point divide $AB$ in the ratio $k:1$, so $P=\left(\dfrac{kx_2+x_1}{k+1},\ \dfrac{ky_2+y_1}{k+1}\right)$.
Write what you know about $P$: on the $x$-axis means its $y$-coordinate is $0$; on the $y$-axis means its $x$-coordinate is $0$; a given point means both coordinates are known.
Solve for $k$. For the $x$-axis: $ky_2+y_1=0$ gives $k=-\dfrac{y_1}{y_2}$. For the $y$-axis: $k=-\dfrac{x_1}{x_2}$.
Check that $k\gt0$. If $k$ is negative, the axis does not cut the segment between $A$ and $B$ — both ends lie on the same side of the axis.
Substitute $k$ back to find the other coordinate of $P$, and write the ratio as whole numbers.

✏️ Example 4 — Where does an axis cut the segment?

(a) In what ratio does the $x$-axis divide the segment joining $A(1, -3)$ and $B(4, 6)$? Find the point. (b) In what ratio does the $y$-axis divide the segment joining $A(-2, 5)$ and $B(6, -3)$? Find the point. (c) In what ratio does $P(2, 1)$ divide the segment joining $A(-2, 3)$ and $B(8, -2)$?

💡 Show solution▶

(a) Ratio Let the ratio be $k:1$. The $y$-coordinate of the point is $\dfrac{6k-3}{k+1}=0$, so $k=\dfrac12$. The ratio is $\dfrac12:1=1:2$.

(a) Point $x=\dfrac{1\times4+2\times1}{3}=2$, so the point is $(2, 0)$. Check the $y$-coordinate: $\dfrac{1\times6+2\times(-3)}{3}=0$ ✓.

(b) The $x$-coordinate is $\dfrac{6k-2}{k+1}=0$, so $k=\dfrac13$ and the ratio is $1:3$. The $y$-coordinate is $\dfrac{1\times(-3)+3\times5}{4}=3$. The point is $(0, 3)$.

(c) Use the $x$-coordinates: $2=\dfrac{8k-2}{k+1}$ gives $2k+2=8k-2$, so $k=\dfrac23$ and the ratio is $2:3$. Use the $y$-coordinates to check: $\dfrac{-2k+3}{k+1}=\dfrac{-4/3+3}{5/3}=\dfrac{5/3}{5/3}=1$ ✓. (If the two coordinates gave different values of $k$, then $P$ would not be on the segment at all.)

The point $Q$ slides from $A$ to $B$. The red number is the coordinate that has to become $0$ for $Q$ to be on the axis. Slide until it reaches $0$ (the verdict turns green), or press Snap to the axis. The ratio $AQ:QB$ at that moment is the ratio found by $k=-y_1/y_2$ (or $-x_1/x_2$). The last pair never crosses the axis: both ends are on the same side.

Segment:position of $Q$:

$Q$—

coordinate to make 0—

$AQ:QB$—

✏️ Example 5 — The fourth corner of a parallelogram

Three vertices of the parallelogram $ABCD$ (taken in this order) are $A(1, 2)$, $B(5, 3)$ and $C(7, 8)$. Find the fourth vertex $D$.

💡 Show solution▶

Idea The diagonals of a parallelogram bisect each other, so the diagonals $AC$ and $BD$ have the same midpoint.

Midpoint of AC $\left(\dfrac{1+7}{2}, \dfrac{2+8}{2}\right)=(4, 5)$.

Find D Let $D=(p, q)$. The midpoint of $BD$ is $\left(\dfrac{5+p}{2}, \dfrac{3+q}{2}\right)=(4, 5)$, so $p=3$ and $q=7$. Hence $D(3, 7)$.

Check $AB^2=16+1=17$ and $DC^2=16+1=17$; $BC^2=4+25=29$ and $AD^2=4+25=29$. Opposite sides are equal ✓.

✏️ Example 6 — The petrol pump

On a map (1 unit = 1 km) towns $A(2, 3)$ and $B(14, 11)$ are joined by a straight road. The pump is to stand $\tfrac34$ of the way from $A$ to $B$. Find its coordinates and its distance from $A$.

