**Finding the co-ordinates of the point dividing the line segment** joining two given points in a given ratio:

Let (x₁, y₁) and (x₂, y₂) be the Cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and the point R divides the line-segment PQ internally in a given ratio m : n (say), i.e., PR : RQ = m : n. We are to find the co-ordinates of R.

Let, (x, y) be the required co-ordinate of R . From P, Q and R, draw PL, QM and RN perpendiculars on OX. Again, draw PT parallel to OX to cut RN at S and QM at T.

Then,

PS = LN = ON - OL = x – x₁;

PT = LM = OM – OL = x₂ - x₁;

RS = RN – SN = RN – PL = y - y₁;

and QT = QM – TM = QM – PL = y₂ – y₁

Again, PR/RQ = m/n

or, RQ/PR = n/m

or, RQ/PR + 1 = n/m + 1

or, (RQ + PR/PR) = (m + n)/m

o, PQ/PR = (m + n)/m

Now, by construction, the triangles PRS and PQT are similar; hence,

PS/PT = RS/QT = PR/PQ

Taking, PS/PT = PR/PQ we get,

(x - x₁)/(x₂ - x₁) = m/(m + n)

or, x (m + n) – x₁ (m + n) = mx₂ – mx₁

or, x ( m + n) = mx₂ - mx₁ + m x₁ + nx₁ = mx₂ + nx₁

Therefore, x = (mx₁ + n x₁)/(m + n)

Again, taking RS/QT = PR/PQ we get,

(y - y₁)/(y₂ - y₁) = m/(m + n)

or, ( m + n) y - ( m + n) y₁ = my₂ – my₁

or, ( m+ n)y = my₂ – my₁ + my₁ + ny₁ = my₂ + ny₁

Therefore, y = my₂ + ny₁)/(m + n)

Therefore, the required co-ordinates of the point R are

((mx₂ + nx₁)/(m + n), (my ₂ + ny₁)/(m + n))

Let (x₁, y₁) and (x₂, y₂) be the cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and the point R divides the line-segment PQ externally in a given ratio m : n (say) i.e., PR : RQ = m : n. We are to find the co-ordinates of R.

Let, (x, y) be the required co-ordinates of R. Draw PL, QM and RN perpendiculars on OX. Again, draw PT parallel to OX to cut RN at S and QM and RN at S and T respectively, Then,

PS = LM = OM - OL = x₂ – x₁;

PT = LN = ON – OL = x – x₁;

QT = QM – SM = QM – PL = y₂ – y₁

and RT = RN – TN = RN – PL = y — y₁

Again, PR/RQ = m/n

or, QR/PR = n/m

or, 1 - QR/PR = 1 - n/m

or, PR - RQ/PR = (m - n)/m

or, PQ/PR = (m - n)/m

Now, by construction, the triangles PQS and PRT are similar; hence,

PS/PT = QS/RT = PQ/PR

Taking, PS/PT = PQ/PR we get,

(x₂ - x₁)/(x - x₁) = (m - n)/m

or, (m – n)x - x₁(m – n) = m (x₂ - x₁)

or, (m - n)x = mx₂ – mx₁ + mx₁ - nx₁ = mx₂ - nx₁.

Therefore, x = (mx₂ - nx₁)/(m - n)

Again, taking QS/RT = PQ/PR we get,

(y₂ - y₁)/(y - y₁) = (m - n)/m

or, (m – n)y - (m – n)y₁ = m(y₂ - y₁)

or, (m - n)y = my₂ – my₁ + my₁ - ny₁ = my₂ - ny₁

Therefore, x = (my₂ - ny₁)/(m - n)

Therefore, the co-ordinates of the point R are

((mx₂ - nx₁)/(m - n), (my₂ - ny₁)/(m - n))

Corollary: To find the co-ordinates of the middle point of a given line segment:

Let (x₁, y₁) and (x₂, y₂) he the co-ordinates of the points P and Q respectively and R, the mid-point of the line segment PQ. To find the co-ordinates R. Clearly, the point R divides the line segment PQ internally in the ratio 1 : 1;

hence, the co-ordinates of R are ((x₁ + x₂)/2, (y₁ + y₂)/2).

Considering the points to be in ratio m and n , where m=n , the co-ordinates or R of ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))].

This formula is also known as midpoint formula. By using this formula we can easily find the midpoint between the two co-ordinates.

1. A diameter of a circle has the extreme points (7, 9) and (-1, -3). What would be the co-ordinates of the centre?

**Solution:**

Clearly, the mid-point of the given diameter is the centre of the circle. Therefore, the required co-ordinates of the centre of the circle = the co-ordinates of the mid-point of the line-segment joining the points (7, 9) and (- 1, - 3)

R (Rx, Ry) = ((7 - 1)/2, (9 - 3)/2) = (3, 3).