**Example Problems Using Section Formula :**

Here we are going to see some example problems using the section formula.

We studied bisection and trisection of a given line segment. These are only particular cases of the general problem of dividing a line segment joining two points (x_{1}, y_{1}) and (x_{2} , y_{2}) in the ratio m : n.

**Section formula (internally) :**

(mx_{2} + nx_{1})/(m + n), (my_{2} + ny_{1})/(m + n)

**Section formula (externally) :**

(mx_{2} - nx_{1})/(m - n), (my_{2} - ny_{1})/(m - n)

**Question 1 :**

Find the coordinates of the point which divides the line segment joining the points A(4,−3) and B(9, 7) in the ratio 3:2.

**Solution :**

= (mx_{2} + nx_{1})/(m + n), (my_{2} + ny_{1})/(m + n)

= (3(9) + 2(4))/(3 + 2), (3(7) + 2(-3))/(3 + 2)

= (27 + 8)/5, (21 - 6))/5

= 35/5, 15/5

= (7, 3)

**Question 2 :**

In what ratio does the point P(2,−5) divide the line segment joining A(−3, 5) and B(4, −9).

**Solution :**

= (mx_{2} + nx_{1})/(m + n), (my_{2} + ny_{1})/(m + n)

= (m(4) + n(-3))/(m + n), (m(-9) + n(5))/(m + n)

(4m - 3n)/(m + n), (-9m + 5n)/(m + n) = (2, -5)

By equating x and y-coordinates, we get

(4m - 3n)/(m + n) = 2

4m - 3n = 2m + 2n

4m - 2m = 2n + 3n

2m = 5n

m/n = 5/2

m : n = 5 : 2

Hence the required ratio is 5 : 2

**Question 3 :**

Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = (2/5) AB.

**Solution :**

AP = (2/5) AB.

AP/AB = 2/5

AB = 5

AP + PB = AB

2 + PB = 5

PB = 5 - 2 = 3

So, the point P divides the line segment in the ratio 2 : 3.

A(1, 2) and B(6, 7)

P = (2(6) + 3(1))/(2 + 3), (2(7) + 3(2))/(2 + 3)

P = (15/5, 20/5)

P = (3, 4)

Hence the required point is (3, 4).

After having gone through the stuff given above, we hope that the students would have understood, "Example Problems Using Section Formula"

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