💡 Show solution▶

Ratio $AP=\tfrac34AB$ and $PB=\tfrac14AB$, so $AP:PB=3:1$.

Coordinates $x=\dfrac{3\times14+1\times2}{4}=11$, $\;y=\dfrac{3\times11+1\times3}{4}=9$. The pump is at $P(11, 9)$.

Distance $AP=\sqrt{(11-2)^2+(9-3)^2}=\sqrt{81+36}=\sqrt{117}\approx10.82$ km.

Check $AB=\sqrt{12^2+8^2}=\sqrt{208}\approx14.42$ km, and $\tfrac34\times14.42\approx10.82$ km ✓.

Exercises

Find the point that divides $AB$ internally in the given ratio.

1 $A(2, 4)$, $B(8, 10)$; the midpoint.
2 $A(-1, 3)$, $B(9, -7)$; ratio $2:3$.
3 $A(4, -2)$, $B(-6, 8)$; ratio $3:2$.
4 $A(0, 0)$, $B(7, -14)$; ratio $5:2$.
5 The midpoint of the segment joining $(3a, 2b)$ and $(a, 4b)$.
6 $A(-5, -2)$, $B(11, 6)$; $AP:PB=1:3$.
7 The two trisection points of the segment joining $A(1, -4)$ and $B(10, 5)$.
8 The three points that divide the segment joining $A(-4, 2)$ and $B(12, 10)$ into four equal parts.

Check: type $A$, $B$ and the ratio $m_1:m_2$ — the dividing point

$x_1$$y_1$$x_2$$y_2$$m_1$$m_2$

$x$—

$y$—

$P$—

💡 Show solutions▶

1 $\left(\dfrac{2+8}{2}, \dfrac{4+10}{2}\right)=(5, 7)$.

2 $x=\dfrac{2\times9+3\times(-1)}{5}=3$, $\;y=\dfrac{2\times(-7)+3\times3}{5}=-1$. $P(3, -1)$.

3 $x=\dfrac{3\times(-6)+2\times4}{5}=-2$, $\;y=\dfrac{3\times8+2\times(-2)}{5}=4$. $P(-2, 4)$.

4 $x=\dfrac{5\times7+2\times0}{7}=5$, $\;y=\dfrac{5\times(-14)+0}{7}=-10$. $P(5, -10)$.

5 $\left(\dfrac{3a+a}{2}, \dfrac{2b+4b}{2}\right)=(2a, 3b)$.

6 $x=\dfrac{1\times11+3\times(-5)}{4}=-1$, $\;y=\dfrac{1\times6+3\times(-2)}{4}=0$. $P(-1, 0)$ (it lies on the $x$-axis).

7 Ratio $1:2$: $\left(\dfrac{10+2}{3}, \dfrac{5-8}{3}\right)=(4, -1)$. Ratio $2:1$: $\left(\dfrac{20+1}{3}, \dfrac{10-4}{3}\right)=(7, 2)$.

8 The middle point is the midpoint $(4, 6)$. The others have ratios $1:3$ and $3:1$: $\left(\dfrac{12+3(-4)}{4}, \dfrac{10+6}{4}\right)=(0, 4)$ and $\left(\dfrac{36-4}{4}, \dfrac{30+2}{4}\right)=(8, 8)$. The three points are $(0, 4)$, $(4, 6)$, $(8, 8)$, equally spaced ✓.

Ratios, axes and missing endpoints.

1 $M(4, -1)$ is the midpoint of $AB$ and $A$ is $(-2, 5)$. Find $B$.
2 The centre of a circle is $(2, -3)$ and one end of a diameter is $(6, 1)$. Find the other end.
3 In what ratio does the $x$-axis divide the segment joining $(-4, -6)$ and $(5, 3)$? Find the point of division.
4 In what ratio does the $y$-axis divide the segment joining $(-3, 4)$ and $(9, -2)$? Find the point of division.
5 In what ratio does $P(5, 4)$ divide the segment joining $A(1, 2)$ and $B(7, 5)$?
6 The point $(5, y)$ lies on the segment joining $A(1, 3)$ and $B(7, 9)$. Find the ratio in which it divides $AB$, and $y$.
7 Three vertices of a parallelogram $ABCD$ are $A(-2, -1)$, $B(1, 0)$ and $C(4, 3)$. Find $D$.
8 Show that the diagonals of the quadrilateral $A(3, -1)$, $B(7, 3)$, $C(5, 7)$, $D(1, 3)$ have the same midpoint, and hence that $ABCD$ is a parallelogram. (Verify with the side lengths.)

