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The Brainliest Answer!
2015-02-08T18:01:44+05:30
Let line divides 2x+3y-5=0 to the segment joining (8,-9) and (2,1) in k:1
let point of intersect be (a,b)
by intersection formula
(a,b)=(2k+8/k+1  , k-9/k+1)
a=2k+8/k+1  and b=k-9/k+1
these point will also lie on line 2x+3y-5=0.hence point will satisfy this equation
2×2k+8/k+1 + 3×k-9/k+1 -5 =0
4k+16+3k-27-5k-5=0
2k-16=0
k=8
ratio will be 8:1
coordinate will be x=2k+8/k+1=24/9 and y=k-9/k+1=-1/9

2 5 2
2015-02-08T21:23:58+05:30
Let the ratio be k:1
 
A(8,-9) B(2,1)
The point of intersection be X(x,y)
so ,
x= \frac{8+2k}{k+1}   \\   y= \frac{k-9}{k+1} 

 
So the coordinates will also satisfy the eqn, as it lies on it.
 So,
2x+3y-5=0
2 X \frac{8+2k}{k+1} + 3 X \frac{k-9}{k+1} - 5=0 \\ \frac{16+4k+3k-27}{k+1}=5 \\ -11+7k=5k+5 \\ 2k=16 \\ k=8

So the ratio=8:1
Co-ordiates are :
x=8+2k/k+1=24/9
y=k-9/k+1=-1/9
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