An Important Question of M.L Aggarwal book of class 10 Based on Section Formula Chapter for ICSE BOARD.

You have to find the solution of this given question.

This is the Question Number 18, Exercise 11 of M.L Aggarwal.

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# If A ( – 4, 3) and B (8, – 6), (i) find the length of AB. (ii) in what ratio is the line joining AB, divided by the x-axis?

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(i) Given points are A(-4,3) and B(8,-6).

Here x

_{1 }= -4, y_{1}= 3x

_{2}= 8, y_{2}= -6By distance formula, d(AB) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]d(AB) = √[(8-(-4))

^{2}+(-6-3)^{2}]d(AB) = √[(12)

^{2}+(-9)^{2}]d(AB) = √(144+81)

d(AB) = √225

d(AB) = 15

Hence the length of AB is 15 units.

(ii)Let m:n be the ratio in which the line AB is divided by the X axis.

Since the line meets X axis, its y co-ordinate is zero.

By Section formula y = (my

_{2}+ny_{1})/(m+n)0 = (m×-6+n×3)/(m+n)

0 = (-6m+3n)/m+n

-6m+3n = 0

-6m = -3n

m/n = -3/-6 = 3/6 = 1/2

Hence the ratio is 1:2.