💡 Show solutions▶

1 $B=2M-A=(8+2, -2-5)=(10, -7)$. Check: midpoint of $(-2, 5)$ and $(10, -7)$ is $(4, -1)$ ✓.

2 The centre is the midpoint of a diameter: $(2\times2-6,\ 2\times(-3)-1)=(-2, -7)$.

3 $k=-\dfrac{y_1}{y_2}=-\dfrac{-6}{3}=2$, so the ratio is $2:1$. The $x$-coordinate is $\dfrac{2\times5+1\times(-4)}{3}=2$. Point $(2, 0)$.

4 $k=-\dfrac{x_1}{x_2}=-\dfrac{-3}{9}=\dfrac13$, so the ratio is $1:3$. The $y$-coordinate is $\dfrac{1\times(-2)+3\times4}{4}=\dfrac52$. Point $\left(0, \dfrac52\right)$.

5 $5=\dfrac{7k+1}{k+1}$ gives $5k+5=7k+1$, so $k=2$: the ratio is $2:1$. Check with $y$: $\dfrac{2\times5+2}{3}=4$ ✓.

6 $5=\dfrac{7k+1}{k+1}$ again gives $k=2$, ratio $2:1$. Then $y=\dfrac{2\times9+3}{3}=7$. The point is $(5, 7)$.

7 Midpoint of $AC$ is $(1, 1)$. This is also the midpoint of $BD$: $\dfrac{1+p}{2}=1$, $\dfrac{0+q}{2}=1$ give $D(1, 2)$. Check: $AB^2=9+1=10=DC^2=9+1$ ✓.

8 Midpoint of $AC$: $\left(\tfrac{3+5}{2}, \tfrac{-1+7}{2}\right)=(4, 3)$; of $BD$: $\left(\tfrac{7+1}{2}, \tfrac{3+3}{2}\right)=(4, 3)$. Same midpoint, so the diagonals bisect each other: a parallelogram. Sides: $AB^2=32$, $BC^2=4+16=20$, $CD^2=16+16=32$, $DA^2=4+16=20$, opposite sides equal ✓.

III

Harder problems.

1 $P(2, 3)$ divides $AB$ in the ratio $1:3$ and $A$ is $(-2, 0)$. Find $B$.
2 Find the point $P$ on the segment joining $A(-3, 4)$ and $B(7, -6)$ with $AP=\tfrac25AB$.
3 The vertices of a right-angled triangle are $O(0, 0)$, $B(8, 0)$ and $C(0, 6)$. Find the midpoint $M$ of the hypotenuse $BC$ and show that $MO=MB=MC$.
4 In what ratio does the $y$-axis divide the segment joining $A(-4, 8)$ and $B(6, -2)$? Find the point.
5 The midpoint of the segment joining $A(p, 4)$ and $B(-2, q)$ is $M(3, 5)$. Find $p$ and $q$.
6 (Challenge) $A(1, 1)$ and $C(7, 5)$ are opposite corners of a square $ABCD$. Find $B$ and $D$.

💡 Show solutions▶

1 Let $B=(p, q)$: $\dfrac{1\times p+3\times(-2)}{4}=2$ gives $p=14$; $\dfrac{1\times q+3\times0}{4}=3$ gives $q=12$. $B(14, 12)$. Check: $\left(\tfrac{14-6}{4}, \tfrac{12+0}{4}\right)=(2, 3)$ ✓.

2 $AP=\tfrac25AB$ means $PB=\tfrac35AB$, so $AP:PB=2:3$. $x=\dfrac{2\times7+3\times(-3)}{5}=1$, $\;y=\dfrac{2\times(-6)+3\times4}{5}=0$. $P(1, 0)$.

