Question

In what ratio is the line segment joining the point (-2,3) and (3,7) divide by the Y-Axis? Also , find the coordinates of the point of division.

Answer 1

Hii friend,

Let the y-axis cut the join of A(-2,3) and B(3,7) at point P in the ratio K:1 .

Then,

By sectional formula , the coordinates of P are : P(3K-2/K+1 , 7K-3/K+1)

P lies on the y-axis . So it's abscissa is 0

Therefore,

3K-2/K+1 = 0

3K-2 = 0

K = 2/3

Putting K=2/3 , we get point P as

P(0,7×2/3-3/2/2 +1)

P(0,1)

Hence,

The point of intersection of AB and The y-axis is P(0,1)

HOPE IT WILL HELP YOU..... :-)

Answer 2

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)