This question is Based on Section Formula Chapter of M.L Aggarwal book for ICSE BOARD for class 10.

Here you have to find the value ok ‘k’ as the line segment joining two points is divided at a point in ratio.

Solve the question as i have so many doubts

This is the Question Number 05, Exercise 11 of M.L Aggarwal.

# (i) The line segment joining the points A (3, 2) and B (5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k. (ii) A point P divides the line segment joining the points A (3, – 5) and B ( – 4, 8) such that AP/PB = k/1 If P lies on the line x + y = 0, then find the value of k.

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(i) Let the co-ordinates of P(x, y) divides AB in the ratio m:n.

A

(3,2)and B(5,1)are the given points.Given m:n = 1:2

x

_{1}= 3 , y_{1}= 2 , x_{2}= 5 , y_{2}= 1 , m = 1 and n = 2By Section formula x = (mx

_{2}+nx_{1})/(m+n)x = (1×5+2×3)/(1+2)

x = (5+6)/3

x = 11/3

By Section formula y = (my

_{2}+ny_{1})/(m+n)y = (1×1+2×2)/(1+2)

y = (1+4)/3

y = 5/3

Given P lies on the line 3x-18y+k = 0

Substitute x and y in above equation

3×(11/3)-18×(5/3)+k = 0

11-30+k = 0

-19+k = 0

k = 19

Hence the value of k is 19.

(ii) Let the co-ordinates of P(x, y) divides AB in the ratio m:n.

A

(3,-5)and B(-4,8)are the given points.Given AP/PB = k/1

m:n = k:1

x

_{1}= 3 , y_{1}= -5 , x_{2}= -4 , y_{2}= 8 , m = k and n = 1By Section formula x = (mx

_{2}+nx_{1})/(m+n)x = (k×-4+1×3)/(k+1)

x = (-4k+3)/(k+1)

x = (-4k+3)/(k+1)

By Section formula y = (my

_{2}+ny_{1})/(m+n)y = (k×8+1×-5)/(k+1)

y = (-4k+3)/(k+1)

Co-ordinate of P is ((-4k+3)/(k+1), (8k-5)/(k+1))

Given P lies on line x+y = 0

Substitute value of x and y in above equation

(-4k+3)/(k+1) + (8k-5)/(k+1) = 0

(-4k+3) + (8k-5) = 0

4k-2 = 0

4k = 2

k = 2/4 = ½

Hence the value of k is ½ .