P(x, y)

Dividing Point of a Line Segment

Learn what a point is, see how it’s defined, then watch the equation of the circle get derived — one step at a time.

Point which Divides a Line Segment in a Ratio

a. PA = 2 cm, PB = 3 cm, AB = 5 cm b. RX = 10 cm, RY = 15 cm, XY = 25 cm

Suppose, AP = k m₁ and PB = k m₂, then the ratio between AP and PB is AP:PB = m₁:m₂.

Dividing AB internally by P.

Dividing XR externally by Y.

Similarly, XY = k m₁ and RY = k m₂, then the ratio between XY and RY is AP:PB = m₁:m₂. Point P divides AB internally because P lies on the interior of AB. Similarly, Y divides XR externally because Y lies on the exterior of XR.

Section Formula – Coordinates of a Point Dividing a Line Segment

Step 1 of 7

Section Formula

Let, in the figure, A(x₁, y₁) and B(x₂, y₂) are two points. P(x, y) divides the line segment AB internally in the ratio m₁:m₂. Calculate the coordinate of P(x, y).

Now, draw AD⊥OX, BE⊥OX, PF⊥OX, AM⊥PF and PN⊥BE

In the figure, AM = DF = OF − OD = x − x₁,

PN = FE = OE − OF = x₂ − x

PM = PF − MF = PF − AD = y − y₁

BN = BE − NE = BE − PF = y₂ − y

And APBP = m₁m₂

Right angled triangles ΔPMA and ΔBNP are similar.

AMPN = APBP [ ∵ The ratio of corresponding sides of similar triangle]

Or, x − x₁x₂ − x = m₁m₂

Or, m₂x − m₂x₁ = m₁x₂ − m₁x

Or, m₂x + m₁x = m₁x₂ + m₂x₁

Or, x(m₁ + m₂) = m₁x₂ + m₂x₁

Or, x = m₁x₂ + m₂ x₁m₁ + m₂

Again, PMBN = APBP [ ∵ The ratio of corresponding sides of similar triangle]

Or, y − y₁y₂ − y = m₁m₂

Or, m₂y − m₂y₁ = m₁y₂ − m₁y

Or, m₂y + m₁y = m₁y₂ + m₂y₁

Or, y(m₁ + m₂) = m₁y₂ + m₂y₁

Or, y = m₁y₂ + m₂y₁m₁ + m₂

∴ (x, y) = (m₁x₂ + m₂ x₁m₁ + m₂ , m₁y₂ + m₂ y₁m₁ + m₂)

Thus, the coordinates of the point dividing line segment AB are

P(x, y) = (m₁x₂ + m₂ x₁m₁ + m₂ , m₁y₂ + m₂ y₁m₁ + m₂)

If P(x, y) divides the line segment joining points A(x₁, y₁) and B(x₂, y₂) externally in the ratio m₁:m₂, then AP:BP = m₁:m₂

Step 1 of 5

External Division

Suppose, in the figure, the line joining A(x₁, y₁) and B(x₂, y₂) is divided externally by P(x, y) in the ratio m₁:m₂.

Or, APBP = m₁m₂

Now, draw AL⊥OX, BM⊥OX, PN⊥OX, AR⊥PN and BS⊥PN

In the figure, AR = LN = ON − OL = x − x₁,

BS = MN = ON − OM = x − x₂

PR = PN − RN = PN − AL = y − y₁

PS = PN − SN = PN − BM = y − y₂

And APPB = m₁m₂

Right angled triangles ΔPAR and ΔBSP are similar.

ARBS = APBP [ ∵ The ratio of corresponding sides of similar triangle]

Or, x − x₁x − x₂ = m₁m₂

Or, m₂(x − x₁) = m₁(x − x₂)

Or, m₂x − m₂x₁ = m₁x − m₁x₂

Or, m₁x₂ − m₂x₁ = m₁x − m₂x

Or, m₁x₂ − m₂x₁ = x(m₁ − m₂)

Or, x = m₁x₂ − m₂x₁m₁ − m₂

Midpoint Formula

Step 1 of 5

If P(x, y) be the midpoint of the line segment joining the points A(x₁, y₁) and B(x₂, y₂), then AP = BP and AP:BP = m₁:m₂ = 1:1.

P(x, y) = (m₁x₂ + m₂x₁m₁ + m₂ , m₁y₂ + m₂y₁m₁ + m₂)

= (1.x₂ + 1.x₁1 + 1 , 1.y₂ + 1.y₁1 + 1)

= (x₁ + x₂2 , y₁ + y₂2)

∴ P(x, y) is called the coordinate of midpoint.

Example 1 of 2

Solve It