Question

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Answer 1

Answer:

Let the required point be P( 0, y)    [B'coz the point lies on y-axis]

Also, let the required ratio be k : 1

Using section formula,

P(0,y) = [(m1*x2+m2*x1) / m1+m2 , (m1*y2 +m2*y1) / m1+ m2]

Substituting values m1=k , m2=1, x1= -2 ,x2= 3 ,  y1=-3, y2=7, We get

P(0.y) = [ (3*k +1*-2)/ k+1 , (7*k +1*-3) /k+1]

P(0,y) = [ (3k-2)/ k+1 , (7k-3)/ k+1]

0= (3k-2)/k+1...(a)  ,  y= (7k-3)/k+1......(b)

(a) ..3k-2 = 0 

3k = 2

k=2/3

Sub k= 2/3 in (b)

y = (7*(2/3) - 3)/ (2/3)+1

   =(14/3 - 3)/ 5/3

   = (14-9)/3 / (5/3)

   = (5/3) /(5/3)

   = 1

Therefore ratio is k :1 = 2/3 :1 = 2:3

and point of division is P( 0,1)

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Answer 2

Answer:

2nd question : 1st one explanation

Step-by-step explanation:

sol; Let the given points be A(5,-6) and B(-7,5).

Let P and Q be the points of trisection of AB.

Then,  AP=PQ=QB=1

Thus 'p' divides AB in the ratio 1:2.

Here, we have x1 = 5, y1= -6, x2= -7, y2=5, m1=1, m2=2

* The coordinates of 'p' is given by

P[ m1x2 + m2x1/ m1+m2 ; m1y2 + m2y1/ m1+m2] ..........

case 2:

Now, 'Q' divides AB in the ratio 2:1

here ,

we  have x1=5, y1= -6, x2= -7 , y2=5, m1=2, m2=1

Therefore,

*The coordinates of 'Q' is given by

Q[ m1x2 + m2x1/ m1+m2 , m1y2 + m2y1 /m1+m2 ]............solve it by the given values to get the required answer....hope u like this:)