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# (i) Calculate the ratio in which the line segment joining (3, 4) and( – 2, 1) is divided by the y-axis. (ii) In what ratio does the line x – y – 2 = 0 divide the line segment joining the points (3, – 1) and (8, 9)? Also, find the coordinates of the point of division.

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ICSE Board Question Based on Section Formula Chapter of M.L Aggarwal for class10
In this question following parts to be solved
(i) Calculate the ratio in which the line segment joining two points is divided by the y-axis
(ii) In what ratio does a line divide the line segment joining the points two points ? Also, find the coordinates of the point of division.
This is the Question Number 19, Exercise 11 of M.L Aggarwal.

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1. (i) Let m:n be the ratio in which the line segment joining (3,4) and (-2,1) is divided by the Y axis.

Since the line meets Y axis, its x co-ordinate is zero.

Here x= 3, y1 = 4

x2 = -2, y2 = 1

By section formula, x = (mx2+nx1)/(m+n)

0 = (m×-2+n×3)/(m+n)

0 = (-2m+3n)/( m+n)

0 = -2m+3n

2m = 3n

m/n = 3/2

Hence the ration m:n is 3:2.

(ii)Let the line x-y-2 = 0 divide the line segment joining the points (3,-1) and (8,9) in the ratio m:n at the point P(x,y)

Here x= 3, y1 = -1

x2 = 8 y2 = 9

By section formula, x = (mx2+nx1)/(m+n)

x = (m×8+n×3)/(m+n)

x = (8m+3n)/( m+n) …. (i)

By Section formula y = (my2+ny1)/(m+n)

y = (m×9+n×-1)/(m+n)

y = (9m-n)/(m+n) …. (ii)

Since the point P(x,y) lies on the line x-y-2 = 0,

eqn (i) and (ii) will satify the equation x-y-2 = 0 …(iii)

Substitute (i) and (ii) in (iii)

[(8m+3n)/( m+n)] – [(9m-n)/(m+n)]-2 = 0

[(8m+3n)/( m+n)] – [(9m-n)/(m+n)]-[2(m+n)/(m+n)] = 0

8m+3n-(9m-n)-2(m+n) = 0

8m+3n-9m+n-2m-2n = 0

-3m+2n = 0

-3m = -2n

m/n = -2/-3 = 2/3

Hence the ratio m:n is 2:3.

Substitute m and n in (i)

x = (8m+3n)/( m+n)

x = (8×2+3×3)/(2+3)

x = (16+9)/5

x = 25/5 = 5

Substitute m and n in (ii)

y = (9m-n)/(m+n)

y = (9×2-3)/(2+3)

y = (18-3)/5

y = 15/5 = 3

Hence the co-ordinates of P are (5,3).

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