3 $M=\left(\tfrac{8+0}{2}, \tfrac{0+6}{2}\right)=(4, 3)$. $MO=\sqrt{16+9}=5$, $MB=\sqrt{16+9}=5$, $MC=\sqrt{16+9}=5$. The midpoint of the hypotenuse is equidistant from all three vertices (and $BC=10$).

4 $k=-\dfrac{x_1}{x_2}=\dfrac46=\dfrac23$: ratio $2:3$. $y=\dfrac{2\times(-2)+3\times8}{5}=4$. Point $(0, 4)$.

5 $\dfrac{p-2}{2}=3$ gives $p=8$; $\dfrac{4+q}{2}=5$ gives $q=6$.

6 The diagonals of a square bisect each other: the centre is $M(4, 3)$, and $B$, $D$ are at the same distance from $M$ as $A$ is, $MA^2=9+4=13$. $B(x, y)$ is equidistant from $A$ and $C$: $(x-1)^2+(y-1)^2=(x-7)^2+(y-5)^2$ gives $12x+8y=72$, i.e. $3x+2y=18$, so $y-3=-\tfrac32(x-4)$. Then $MB^2=(x-4)^2\left(1+\tfrac94\right)=13$ gives $(x-4)^2=4$, so $x=6$ or $2$, with $y=0$ or $6$. Take $B(2, 6)$; then $D=2M-B=(6, 0)$. Check: $AB^2=1+25=26$, $BC^2=25+1=26$, $AC^2=BD^2=52$ ✓.

Applications.

1 Rest stop. A highway runs from $A(10, 20)$ to $B(70, 100)$ on a map (1 unit = 1 km). A rest stop is to be built $\tfrac25$ of the way from $A$ to $B$. Find its position.
2 Cricket field. Taking the batter as origin (units in metres), mid-off stands at $(-18, 6)$ and mid-on at $(12, -14)$. A third fielder stands halfway between them. Where?
3 Cable car. A cable runs from the lower station $A(0, 50)$ to the upper station $B(1200, 450)$ ($x$ = horizontal distance, $y$ = height, both in metres). Three support pylons divide the cable into four equal parts. Find their positions.
4 Garden. A rectangular garden has corners $A(0, 0)$, $B(20, 0)$, $C(20, 12)$, $D(0, 12)$ (metres). A sprinkler is at the point where the diagonals meet, and a tap is at the midpoint of $BC$. Find both points and the distance between them.
5 Triangular plot. A plot has corners $A(0, 0)$, $B(8, 2)$, $C(4, 10)$. Find the midpoints $D$ of $BC$, $E$ of $CA$ and $F$ of $AB$, and verify that $DE=\tfrac12AB$.
6 Bus stop. A new bus stop $S$ is to be placed on the road from $P(-4, -1)$ to $Q(8, 5)$ so that $PS:SQ=5:1$. Find $S$.

💡 Show solutions▶

1 $AP:PB=2:3$. $x=\dfrac{2\times70+3\times10}{5}=34$, $\;y=\dfrac{2\times100+3\times20}{5}=52$. The rest stop is at $(34, 52)$.

2 Midpoint: $\left(\tfrac{-18+12}{2}, \tfrac{6-14}{2}\right)=(-3, -4)$, i.e. $3$ m to the left of and $4$ m behind the batter.

3 Ratios $1:3$, $2:2$, $3:1$. The pylons are at $\left(\tfrac{1200}{4}, 50+\tfrac{400}{4}\right)=(300, 150)$, $(600, 250)$ and $(900, 350)$, each $300$ m further along and $100$ m higher than the one before.

4 The diagonals of a rectangle bisect each other, so the sprinkler is at the midpoint of $AC$: $(10, 6)$. The tap is at $\left(\tfrac{20+20}{2}, \tfrac{0+12}{2}\right)=(20, 6)$. Distance $=\sqrt{10^2+0^2}=10$ m.

