# ML Aggarwal Solutions Class 10 Mathematics Solutions for Section Formula Exercise 11 in Chapter 11 - Section Formula

Question 41 Section Formula Exercise 11

A (2, 5), B ( – 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.

Find the co-ordinates of P and Q.

Solution:

The section formula is used to determine the coordinates of a point that splits a line segment externally or internally in a certain ratio. It's a useful tool for determining the coordinates of the point that divides a line segment into a certain number of segments.

Given vertices of the ABC are A(2,5), B(-1,2) and C(5,8).

P and Q are points on AB and AC respectively such that AP:PB = AQ :QC = 1:2.

P(x,y) divides AB in the ratio 1:2.

x1= 2, y1 = 5

x2 = -1, y2 = 2

m:n = 1:2

By\ section\ formula,\\x=\frac{\left[(1\times−1+2\times2)\right]}{(1+2)}\\x=\frac{(−1+4)}{3}\\ x=\frac{3}{3}=1

By section formula,

y=\frac{\left(my_2+ny_1\right)}{(m+n)}\\y=\frac{\left(1\times2+2\times5\right)}{\left(1+2\right)}\\y=\frac{\left(2+10\right)}{3}\\y=\frac{12}{3}=4

Co-ordinates of P are (1,4).

Q(x,y) divides AC in the ratio 1:2.

x1= 2, y1 = 5

x2 = 5, y2 = 8

m:n = 1:2

By section formula,

x=\frac{\left(mx_2+nx_1\right)}{(m+n)}\\x=\frac{\left(1\times5+2\times2\right)}{(1+2)}\\x=\frac{\left(5+4\right)}{3}\\x=\frac{9}{3}=3

By section formula,

y=\frac{\left(my_2+ny_1\right)}{(m+n)}\\y=\frac{\left(1\times8+2\times5\right)}{(1+2)}\\y=\frac{\left(8+10\right)}{3}\\y=\frac{18}{3}=6

The coordinates of Q are (3,6).

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