Consider a point P between A and B is such a way that the distance of the point P from B is twice its distance from A. If P lies on AB, it will divide AB in the ratio 1 : 2 (see Fig.). If we take A as the origin O, and 1 km as one unit on both the axis, the coordinates of B will be (36, 15). Find the position of point P?

A(0,0) B(36,15) P(x,y) X Y 1 2 O D E C
Let the coordinates of P be (x, y). perpendiculars from P and B to the x-axis, meeting it in D and E, respectively. Draw PC perpendicular to BE. Then, by the AA(Angle-Angle) similarity criterion, ∆ POD and ∆ BPC are similar.

Therefore , ${OD}/{PC}={OP}/{PB}=1/2$ and ${PD}/{BC}={OP}/{PB}=1/2$

So, ${x}/{36-x}=1/2$ and ${y}/{15-y}=1/2$

These equations give x = 12 and y = 5.

You can check that P(12, 5) meets the condition that OP : PB = 1 : 2.

General Formula

B(x2,y2) A(x1,y1) P(x,y) R S T C Q m1 m2
Consider any two points A$(x_1, y_1)$ and B$(x_2, y_2)$ and assume that $P (x, y) $ divides AB internally in the ratio $m_1$: $m_2$ i.e

${PA}/{PB}={m_1}/{m_2}$

Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC parallel to the x-axis. Then, by the AA similarity criterion,
∆ PAQ ~ ∆ BPC

Therefore, ${PA}/{BP}={AQ}/{PC}={PQ}/{BC}$ ..........(1)

Now, AQ = RS = OS – OR =$x – x_1$

PC = ST = OT – OS = $x_2 – x$

PQ = PS – QS = PS – AR =$y – y_1$

BC = BT– CT = BT – PS = $y_2 – y$

Substituting these values in (1), we get

${m_1}/{m_2}={x-x_1}/{x_2-x}={y-y_1}/{y_2-y}$

Taking ${m_1}/{m_2}={x-x_1}/{x_2-x}$ we get $x={m_1 x_2 + m_2 x_1}/{m_1 + m_2}$

Taking ${m_1}/{m_2}={y-y_1}/{y_2-y}$ we get $y={m_1 y_2 + m_2 y_1}/{m_1 + m_2}$

So, the coordinates of the point $P(x, y)$ which divides the line segment joining the points $ A(x_1, y_1)$ and $B(x_2, y_2)$, internally, in the ratio $m_1 : m_2$ are

$$({m_1 x_2 + m_2 x_1}/{m_1 + m_2},{m_1 y_2 + m_2 y_1}/{m_1 + m_2})$$

This is known as the section formula.

This can also be derived by drawing perpendiculars from A, P and B on they-axis and proceeding as above.

If the ratio in which P divides AB is $k : 1$, then the coordinates of the point P will be

$$({k x_2 + x_1}/{k + 1},{k y_2 + y_1}/{k + 1})$$

Special Case : The mid-point of a line segment divides the line segment in the ratio 1 : 1. Therefore, the coordinates of the mid-point P of the join of the points $ A(x_1, y_1)$ and $B(x_2, y_2)$ is
$$({ x_2 + x_1}/{2},{y_2 + y_1}/{2})$$

Example: Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.

Solution : Let P(x, y) be the required point. Using the section formula, $({k x_2 + x_1}/{k + 1},{k y_2 + y_1}/{k + 1})$

$x={3(8) + 4}/{3 + 1} = 7$

$y={3(5) + (-3)}/{3 + 1}=3$

Ans:Therefore, (7, 3) is the required point.

Example: In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?

Solution : Let (– 4, 6) divide AB internally in the ratio $m_1:m_2$. Using the section formula, i.e;
$x={m_1 x_2 + m_2 x_1}/{m_1 + m_2}$ and $y={m_1 y_2 + m_2 y_1}/{m_1 + m_2}$

$(x,y)=({m_1 x_2 + m_2 x_1}/{m_1 + m_2},{m_1 y_2 + m_2 y_1}/{m_1 + m_2})$

$(-4,6)=({m_1 3 + m_2 (-6)}/{m_1 + m_2},{m_1 (-8) + m_2 (10)}/{m_1 + m_2})$

$-4={3m_1 -6m_2}/{m_1 + m_2}$ and $6={-8m_1 +10 m_2}/{m_1 + m_2}$

$-4m_1 -4m_2=3m_1 -6m_2$

$7m_1=2m_2$

$m_1:m_2=2:7$

and $6m_1 + 6m_2=-8m_1 +10 m_2$

i.e $14m_1 = 4 m_2$

Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.


Example: Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).

Solution : Let P and Q be the points of trisection of AB i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are
$({1(-7) + 2(2)}/{1+2},{1(4) + 2(-2)}/{1+2})$
i.e $(-1,0)$

Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
$({2(-7) + 1(2)}/{1+2},{2(4) + 1(-2)}/{1+2})$
i.e $(-4,2)$

Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).

Note : We could also have obtained Q by noting that it is the mid-point of PB. So, we could have obtained its coordinates using the mid-point formula.


Example: Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.

Solution : Let the ratio be $k : 1$. Then by the section formula, the coordinates of the point which divides AB in the ratio $k : 1$ are

$(x,y)=({-k + 5}/{k + 1},{-4k -6}/{k + 1})$

This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.

Therefore, ${-k + 5}/{k + 1}=0$
So, $k=5$

That is, the ratio is 5 : 1. Putting the value of $k = 5$, we get the point of intersection as
$(0,{-13/3})$


Example: If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Solution : We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD
i.e $({6+9}/{2},{1+4}/{2})=({8+p}/{2},{2+3}/{2})$

i.e $({15}/{2},{5}/{2})=({8+p}/{2},{5}/{2})$

i.e ${15/2}={8+p}/2$

i.e $p=7$