1. Point of Division

2. Mid-point
If M is a point on the line segment AB such that M bisects AB
(i.e. AM = MB),
then M is called the mid-point of AB. x y
(x1, y1)
(x2, y2) A M B
Let’s try to find the coordinates of the mid-point M of AB.

3. Draw a horizontal line segment AE and a vertical line segment ME.y
Draw a horizontal line segment AE and a vertical line segment ME.
B(x2, y2)
x2 – x
y2 – y
M(x, y)
Obviously, AEM = 90.
y – y1 F
(x2, y)
Similarly, draw a horizontal line segment MF and a vertical line segment BF.
A(x1, y1) E
(x, y1)
x – x1 x
Obviously, MFB = 90.
Coordinates of E =
Now, we need to find the coordinates of E and F. Then, calculate the lengths of AE, ME, MF and BF.
(x, y1)
Coordinates of F =
(x2, y)
AE =
x  x1
ME =
y  y1
MF =
x2  x
BF =
y2  y

4. ∵ △AEM  △MFB (AAS) ∴ (corr. sides,  △s) MF AE = x - = 2 x + =y
B(x2, y2)
∵ △AEM  △MFB (AAS)
x2 – x
y2 – y ∴
(corr. sides,  △s) MF AE =
M(x, y)
y – y1 F x 2 1 - = 2 x 1 + =
A(x1, y1) E
x – x1 x
(corr. sides,  △s) BF ME =
and y 2 1 - = 2 y 1 + = ∴ ç è æ = , of s
Coordinate M + 2 x 1 + 2 1 y

5. This is known as the mid-point formula.
If M(x, y) is the mid-point of the line segment joining A(x1, y1) and B(x2, y2), then
A(x1, y1)
M(x, y)
B(x2, y2) x y
and
This is known as the mid-point formula.

6. Find the coordinates of the mid-point M of AB in the figure.x y
B(6, 1)
A(–4, 3) M
Let (x, y) be the coordinates of M.
By the mid-point formula, we have 2 6 4 + - = x 2 1 3 + = y
and 1 = 2 =
(1, 2) of
Coordinates = ∴ M

7. \ Follow-up question 2 8) ( 12 - + = x 2 10) ( - + = y 2 = 6 - =
In the figure, if M is the mid-point of AB, find the coordinates of M. x y
A (12, –2)
B (–8, –10) M
Solution
Let (x, y) be the coordinates of M.
By the mid-point formula, we have 2 8) ( 12 - + = x 2
10) ( - + = y
and 2 = 6 - = 6)
(2, of
Coordinates - = \ M

8. Internal Point of Division
If a point P divides AB into AP and PB such that AP : PB = : , r s
then P is called the internal point of division of AB. r s A P B x y :
Also, we can say that ‘P divides AB internally’.

9. Using the property of similar triangles, we can find the coordinates of P.
Then, we find the coordinates of E and F,
vertical line segments PE and BF.
Similar to the case for the mid-point, we can also derive the coordinates of internal point of division.
First, we construct horizontal line segments AE and PF, and
and the lengths of AE, PE,
PF and BF.

10. ∵ △AEP ~ △PFB (AAA)
∴ Consider their corresponding sides.

11. \ + = s r ry sy rx sx P , ç è æ of Coordinate ∵ △AEP ~ △PFB (AAA)
∴ Consider their corresponding sides. + = \ s r ry sy rx sx P 2 1 , ç è æ of
Coordinate

12. This is known as the section formula for internal division.
If P(x, y) is the internal point of division of the line segment joining A(x1, y1) and B(x2, y2) such that AP : PB = r : s, then r s
A(x1, y1)
P(x, y)
B(x2, y2) :
and .
This is known as the section formula for internal division.

13. In the figure, if P(x, y) is a point on the line segment AB such that AP : PB = 2 : 3, find the coordinates of P.
By the section formula for internal division, we have x y
B(2, 4)
A(–9, 1)
P(x, y) 3 : 2 9) 3( + -
2(2)
3(1) +
2(4) = x
and = y 3 2 + 3 2 +
4.6 - =
2.2 =
2.2)
4.6, ( of s
Coordinate - = \ P

14. Follow-up question
The figure shows two points A(13, 5) and B(3, 2). If P(x, y) lies on AB such that AP : PB = 1 : 4, find the coordinates of P.
Solution x y
B(3, –2)
A(–13, –5)
P(x, y)
By the section formula for internal division, we have
Coordinates of P = (–9.8, –4.4) \