5 $D=(6, 6)$, $E=(2, 5)$, $F=(4, 1)$. $DE=\sqrt{16+1}=\sqrt{17}$ and $AB=\sqrt{64+4}=\sqrt{68}=2\sqrt{17}$, so $DE=\tfrac12AB$ ✓.

6 $m_1=5$, $m_2=1$: $x=\dfrac{5\times8+1\times(-4)}{6}=6$, $\;y=\dfrac{5\times5+1\times(-1)}{6}=4$. $S(6, 4)$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main

How this is examined: expect a one-mark midpoint, a two-mark "find the point dividing in the ratio", a three-mark "in what ratio does an axis divide" with the point of division, and a case study on a map. Write the formula with the values of $m_1, m_2$ marked, then substitute.

E1. Board · 1 mark The midpoint of the segment joining $(-3, 7)$ and $(9, -1)$ is (A) $(3, 3)$ (B) $(6, 6)$ (C) $(12, -8)$ (D) $(-12, 8)$

💡 Show solution▶

Idea $\left(\dfrac{-3+9}{2}, \dfrac{7-1}{2}\right)=(3, 3)$.

Traps (B) forgets to divide by $2$. (C) is the difference $B-A$, the vector from $A$ to $B$, not a point on the segment.

Answer (A) $(3, 3)$.

E2. Board · 1 mark Assertion (A): The $x$-axis divides the segment joining $(2, -3)$ and $(5, 6)$ in the ratio $1:2$. Reason (R): A point on the $x$-axis has $y$-coordinate $0$.
(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check A With ratio $k:1$, the $y$-coordinate is $\dfrac{6k-3}{k+1}$. Setting it to $0$ gives $k=\dfrac12$, so the ratio is $1:2$. A is true.

Check R R is true, and it is exactly the fact used to set the $y$-coordinate equal to $0$ in A.

Answer (A).

E3. Board · 2 marks Find the coordinates of the point $P$ that divides the segment joining $A(-8, 6)$ and $B(4, -6)$ in the ratio $1:3$.

💡 Show solution▶

Formula $m_1=1$, $m_2=3$: $\;x=\dfrac{m_1x_2+m_2x_1}{m_1+m_2}=\dfrac{1\times4+3\times(-8)}{4}=\dfrac{-20}{4}=-5$. (1 mark)

y $y=\dfrac{1\times(-6)+3\times6}{4}=\dfrac{12}{4}=3$. So $P(-5, 3)$. (1 mark)

E4. Board · 3 marks Find the ratio in which the $y$-axis divides the segment joining $A(-5, -7)$ and $B(10, 8)$. Also find the point of division.

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Ratio Let the ratio be $k:1$. On the $y$-axis the $x$-coordinate is $0$: $\dfrac{10k-5}{k+1}=0$, so $k=\dfrac12$ and the ratio is $1:2$. (1 mark)

Point $y=\dfrac{1\times8+2\times(-7)}{3}=\dfrac{-6}{3}=-2$. (1 mark)

Answer The $y$-axis divides $AB$ in the ratio $1:2$ at the point $(0, -2)$. (1 mark) Check: the $x$-coordinate is $\dfrac{1\times10+2\times(-5)}{3}=0$ ✓.

E5. Board · Case study A city plans a metro line in a straight line from station $A(1, 2)$ to station $B(13, 11)$ on a map where $1$ unit $=1$ km. Two new stations $C$ and $D$ are to be built so that $A$, $C$, $D$, $B$ are equally spaced along the line ($C$ nearer $A$).
(i) Find the coordinates of $C$ and $D$. (2 marks)
(ii) Find the length $AB$, and hence the distance between neighbouring stations. (1 mark)
(iii) An interchange is to be built at the midpoint of $AB$. Show that it is also the midpoint of $CD$. (1 mark)

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(i) C $AC:CB=1:2$: $\;x=\dfrac{1\times13+2\times1}{3}=5$, $\;y=\dfrac{1\times11+2\times2}{3}=5$. $C(5, 5)$.

(i) D $AD:DB=2:1$: $\;x=\dfrac{2\times13+1\times1}{3}=9$, $\;y=\dfrac{2\times11+1\times2}{3}=8$. $D(9, 8)$.

(ii) $AB=\sqrt{12^2+9^2}=\sqrt{225}=15$ km, so neighbouring stations are $15\div3=5$ km apart. (Check: $AC=\sqrt{16+9}=5$ ✓.)

(iii) Midpoint of $AB$: $\left(\tfrac{1+13}{2}, \tfrac{2+11}{2}\right)=(7, 6.5)$. Midpoint of $CD$: $\left(\tfrac{5+9}{2}, \tfrac{5+8}{2}\right)=(7, 6.5)$. They are the same point.

E6. JEE Main The point $P$ divides the segment joining $A(2, -1)$ and $B(8, 5)$ internally in the ratio $k:1$ and lies on the line $x+y=10$. Then $k$ equals (A) $2$ (B) $3$ (C) $\tfrac13$ (D) $4$

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Idea $P=\left(\dfrac{8k+2}{k+1}, \dfrac{5k-1}{k+1}\right)$, so $x+y=\dfrac{13k+1}{k+1}$.

Solve $\dfrac{13k+1}{k+1}=10$ gives $13k+1=10k+10$, so $3k=9$ and $k=3$. Then $P=\left(\tfrac{26}{4}, \tfrac{14}{4}\right)=(6.5, 3.5)$ and $6.5+3.5=10$ ✓.

Answer (B) $3$. Option (C) is the reciprocal: it is what you get if you read the ratio as $BP:PA$ instead of $AP:PB$.

🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. Two friends live in different parts of your town and want to meet at a café. If they walk at the same speed, the fair meeting point is the midpoint. But what if one of them walks twice as fast as the other, or cycles? Where should the meeting point be so that both arrive at the same time?

Why it matters. Ride-share apps, emergency services and delivery companies constantly decide where two moving things should meet or where a facility should go. If two travellers go at different speeds, the time-fair meeting point divides the straight line between them in the ratio of their speeds: exactly the section formula.

What to do.

Draw a map of your neighbourhood on squared paper (1 square = 100 m). Mark two homes $A$ and $B$ and read their coordinates.
Use the midpoint formula for the meeting point of two equally fast walkers. Compute the distance each walks with the distance formula.
Now suppose friend $B$ cycles at twice the speed of friend $A$. In what ratio $AP:PB$ must the meeting point $P$ divide the line? Use the section formula to find $P$, then use the distance formula to check that $PB=2\times AP$.
Measure your own walking speed (time $100$ m with a phone stopwatch) and recompute with real speeds.
Compare the straight-line meeting point with a real street route on a map app. Where do they differ?

Tools: Squared paper or a spreadsheet, a phone stopwatch, a map app

Hand in: The map with coordinates, the two meeting points with your calculations, your measured speed, and two sentences on how the ratio relates to the speeds.

Going further. Three friends live at three different places and want one meeting point. Can you find a point that is equally far from all three? (Use the idea of Exercise III in Section 7.1.)

Teacher/parent note: with $A(4, 2)$ and $B(16, 10)$ (metres), $AB=\sqrt{144+64}\approx14.42$ m. Equal speeds: midpoint $(10, 6)$, each walks $\approx7.21$ m. If $B$ goes twice as fast, $B$ covers twice the distance in the same time, so $AP:PB=1:2$ and $P=\left(\tfrac{16+2\times4}{3}, \tfrac{10+2\times2}{3}\right)=(8, 4.67)$: $AP\approx4.81$ m and $PB\approx9.62$ m.

🧪 Self-check quiz

0/5 correct

Q1. The midpoint of the segment joining $(-2, 8)$ and $(6, -4)$ is

Q2. The point dividing the segment from $(0, 0)$ to $(10, 5)$ in the ratio $2:3$ is

Q3. $M(2, 3)$ is the midpoint of $AB$ and $A$ is $(-2, 1)$. Then $B$ is

Q4. The $y$-axis divides the segment joining $(-4, 2)$ and $(12, -6)$ in the ratio

Q5. Of the two trisection points of the segment from $A(0, 0)$ to $B(9, -6)$, the one nearer to $B$ is

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